Galois Theory Tom Leinster, University of Edinburgh Version of 3 May 2021 1 2 3 4 5 6 7 8 9 10 Note to the reader Overview of Galois theory 1.1 The view of C from Q . . . . . . . . . . . . . 1.2 Every polynomial has a symmetry group. . . . 1.3 . . . which determines whether it can be solved Rings and fields 2.1 Rings . . . . . . . . . . . . . . . . . . . . . 2.2 Fields . . . . . . . . . . . . . . . . . . . . . Polynomials 3.1 The ring of polynomials . . . . . . . . . . . . 3.2 Factorizing polynomials . . . . . . . . . . . 3.3 Irreducible polynomials . . . . . . . . . . . . Field extensions 4.1 Definition and examples . . . . . . . . . . . 4.2 Algebraic and transcendental elements . . . . 4.3 Simple extensions . . . . . . . . . . . . . . . Degree 5.1 Degrees of extensions and polynomials . . . . 5.2 The tower law . . . . . . . . . . . . . . . . . 5.3 Algebraic extensions . . . . . . . . . . . . . 5.4 Ruler and compass constructions . . . . . . . Splitting fields 6.1 Extending homomorphisms . . . . . . . . . . 6.2 Existence and uniqueness of splitting fields . 6.3 The Galois group . . . . . . . . . . . . . . . Preparation for the fundamental theorem 7.1 Normality . . . . . . . . . . . . . . . . . . . 7.2 Separability . . . . . . . . . . . . . . . . . . 7.3 Fixed fields . . . . . . . . . . . . . . . . . . The fundamental theorem of Galois theory 8.1 Introducing the Galois correspondence . . . . 8.2 The theorem . . . . . . . . . . . . . . . . . . 8.3 A specific example . . . . . . . . . . . . . . Solvability by radicals 9.1 Radicals . . . . . . . . . . . . . . . . . . . . 9.2 Solvable polynomials have solvable groups . . 9.3 An unsolvable polynomial . . . . . . . . . . Finite fields 10.1 πth roots in characteristic π . . . . . . . . . . 10.2 Classification of finite fields . . . . . . . . . 10.3 Multiplicative structure . . . . . . . . . . . . 10.4 Galois groups for finite fields . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 4 4 9 11 14 14 19 24 24 28 33 38 38 43 47 51 51 56 58 60 67 68 70 76 82 83 91 96 100 100 104 109 115 116 119 126 130 131 133 135 136 Note to the reader These are the 2020–21 course notes for Galois Theory (MATH10080). Structure Each chapter corresponds to one week of the semester. You should read Chapter 1 in Week 1, Chapter 2 in Week 2, and so on. I’m writing the notes as we go along, so the chapters will appear one by one: keep your eye on Learn. Exercises looking like this are sprinkled through the notes. The idea is that you try them immediately, before you continue reading. Most of them are meant to be quick and easy, much easier than assignment or workshop questions. If you can do them, you can take it as a sign that you’re following. For those that defeat you, talk with your group or ask on Piazza; or if you’re really stuck, ask me. I promise you that if you make a habit of trying every exercise, you’ll enjoy the course more and understand it better than if you don’t bother. Digressions like this are optional and not examinable, but might interest you. They’re usually on points that I find interesting, and often describe connections between Galois theory and other parts of mathematics. Here you’ll see titles of relevant videos References to theorem numbers, page numbers, etc., are clickable links. What to prioritize You know by now that the most important things in almost any course are the definitions and the results called Theorem. But I also want to emphasize the proofs. This course presents a wonderful body of theory, and the idea is that you learn it all, including the proofs that are its beating heart. A closed-book exam would test that by asking you to reproduce some proofs. Your exam will be open book, so it can’t ask you to reproduce proofs, but it will test something arguably harder: that you understand them. So the proofs will need your attention and energy. 2 Prerequisites You are required to have taken these two courses: • Honours Algebra: We’ll need some linear algebra, corresponding to Chapter 1 of that course. For example, you should be able to convince yourself that an endomorphism of a finite-dimensional vector space is injective if and only if it is surjective. We’ll also need everything from Honours Algebra about rings and polynomials (Chapter 3 there), including ideals, quotient rings (factor rings), the universal property of quotient rings, and the first isomorphism theorem for rings. • Group Theory: From that course, we’ll need fundamentals such as normal subgroups, quotient groups, the universal property of quotient groups, and the first isomorphism theorem for groups. You should know lots about the symmetric groups ππ , alternating groups π΄π , cyclic groups πΆπ and dihedral groups π· π , and I hope you can list all of the groups of order < 8 without having to think too hard. Chapter 8 of Group Theory, on solvable groups, will be crucial for us! If you skipped it, you’ll need to go back and fix that. For example, you’ll need to understand the statement that π4 is solvable but π΄5 is not. We won’t need anything on free groups, the Sylow theorems, or the Jordan– Hölder theorem. If you’re a visiting student and didn’t take those courses, please get in touch so we can decide whether your background is suitable. Mistakes I’ll be grateful to hear of any mistakes (Tom.Leinster@ed.ac.uk), even if it’s something very small and even if you’re not sure. 3 Chapter 1 Overview of Galois theory Introduction to Week 1 This chapter stands apart from all the others, Modern treatments of Galois theory take advantage of several well-developed branches of algebra: the theories of groups, rings, fields, and vector spaces. This is as it should be! However, assembling all the algebraic apparatus will take us some time, during which it’s easy to lose sight of what it’s all for. Galois theory came from two basic insights: • every polynomial has a symmetry group; • this group determines whether the polynomial can be solved by radicals (in a sense I’ll define). In this chapter, I’ll explain these two ideas in as short and low-tech a way as I can manage. In Chapter 2 we’ll start again, beginning the modern approach that will take up the rest of the course. But I hope that all through that long build-up, you’ll keep in mind the fundamental ideas that you learn in this chapter. 1.1 The view of C from Q Imagine you lived several centuries ago, before the discovery of complex numbers. Your whole mathematical world is the real numbers, and there is no square root of −1. This situation frustrates you, and you decide to do something about it. So, you invent a new symbol π (for ‘imaginary’) and decree that π 2 = −1. You still want to be able to do all the usual arithmetic operations (+, ×, etc.), and you want to keep all the rules that govern them (associativity, commutativity, etc.). So you’re also forced to introduce new numbers such as 2 + 3 × π, and you end up with what today we call the complex numbers. So far, so good. But then you notice something strange. When you invented the complex numbers, you only intended to introduce one square root of −1. But 4 accidentally, you introduced a second one at the same time: −π. (You wait centuries for a square root of −1, then two arrive at once.) Maybe that’s not so strange in itself; after all, positive reals have two square roots too. But then you realize something genuinely weird: There’s nothing you can do to distinguish π from −π. Try as you might, you can’t find any reasonable statement that’s true for π but not −π. For example, you notice that π is a solution of π§ 3 − 3π§2 − 16π§ − 3 = 17 , π§ but then you realize that −π is too. Of course, there are unreasonable statements that are true for π but not −π, such as ‘π§ = π’. We should restrict to statements that only refer to the known world of real numbers. More precisely, let’s consider statements of the form π 1 (π§) π 3 (π§) = , π 2 (π§) π 4 (π§) (1.1) where π 1 , π 2 , π 3 , π 4 are polynomials with real coefficients. Any such equation can be rearranged to give π(π§) = 0, where again π is a polynomial with real coefficients, so we might as well just consider statements of that form. The point is that if π(π) = 0 then π(−π) = 0. Let’s make this formal. Definition 1.1.1 Two complex numbers π§ and π§0 are indistinguishable when seen from R, or conjugate over R, if for all polynomials π with coefficients in R, π(π§) = 0 ⇐⇒ π(π§0) = 0. Warning 1.1.2 The standard term is ‘conjugate over R’. ‘Indistinguishable’ is a term I invented for this chapter only, to express the idea of not being able to tell the two numbers apart. For example, π and −π are indistinguishable when seen from R. This follows from a more general result: Lemma 1.1.3 Let π§, π§0 ∈ C. Then π§ and π§0 are indistinguishable when seen from R if and only if π§0 = π§ or π§0 = π§. 5 Proof ‘Only if’: suppose that π§ and π§0 are indistinguishable when seen from R. Write π§ = π₯ + ππ¦ with π₯, π¦ ∈ R. Then (π§ − π₯) 2 = −π¦ 2 . Since π₯ and π¦ are real, indistinguishability implies that (π§0 − π₯) 2 = −π¦ 2 , so π§0 − π₯ = ±ππ¦, so π§0 = π₯ ± ππ¦. ‘If’: obviously π§ is indistinguishable from itself, so it’s enough to prove that π§ is indinguishable from π§. I’ll give two proofs. Each one teaches us a lesson that will be valuable later. First proof: recall that complex conjugation satisfies π€1 + π€2 = π€1 + π€2, π€1 · π€2 = π€1 · π€2 for all π€ 1 , π€ 2 ∈ C, and π = π for all π ∈ R. It follows by induction that for any polynomial π over R, π(π€) = π(π€) for all π€ ∈ C. So π(π§) = 0 ⇐⇒ π(π§) = 0 ⇐⇒ π(π§) = 0. Second proof: write π§ = π₯ + ππ¦ with π₯, π¦ ∈ R. Let π be a polynomial over R such that π(π§) = 0. We will prove that π(π§) = 0. This is trivial if π¦ = 0, so suppose that π¦ ≠ 0. Consider the polynomial π(π‘) = (π‘ − π₯) 2 + π¦ 2 . Then π(π§) = 0. You know from Honours Algebra that π(π‘) = π(π‘)π(π‘) + π (π‘) (1.2) for some polynomials π and π with deg(π) < deg(π) = 2 (so π is either a constant or of degree 1). Putting π‘ = π§ in (1.2) gives π (π§) = 0. But it’s easy to see that this is impossible unless π is the zero polynomial (using the assumption that π¦ ≠ 0). So π(π‘) = π(π‘)π(π‘). But π(π§) = 0, so π(π§) = 0, as required. We have just shown that for all polynomials π over R, if π(π§) = 0 then π(π§) = 0. Exchanging the roles of π§ and π§ proves the converse. Hence π§ and π§ are indistinguishable when seen from R. Exercise 1.1.4 Both proofs of ‘if’ contain little gaps: ‘It follows by induction’ in the first proof, and ‘it’s easy to see’ in the second. Fill them. Digression 1.1.5 With complex analysis in mind, we could imagine a stricter definition of indistinguishability in which polynomials are replaced by arbitrary convergent power series (still with coefficients in R). This would allow functions such as exp, cos and sin, and equations such as exp(ππ) = −1. 6 But this apparently different definition of indistinguishability is, in fact, equivalent. A complex number is still indistinguishable from exactly itself and its complex conjugate. (For example, exp(−ππ) = −1 too.) Do you see why? Example 1.1.6 Lemma 1.1.3 tells us that indistinguishability over R is rather simple. But the same idea becomes much more interesting if we replace R by Q. And in this course, we will mainly focus on polynomials over Q. Define indistinguishability seen from Q, or officially conjugacy over Q, by replacing R by Q in Definition 1.1.1. From now on, I will usually just say ‘indistinguishable’, dropping the ‘seen from Q’. √ √ Example 1.1.6 I claim that 2 and − 2 are indistinguishable. And I’ll give you two different proofs, closely analogous to the two proofs of the ‘if’ part of Lemma 1.1.3. First proof: write √ √ Q( 2) = {π + π 2 : π, π ∈ Q}. √ √ √ For π€ ∈ Q( 2), there are unique π, π ∈ Q such that π€ = π + π 2, because 2 is irrational. So it is logically valid to define √ √ π€ e = π − π 2 ∈ Q( 2). It is straightforward to check that π€Β f1 + π€ f2 , π€ Β f1 · π€ f2 1 + π€2 = π€ 1 · π€2 = π€ √ for all π€ 1 , π€ 2 ∈ Q( 2), and that e π = π for all π ∈ Q. Just as in the proof √ of Lemma 1.1.3,√it follows that π€ and π€ e are indistinguishable for every π€ ∈ Q( 2). √ In particular, 2 is indistinguishable from − 2. √Second proof: let π = π(π‘) be a polynomial with coefficients in Q such that π( 2) = 0. You know from Honours Algebra that π(π‘) = (π‘ 2 − 2)π(π‘) + π (π‘) √ for√some polynomials π(π‘) and π (π‘) over Q with deg π < 2. Putting π‘ = 2 gives √ π ( 2) = 0. But 2 is irrational and π (π‘) is of the form ππ‘ + π with π,√π ∈ Q, so π must be the zero polynomial. Hence π(π‘) = (π‘ 2 − 2)π(π‘), giving π(− √ 2) = 0. We √ have just shown that for all polynomials π √over Q, if √ π( 2) = 0 then π(− 2) = 0. The same √ argument with the roles of 2 and − 2 reversed proves the converse. Hence ± 2 are indistinguishable. 7 π π2 1 π3 π4 Figure 1.1: The 5th roots of unity. Exercise 1.1.7 Let π§ ∈ Q. Show that π§ is distinguishable from π§0 for any complex number π§0 ≠ π§. One thing that makes indistinguishability more subtle over Q than over R is that over Q, more than two numbers can be indistinguishable: Example 1.1.8 The 5th roots of unity are 1, π, π2 , π3 , π4 , where π = π 2ππ/5 (Figure 1.1). Now 1 is distinguishable from the rest, since it is a root of the polynomial π‘ − 1 and the others are not. (See also Exercise 1.1.7.) But it turns out that π, π2 , π3 , π4 are all indistinguishable from each other. Complex conjugates are indistinguishable when seen from R, so they’re certainly indistinguishable when seen from Q. Since π4 = 1/π = π, it follows that π and π4 are indistinguishable when seen from Q. By the same argument, π2 and π3 are indistinguishable. What’s not so obvious is that π and π2 are indistinguishable. I know two proofs, which are like the two proofs of Lemma 1.1.3 and Example 1.1.6. But we’re not equipped to do either yet. Example 1.1.9 More generally, let π be any prime and put π = π 2ππ/π . Then π, π2 , . . . , π π−1 are indistinguishable. So far, we have asked when one complex number can be distinguished from another, using only polynomials over Q. But what about more than one? Definition 1.1.10 Let π ≥ 0 and let (π§1 , . . . , π§ π ) and (π§01 , . . . , π§0π ) be π-tuples of complex numbers. We say that (π§ 1 , . . . , π§ π ) and (π§01 , . . . , π§0π ) are indistinguishable if for all polynomials π(π‘1 , . . . , π‘ π ) over Q in π variables, π(π§1 , . . . , π§ π ) = 0 ⇐⇒ π(π§01 , . . . , π§0π ) = 0. 8 When π = 1, this is just the previous definition of indistinguishability. Exercise 1.1.11 Suppose that (π§1 , . . . , π§ π ) and (π§01 , . . . , π§0π ) are indistinguishable. Show that π§π and π§0π are indistinguishable, for each π ∈ {1, . . . , π }. Example 1.1.12 For any π§1 , . . . , π§ π ∈ C, the π-tuples (π§1 , . . . , π§ π ) and (π§1 , . . . , π§ π ) are indistinguishable. For let π(π‘1 , . . . , π‘ π ) be a polynomial over Q. Then π(π§1 , . . . , π§ π ) = π(π§1 , . . . , π§ π ) since the coefficients of π are real, by a similar argument to the one in the first proof of Lemma 1.1.3. Hence π(π§1 , . . . , π§ π ) = 0 ⇐⇒ π(π§1 , . . . , π§ π ) = 0, which is what we had to prove. Example 1.1.13 Let π = π 2ππ/5 , as in Example 1.1.8. Then (π, π2 , π3 , π4 ) and (π4 , π3 , π2 , π) are indistinguishable, by Example 1.1.12. It can also be shown that (π, π2 , π3 , π4 ) and (π2 , π4 , π, π3 ) are indistinguishable, although the proof is beyond us for now. But (π, π2 , π3 , π4 ) and (π2 , π, π3 , π4 ) are distinguishable, since if we put π(π‘ 1 , π‘2 , π‘3 , π‘4 ) = π‘2 − π‘12 then π(π, π2 , π3 , π4 ) = 0 ≠ π(π2 , π, π3 , π4 ). So the converse of Exercise 1.1.11 is false: just because π§π and π§0π are indistinguishable for all π, it doesn’t follow that (π§ 1 , . . . , π§ π ) and (π§01 , . . . , π§0π ) are indistinguishable. 1.2 Every polynomial has a symmetry group. . . We are now ready to describe the first main idea of Galois theory: every polynomial has a symmetry group. 9 Definition 1.2.1 Let π be a polynomial with coefficients in Q. Write πΌ1 , . . . , πΌ π for its distinct roots in C. The Galois group of π is Gal( π ) = {π ∈ π π : (πΌ1 , . . . , πΌ π ) and (πΌπ(1) , . . . , πΌπ(π) ) are indistinguishable}. ‘Distinct roots’ means that we ignore any repetition of roots: e.g. if π (π‘) = π‘ 5 (π‘ − 1) 9 then π = 2 and {πΌ1 , πΌ2 } = {0, 1}. Exercise 1.2.2 Show that Gal( π ) is a subgroup of π π . (This one is maybe a bit tricky notationally. Stick to π = 3 if you like.) Exercise 1.2.2 Digression 1.2.3 I brushed something under the carpet. The definition of Gal( π ) depends on the order in which the roots are listed. Different orderings gives different subgroups of π π . However, these subgroups are all conjugate to each other, and therefore isomorphic as abstract groups. So Gal( π ) is well-defined as an abstract group, independently of the choice of ordering. Example 1.2.4 Let π be a polynomial over Q all of whose complex roots πΌ1 , . . . , πΌ π are rational. If π ∈ Gal( π ) then πΌπ(π) and πΌπ are indistinguishable for each π, by Exercise 1.1.11. But since they are rational, that forces πΌπ(π) = πΌπ (by Exercise 1.1.7), and since πΌ1 , . . . , πΌ π are distinct, π(π) = π. Hence π = id. So the Galois group of π is trivial. Example 1.2.5 Let π be a quadratic over Q. If π has rational roots then as we have just seen, Gal( π ) is trivial. If π has two non-real roots then they are complex conjugate, so Gal( π ) = π2 by Example 1.1.12. The remaining case is where π has two distinct roots that are real but not rational, and it can be shown that in that case too, Gal( π ) = π2 . Warning 1.2.6 On terminology: note that just now I said ‘non-real’. Sometimes people casually say ‘complex’ to mean ‘not real’. But try not to do this yourself. It makes as little sense as saying ‘real’ to mean ‘irrational’, or ‘rational’ to mean ‘not an integer’. Example 1.2.7 Let π (π‘) = π‘ 4 + π‘ 3 + π‘ 2 + π‘ + 1. Then (π‘ − 1) π (π‘) = π‘ 5 − 1, so π has roots π, π2 , π3 , π4 where π = π 2ππ/5 . We saw in Example 1.1.13 that 1 2 3 4 1 2 3 4 1 2 3 4 , ∈ Gal( π ), ∉ Gal( π ). 4 3 2 1 2 4 1 3 2 1 3 4 In fact, it can be shown that Gal( π ) = 1 2 3 4 2 4 1 3 10 πΆ4 . Example 1.2.8 Let π (π‘) = π‘ 3 + ππ‘ 2 + ππ‘ + π be a cubic over Q with no rational roots. Then ( √ π΄3 if −27π 2 + 18πππ − 4π3 − 4π 3 π + π 2 π2 ∈ Q, Gal( π ) π3 otherwise. This appears as Proposition 22.4 in Stewart, but is way beyond us for now: calculating Galois groups is hard. Galois groups, informally 1.3 . . . which determines whether it can be solved Here we meet the second main idea of Galois theory: the Galois group of a polynomial determines whether it can be solved. More exactly, it determines whether the polynomial can be ‘solved by radicals’. To explain what this means, let’s begin with the quadratic formula. The roots of a quadratic ππ‘ 2 + ππ‘ + π are √ −π ± π 2 − 4ππ . 2π After much struggling, it was discovered that there is a similar formula for cubics ππ‘ 3 + ππ‘ 2 + ππ‘ + π: the roots are given by q √ √ 3 q 3 −27π 2 π+9πππ−2π 3 +3π 3(27π 2 π 2 −18ππππ+4ππ3 +4π 3 π−π 2 π2 ) + −27π 2 π+9πππ−2π 3 −3π 3(27π 2 π 2 −18ππππ+4ππ3 +4π 3 π−π 2 π2 ) . √ 3 3 2π (No, you don’t need to memorize that!) This is a complicated formula, and there’s also something strange about it. Any nonzero complex number has three cube roots, √3 √3 and there are two signs in the formula (ignoring the 2 in the denominator), so it looks as if the formula gives nine roots for the cubic. But a cubic can only have three roots. What’s going on? It turns out that some of the nine aren’t roots of the cubic at all. You have to choose your cube roots carefully. Section 1.4 of Stewart’s book has much more on this point, as well as an explanation of how the cubic formula was obtained. (We won’t be going into this ourselves.) As Stewart also explains, there is a similar but even more complicated formula for quartics (polynomials of degree 4). Digression 1.3.1 Stewart doesn’t actually write out the explicit formula for the cubic, let alone the much worse one for the quartic. He just describes algorithms by which they can be solved. But if you unwind the algorithm for the cubic, you get the formula above. I have done this exercise once and do not recommend it. 11 Once mathematicians discovered how to solve quartics, they naturally looked for a formula for quintics (polynomials of degree 5). But it was eventually proved by Abel and Ruffini, in the early 19th century, that there is no formula like the quadratic, cubic or quartic formula for polynomials of degree ≥ 5. A bit more precisely, there is no formula for the roots in terms of the coefficients that uses only the usual arithmetic operations (+, −, ×, ÷) and πth roots (for integers π). Spectacular as this result was, Galois went further, and so will we. Informally, let us say that a complex number is radical if it can be obtained from the rationals using only the usual arithmetic operations and πth roots. For example, p3 √ √2 7 1 + 2 − 7 2 r q 4 6 + 5 23 is radical, whichever square root, cube root, etc., we choose. A polynomial over Q is solvable (or soluble) by radicals if all of its complex roots are radical. Example 1.3.2 Every quadratic √ over Q is solvable by radicals. This follows from the quadratic formula: (−π ± π 2 − 4ππ)/2π is visibly a radical number. Example 1.3.3 Similarly, the cubic formula shows that every cubic over Q is solvable by radicals. The same goes for quartics. Example 1.3.4 Some quintics are solvable by radicals. For instance, (π‘ − 1)(π‘ − 2)(π‘ − 3)(π‘ − 4)(π‘ − 5) is solvable by radicals, since all its roots are rational and, therefore, radical. A bit less trivially, (π‘ − 123) 5 √ + 456 is solvable by radicals, since its roots are the five 5 complex numbers 123 + −456, which are all radical. What determines whether a polynomial is solvable by radicals? Galois’s amazing achievement was to answer this question completely: Theorem 1.3.5 (Galois) Let π be a polynomial over Q. Then π is solvable by radicals ⇐⇒ Gal( π ) is a solvable group. Example 1.3.6 Definition 1.2.1 implies that if π has degree π then Gal( π ) is isomorphic to a subgroup of ππ . You saw in Group Theory that π4 is solvable, and that every subgroup of a solvable group is solvable. Hence the Galois group of any polynomial of degree ≤ 4 is solvable. It follows from Theorem 1.3.5 that every polynomial of degree ≤ 4 is solvable by radicals. 12 Example 1.3.7 Put π (π‘) = π‘ 5 −6π‘ +3. Later we’ll show that Gal( π ) = π5 . You saw in Group Theory that π5 is not solvable (or at least, you saw that π΄5 isn’t solvable, which implies that π5 isn’t either, as π5 contains π΄5 as a subgroup). Hence π is not solvable by radicals. If there was a quintic formula then all quintics would be solvable by radicals, for the same reason as in Examples 1.3.2 and 1.3.3. But since this is not the case, there is no quintic formula. Galois’s result is much sharper than Abel and Ruffini’s. They proved that there is no formula providing a solution by radicals of every quintic, whereas Galois found a way of determining which quintics (and higher) can be solved by radicals and which cannot. Digression 1.3.8 From the point of view of modern numerical computation, this is all a bit odd. Computationally speaking, there is probably not much difference between solving π‘ 5 + 3 = 0 to 100 decimal places (that is, finding √5 −3) and solving π‘ 5 − 6π‘ + 3 = 0 to 100 decimal places (that is, solving a polynomial that isn’t solvable by radicals). Numerical computation and abstract algebra have different ideas about what is easy and what is hard! ∗ ∗ ∗ This completes our overview of Galois theory. What’s next? Mathematics increasingly emphasizes abstraction over calculation. Individual mathematicians’ tastes vary, but the historical trend is clear. In the case of Galois theory, this means dealing with abstract algebraic structures, principally fields, instead of manipulating explicit algebraic expressions such as polynomials. The cubic formula already gave you a taste of how hairy that can get. Developing Galois theory using abstract algebraic structures helps us to see its connections to other parts of mathematics, and also has some fringe benefits. For example, we’ll solve some notorious geometry problems that perplexed the ancient Greeks and remained unsolved for millennia. For that and many other things, we’ll need some ring theory and some field theory—and that’s what’s next. 13 Chapter 2 Rings and fields We now start again. This chapter is a mixture of revision and material that is likely to be new to you. The revision is from Honours Algebra and Introduction to Number Theory (if you took it, which I won’t assume). Introduction to Week 2 2.1 Rings We’ll begin with some stuff you know—but with a twist. In this course, the word ring means commutative ring with 1 (multiplicative identity). Noncommutative rings and rings without 1 are important in some parts of mathematics, but since we’ll be focusing on commutative rings with 1, it will be easier to just call them ‘rings’. Given rings π and π, a homomorphism from π to π is a function π : π → π satisfying the equations π(π + π 0) = π(π) + π(π 0), π(ππ 0) = π(π)π(π 0), π(0) = 0, π(1) = 1 (note this!) π(−π) = −π(π), for all π, π 0 ∈ π . For example, complex conjugation is a homomorphism C → C. It is a very useful lemma that if π(π + π 0) = π(π) + π(π 0), π(ππ 0) = π(π)π(π 0), π(1) = 1 for all π, π 0 ∈ π then π is a homomorphism. In other words, to show that π is a homomorphism, you only need to check it preserves +, · and 1; preservation of 0 and negatives then comes for free. But you do need to check it preserves 1. That doesn’t follow from the other conditions. A subring of a ring π is a subset π ⊆ π that contains 0 and 1 and is closed under addition, multiplication and negatives. Whenever π is a subring of π , the inclusion π : π → π (defined by π(π ) = π ) is a homomorphism. 14 Warning 2.1.1 In Honours Algebra, rings had 1s but homomorphisms were not required to preserve 1. Similarly, subrings of π had to have a 1, but it was not required to be the same as the 1 of π . For example, take the ring C of complex numbers, the noncommutative ring π of 2 × 2 matrices over C, and the function π : C → π defined by π§ 0 π(π§) = . 0 0 In the terminology of Honours Algebra, π is a homomorphism and its image im π is a subring of π. But in our terminology, π is not a homomorphism (as π(1) ≠ πΌ) and im π is not a subring of π (as πΌ ∉ im π). Exercise 2.1.2 Let π be a ring and let S be any set (perhaps infinite) Ñ of subrings of π . Prove that their intersection π∈S π is also a subring of π . (In contrast, in the Honours Algebra setup, even the intersection of two subrings need not be a subring.) Example 2.1.3 For any ring π , there is exactly one homomorphism Z → π . Here is a sketch of the proof. To show there is at least one homomorphism π : Z → π , we will construct one. Define π on nonnegative integers π inductively by π(0) = 0 and π(π+1) = π(π)+1. (Thus, π(π) = 1 π + · · · + 1 π .) Define π on negative integers π by π(π) = −π(−π). A series of tedious checks shows that π is indeed a ring homomorphism. To show there is only one homomorphism Z → π , let π be any homomorphism Z → π ; we have to prove that π = π. Certainly π(0) = 0 = π(0). Next prove by induction on π that π(π) = π(π) for nonnegative integers π. I leave the details to you, but the crucial point is that because homomorphisms preserve 1, we must have π(π + 1) = π(π) + π(1) = π(π) + 1 for all π ≥ 0. Once we have shown that π and π agree on the nonnegative integers, it follows that for negative π, π(π) = −π(−π) = −π(−π) = π(π). The meaning of ‘π · 1’, and Exercise 2.2.7 Hence π(π) = π(π) for all π ∈ Z; that is, π = π. Usually we write π(π) as π · 1 π , or simply as π if it is clear from the context that π is to be interpreted as an element of π . 15 Every ring homomorphism π : π → π has an image im π, which is a subring of π, and a kernel ker π, which is an ideal of π . Warning 2.1.4 Subrings in ring theory are analogous to subgroups in group theory, and ideals in ring theory are analogous to normal subgroups in group theory. But whereas normal subgroups are a special kind of subgroup, ideals are not a special kind of subring! Subrings must contain 1, but most ideals don’t. Exercise 2.1.5 Prove that the only subring of a ring π that is also an ideal is π itself. Quotient rings Given a ring π and an ideal πΌ P π , we obtain the quotient ring or factor ring π /πΌ and the canonical homomorphism π πΌ : π → π /πΌ, which is surjective and has kernel πΌ. As explained in Honours Algebra, the quotient ring together with the canonical homomorphism has a ‘universal property’: given any ring π and any homomorphism π : π → π satisfying ker π ⊇ πΌ, there is exactly one homomorphism π¯ : π /πΌ → π such that this diagram commutes: π ππΌ / π π /πΌ π¯ π. (For a diagram to commute means that whenever there are two different paths from one object to another, the composites along the two paths are equal. Here, it means that π = π¯ β¦ π πΌ .) The first isomorphism theorem says that if π is surjective and has kernel equal to πΌ then π¯ is an isomorphism. So π πΌ : π → π /πΌ is essentially the only surjective homomorphism out of π with kernel πΌ. Digression 2.1.6 Loosely, the ideals of a ring π correspond one-to-one with the surjective homomorphisms out of π . This means four things: • given an ideal πΌ P π , we get a surjective homomorphism out of π (namely, π πΌ : π → π /πΌ); • given a surjective homomorphism π out of π , we get an ideal of π (namely, ker π); • if we start with an ideal πΌ of π , take its associated surjective homomorphism π πΌ : π → π /πΌ, then take its associated ideal, we end up where we started (that is, ker(π πΌ ) = πΌ); 16 • if we start with a surjective homomorphism π : π → π, take its associated ideal ker π, then take its associated surjective homomorphism πker π : π → π /ker π, we end up where we started (at least ‘up to isomorphism’, in that we have the isomorphism π¯ : π /ker π → π making the triangle commute). This is the first isomorphism theorem. Analogous stories can be told for groups and for modules. An integral domain is a ring π such that 0 π ≠ 1 π and for π, π 0 ∈ π , ππ 0 = 0 ⇒ π = 0 or π 0 = 0. Exercise 2.1.7 The trivial ring or zero ring is the one-element set with its only possible ring structure. Show that the only ring in which 0 = 1 is the trivial ring. Equivalently, an integral domain is a nontrivial ring in which cancellation is valid: π π = π 0 π implies π = π 0 or π = 0. Digression 2.1.8 Why is the condition 0 ≠ 1 in the definition of integral domain? My answer begins with a useful general point: the sum of no things should always be interpreted as 0. (The amount you pay in a shop is the sum of the prices of the individual things. If you buy no things, you pay £0.) Similarly, the product of no things should be interpreted as 1. Now consider the following condition on a ring π : for all π ≥ 0 and π 1 , . . . , π π ∈ π , π 1π 2 · · · π π = 0 ⇒ there exists π ∈ {1, . . . , π} such that π π = 0. (2.1) In the case π = 0, this says: if 1 = 0 then there exists π ∈ ∅ such that π π = 0. But any statement beginning ‘there exists π ∈ ∅’ is false! So in the case π = 0, condition (2.1) states that 1 ≠ 0. And for π = 2, it’s the main condition in the definition of integral domain. So ‘1 ≠ 0’ is the 0-fold analogue of the main condition. On the other hand, if (2.1) holds for π = 0 and π = 2 then a simple induction shows that it holds for all π ≥ 0. Conclusion: an integral domain can equivalently be defined as a ring in which (2.1) holds for all π ≥ 0. Let π be a subset of a ring π . The ideal hπi generated by π is the intersection of all the ideals of π containing π. You can show that any intersection of ideals is an ideal (much as you did for subrings in Exercise 2.1.2), so hπi is an ideal. It is 17 the smallest ideal of π containing π. That is, hπi is an ideal containing π, and if πΌ is another ideal containing π then hπi ⊆ πΌ. When π is a finite set {π 1 , . . . , π π }, we write hπi as hπ 1 , . . . , π π i, and it satisfies hπ 1 , . . . , π π i = {π 1π 1 + · · · + π π π π : π 1 , . . . , π π ∈ π }. (2.2) In particular, when π = 1, hπi = {ππ : π ∈ π }. Ideals of the form hπi are called principal ideals. A principal ideal domain is an integral domain in which every ideal is principal. Example 2.1.9 Z is a principal ideal domain. Indeed, if πΌ P Z then either πΌ = {0}, in which case πΌ = h0i, or πΌ contains some positive integer, in which case we can define π to be the least positive integer in πΌ and use the division algorithm to show that πΌ = hπi. Exercise 2.1.10 Fill in the details of Example 2.1.9. Let π and π be elements of a ring π . We say that π divides π , and write π | π , if there exists π ∈ π such that π = ππ. This condition is equivalent to π ∈ hπi, and to hπ i ⊆ hπi. An element π’ ∈ π is a unit if it has a multiplicative inverse, or equivalently if hπ’i = π . The units form a group π × under multiplication. For instance, Z× = {1, −1}. Exercise 2.1.11 Let π and π be elements of an integral domain. Show that π | π | π ⇐⇒ hπi = hπ i ⇐⇒ π = π’π for some unit π’. Elements π and π of a ring are coprime if for π ∈ π , π | π and π | π ⇒ π is a unit. Proposition 2.1.12 Let π be a principal ideal domain and π, π ∈ π . Then π and π are coprime ⇐⇒ ππ + ππ = 1 for some π, π ∈ π . Proof ⇒: suppose that π and π are coprime. Since π is a principal ideal domain, hπ, π i = hπ’i for some π’ ∈ π . Since π ∈ hπ, π i = hπ’i, we must have π’ | π, and similarly π’ | π . But π and π are coprime, so π’ is a unit. Hence 1 ∈ hπ’i = hπ, π i. But by equation (2.2), hπ, π i = {ππ + ππ : π, π ∈ π }, and the result follows. ⇐: suppose that ππ + ππ = 1 for some π, π ∈ π . If π’ ∈ π with π’ | π and π’ | π then π’ | (ππ + ππ ) = 1, so π’ is a unit. Hence π and π are coprime. 18 2.2 Fields A field is a ring πΎ in which 0 ≠ 1 and every nonzero element is a unit. Equivalently, it is a ring such that πΎ × = πΎ \ {0}. Every field is an integral domain. Exercise 2.2.1 Write down all the examples of fields that you know. A field πΎ has exactly two ideals: {0} and πΎ. For if {0} ≠ πΌ P πΎ then π’ ∈ πΌ for some π’ ≠ 0; but then π’ is a unit, so hπ’i = πΎ, so πΌ = πΎ. Lemma 2.2.2 Every homomorphism between fields is injective. Proof Let π : πΎ → πΏ be a homomorphism between fields. Then ker π P πΎ, so ker π is either {0} or πΎ. If ker π = πΎ then π(1) = 0; but π(1) = 1 by definition of homomorphism, so 0 = 1 in πΏ, contradicting the assumption that πΏ is a field. Hence ker π = {0}, that is, π is injective. Warning 2.2.3 With the Honours Algebra definition of homomorphism, Lemma 2.2.2 would be false, since the map with constant value 0 would be a homomorphism. Let π be any ring. By Example 2.1.3, there is a unique homomorphism π : Z → π . Its kernel is an ideal of the principal ideal domain Z. Hence ker π = hπi for a unique integer π ≥ 0. This π is called the characteristic of π , and written as char π . Explicitly, ( the least π > 0 such that π · 1 π = 0 π , if such an π exists; (2.3) char π = 0, otherwise. Another way to say it: for π ∈ Z, we have π · 1 π = 0 if and only if π is a multiple of char π . The concept of characteristic is mostly used in the case of fields. Examples 2.2.4 i. Q, R and C all have characteristic 0. ii. For a prime number π, we write F π for the field Z/hπi of integers modulo π. Then char F π = π. Lemma 2.2.5 The characteristic of an integral domain is 0 or a prime number. 19 Proof Let π be an integral domain and write π = char π . Suppose that π > 0; we must prove that π is prime. Since 1 ≠ 0 in an integral domain, π ≠ 1. (Remember that 1 is not a prime! So that step was necessary.) Now let π, π > 0 with ππ = π. Writing π for the unique homomorphism Z → π , we have π(π) π(π) = π(ππ) = π(π) = 0, and π is an integral domain, so π(π) = 0 or π(π) = 0. WLOG, π(π) = 0. But ker π = hπi, so π | π, so π = π. Hence π is prime. Examples 2.2.4 show that there exist fields of every possible characteristic. But there is no way of mapping between fields of different characteristics: Lemma 2.2.6 Let π : πΎ → πΏ be a homomorphism of fields. Then char πΎ = char πΏ. Proof Write ππΎ and ππΏ for the unique homomorphisms from Z to πΎ and πΏ, respectively. Since ππΏ is the unique homomorphism Z → πΏ, the triangle Z ππΎ ππΏ πΎ π / πΏ commutes. (Concretely, this says that π(π · 1πΎ ) = π · 1 πΏ for all π ∈ Z.) Hence ker(π β¦ ππΎ ) = ker ππΏ . But π is injective by Lemma 2.2.2, so ker(π β¦ ππΎ ) = ker ππΎ . Hence ker ππΎ = ker ππΏ , or equivalently, char πΎ = char πΏ. For example, the inclusion Q → R is a homomorphism of fields, and both have characteristic 0. Exercise 2.2.7 This proof of Lemma 2.2.6 is quite abstract. Find a more concrete proof, taking equation (2.3) as your definition of characteristic. (You will still need the fact that π is injective.) The meaning of ‘π · 1’, and Exercise 2.2.7 A subfield of a field πΎ is a subring that is a field. The prime subfield of πΎ is the intersection of all the subfields of πΎ. It is straightforward to show that any intersection of subfields is a subfield (just as you showed in Exercise 2.1.2 that any intersection of subrings is a subring). Hence the prime subfield is a subfield. It is the smallest subfield of πΎ, in the sense that any other subfield of πΎ contains it. Concretely, the prime subfield of πΎ is π · 1πΎ : π, π ∈ Z with π · 1πΎ ≠ 0 . π · 1πΎ 20 To see this, first note that this set is a subfield of πΎ. It is the smallest subfield of πΎ: for if πΏ is a subfield of πΎ then 1πΎ ∈ πΏ by definition of subfield, so π · 1πΎ ∈ πΏ for all integers π, so (π · 1πΎ )/(π · 1πΎ ) ∈ πΏ for all integers π and π such that π · 1πΎ ≠ 0. Examples 2.2.8 i. The field Q has no proper subfields, so the prime subfield of Q is Q itself. ii. Let π be a prime. The field F π has no proper subfields, so the prime subfield of F π is F π itself. Exercise 2.2.9 What is the prime subfield of R? Of C? The prime subfields appearing in Examples 2.2.8 were Q and F π . In fact, these are the only prime subfields of anything: Lemma 2.2.10 Let πΎ be a field. i. If char πΎ = 0 then the prime subfield of πΎ is Q. ii. If char πΎ = π > 0 then the prime subfield of πΎ is F π . In the statement of this lemma, as so often in mathematics, the word ‘is’ means ‘is isomorphic to’. I hope you’re comfortable with that by now. Proof For (i), suppose that char πΎ = 0. By definition of characteristic, π · 1πΎ ≠ 0 for all integers π ≥ 0. One can check that there is a well-defined homomorphism π : Q → πΎ defined by π/π β¦→ (π · 1πΎ )/(π · 1πΎ ). (The check uses the fact that π : π → π · 1πΎ is a homomorphism.) Now π is injective, being a homomorphism of fields, so im π Q. But im π is a subring of πΎ, and in fact a subfield since it is isomorphic to Q. It is the prime subfield, since Q has no proper subfields. For (ii), suppose that char πΎ = π > 0. By Lemma 2.2.5, π is prime. The unique homomorphism π : Z → πΎ has kernel hπi, by definition. By the first isomorphism theorem, im π Z/hπi = F π . But im π is a subring of πΎ, and in fact a subfield since is it isomorphic to F π . It is the prime subfield, since F π has no proper subfields. Lemma 2.2.11 Every finite field has positive characteristic. Proof By Lemma 2.2.10, a field of characteristic 0 contains a copy of Q and is therefore infinite. 21 Warning 2.2.12 There are also infinite fields of positive characteristic! We haven’t met one yet, but we will soon. So far, we have rather few examples of fields. The following construction will allow us to manufacture many, many more. An element π of a ring π is irreducible if π is not 0 or a unit, and if for π, π ∈ π , Building blocks π = ππ ⇒ π or π is a unit. For example, the irreducibles in Z are ±2, ±3, ±5, . . .. An element of a ring is reducible if it is not 0, a unit, or irreducible. So 0 and units count as neither reducible nor irreducible. Exercise 2.2.13 What are the irreducible elements of a field? Proposition 2.2.14 Let π be a principal ideal domain and 0 ≠ π ∈ π . Then π is irreducible ⇐⇒ π /hπi is a field. Proof Write π for the canonical homomorphism π → π /hπi. ⇒: suppose that π is irreducible. To show that 1 π /hπi ≠ 0 π /hπi , note that since π is not a unit, 1 π ∉ hπi = ker π, so 1 π /hπi = π(1 π ) ≠ 0 π /hπi . Next we have to show that every nonzero element of π /hπi is a unit, or equivalently that π(π ) is a unit whenever π ∈ π with π ∉ hπi. We have π - π , and π is irreducible, so π and π are coprime. Hence by Proposition 2.1.12 (and the assumption that π is a principal ideal domain), we can choose π, π ∈ π such that ππ + ππ = 1 π . Applying π to each side gives π(π)π(π) + π(π)π(π ) = 1 π /hπi . But π(π) = 0, so π(π)π(π ) = 1, so π(π ) is a unit. ⇐: suppose that π /hπi is a field. Then 1 π /hπi ≠ 0 π /hπi , that is, π(1 π ) ∉ ker π = hπi, that is, π - 1 π . Hence π is not a unit. Next we have to show that if π, π ∈ π with π = ππ then π or π is a unit. We have 0 = π(π) = π(π)π(π) 22 and π /hπi is an integral domain, so WLOG π(π) = 0. Then π ∈ ker π = hπi, so π = ππ0 for some π0 ∈ π . This gives π = ππ = ππ0 π. But π ≠ 0 by hypothesis, and π is an integral domain, so π0 π = 1. Hence π is a unit. Example 2.2.15 When π is an integer, Z/hπi is a field if and only if π is irreducible (that is, ± a prime number). Proposition 2.2.14 enables us to construct fields from irreducible elements. . . but irreducible elements of a principal ideal domain. Right now that’s not much help, because we don’t have many examples of principal ideal domains. But we will soon. 23 Chapter 3 Polynomials This chapter revisits and develops some themes you met in Honours Algebra. Although it’s long, it contains material you’ve seen before. Before you begin, it may help you to reread Section 3.3 (Polynomials) of the Honours Algebra notes. Introduction to Week 3 3.1 The ring of polynomials You already know the definition of polynomial, but I want to make a point by phrasing it in an unfamiliar way. Definition 3.1.1 Let π be a ring. A polynomial over π is an infinite sequence (π 0 , π 1 , π 2 , . . .) of elements of π such that {π : ππ ≠ 0} is finite. The set of polynomials over π forms a ring as follows: (π 0 , π 1 , . . .) + (π 0 , π 1 , . . .) = (π 0 + π 0 , π 1 + π 1 , . . .), (π 0 , π 1 , . . .) · (π 0 , π 1 , . . .) = (π 0 , π 1 , . . .) Õ where π π = ππ π π , (3.1) (3.2) (3.3) π, π : π+ π=π the zero of the ring is (0, 0, . . .), and the multiplicative identity is (1, 0, 0, . . .). Of course, we almost always write (π 0 , π 1 , π 2 , . . .) as π 0 + π 1 π‘ + π 2 π‘ 2 + · · · , or the same with some other symbol in place of π‘. In that notation, formulas (3.1) and (3.2) look like the usual formulas for addition and multiplication of polynomials. Nevertheless: Warning 3.1.2 A polynomial is not a function! A polynomial gives rise to a function, as we’ll recall in a moment. But a polynomial itself is a purely formal object. 24 Why study polynomials? The set of polynomials over π is written as π [π‘] (or π [π’], π [π₯], etc.). Since π = π [π‘] is itself a ring, we can consider the ring π[π’] = (π [π‘]) [π’], usually written as π [π‘, π’], and similarly π [π‘, π’, π£] = (π [π‘, π’]) [π£], etc. We use π , π, β, . . . and π (π‘), π(π‘), β(π‘), . . . interchangeably to denote elements of π [π‘]. A polynomial π = (π 0 , π 1 , . . .) over π gives rise to a function π → π π β¦→ π 0 + π 1π + π 2π 2 + · · · . (The sum on the right-hand side makes sense because only finitely many ππ s are nonzero.) This function is usually called π too. But calling it that is slightly dangerous, because: Warning 3.1.3 Different polynomials can give rise to the same function. For example, consider π‘, π‘ 2 ∈ F2 [π‘]. They are different polynomials: going back to Definition 3.1.1, they’re alternative notation for the sequences (0, 1, 0, 0, . . .) and (0, 0, 1, 0, . . .), which are plainly not the same. On the other hand, they induce the same function F2 → F2 , because π = π 2 for all (both) π ∈ F2 . Exercise 3.1.4 Show that whenever π is a finite nontrivial ring, it is possible to find distinct polynomials over π that induce the same function π → π . (Hint: are there finitely or infinitely many polynomials over π ? Functions π → π ?) The ring of polynomials has a universal property: a homomorphism from π [π‘] to some other ring π΅ is determined by its effect on constant polynomials and on π‘ itself, in the following sense. The universal property of π [π‘] Lemma 3.1.5 (Universal property of the polynomial ring) Let π and π΅ be rings. For every homomorphism π : π → π΅ and every π ∈ π΅, there is exactly one homomorphism π : π [π‘] → π΅ such that π (π) = π(π) π (π‘) = π. for all π ∈ π , (3.4) (3.5) On the left-hand side of (3.4), the ‘π’ means the polynomial π + 0π‘ + 0π‘ 2 + · · · . 25 Proof To show there is at most one such π, take any homomorphism π : π [π‘] → π΅ Í satisfying (3.4) and (3.5). Then for every polynomial π ππ π‘ π over π , Õ Õ ππ π‘ π = π (ππ )π (π‘) π since π is a homomorphism π π = π Õ π(ππ )ππ by (3.4) and (3.5). π So π is uniquely determined. To show there is at least one such π, define a function π : π [π‘] → π΅ by Õ Õ ππ π‘ π = π(ππ )ππ π π π Í ( π ππ π‘ π ∈ π [π‘]). Then π clearly satisfies conditions (3.4) and (3.5). It remains to check that π is a homomorphism. I will do the worst part of this, which is to check that π preserves multiplication, and leave the rest to you. Í Í So, take polynomials π (π‘) = π ππ π‘ π and π(π‘) = π π π π‘ π . Then π (π‘)π(π‘) = Í π π π π π‘ , where π π is as defined in equation (3.3). We have Õ ππ π‘ π π ( π π) = π Õπ = π(π π )π π by definition of π π Õ Õ = π ππ π π π π by definition of π π π, π : π+ π=π π = Õ Õ π(ππ )π(π π )π π since π is a homomorphism π π, π : π+ π=π = Õ = Õ π(ππ )π(π π )ππ+ π π, π π(ππ )π π π Õ π(π π )π π π = π ( π )π (π) by definition of π. Here are three uses for the universal property of the ring of polynomials. First: Definition 3.1.6 Let π : π → π be a ring homomorphism. The induced homomorphism π∗ : π [π‘] → π[π‘] is the unique homomorphism π [π‘] → π[π‘] such that π∗ (π) = π(π) for all π ∈ π and π∗ (π‘) = π‘. 26 The universal property guarantees that there is one and only one homomorphism π∗ with these properties. Concretely, Õ Õ π π(ππ )π‘ π ππ π‘ = π∗ π π Í for all π ππ π‘ π ∈ π [π‘]. Second, let π be a ring and π ∈ π . By the universal property, there is a unique homomorphism evπ : π [π‘] → π such that evπ (π) = π for all π ∈ π and evπ (π‘) = π. Concretely, Õ Õ ππ π‘ π = ππ π π evπ π Í π for all π ππ ∈ π [π‘]. This map evπ is called evaluation at π. Í (The notation ππ π‘ π for what is officially (π 0 , π 1 , . . .) makes it look obvious that we can evaluate a polynomial at an element and that this gives a homomorphism: of course ( π · π)(π) = π (π)π(π), for instance! But that’s only because of the notation: there was actually something to prove here.) Third, let π be a ring and π ∈ π . For any π (π‘) ∈ π [π‘], we can ‘substitute π‘ = π’ + π’ to get a polynomial in π’. What exactly does this mean? Formally, there is a unique homomorphism π : π [π‘] → π [π’] such that π (π) = π for all π ∈ π and π (π‘) = π’ + π. Concretely, Õ Õ ππ (π’ + π) π . π ππ π‘ π = π‘π π π This particular substitution is invertible. Informally, the inverse is ‘substitute π’ = π‘ − π’. Formally, there is a unique homomorphism π 0 : π [π’] → π [π‘] such that π 0 (π) = π for all π ∈ π and π 0 (π’) = π‘ − π. These maps π and π 0 carrying out the substitutions are inverse to each other, as you can deduce this from either the universal property or the concrete descriptions. So, the substitution maps π [π‘] o π π0 / π [π’] (3.6) define an isomorphism between π [π‘] and π [π’]. For example, since isomorphism preserve irreducibility (and everything else that matters!), π (π‘) is irreducible if and only if π (π‘ − π) is irreducible. Exercise 3.1.7 What happens to everything in the previous paragraph if we substitute π‘ = π’ 2 + π instead? The rest of this section is about degree. 27 Í Definition 3.1.8 The degree deg( π ) of a nonzero polynomial π (π‘) = ππ π‘ π is the largest π ≥ 0 such that π π ≠ 0. By convention, deg(0) = −∞, where −∞ is a formal symbol which we give the properties −∞ < π, (−∞) + π = −∞, (−∞) + (−∞) = −∞ for all integers π. Digression 3.1.9 Defining deg(0) like this is helpful because it allows us to make statements about all polynomials without having to make annoying exceptions for the zero polynomial (e.g. Lemma 3.1.10(i)). But putting deg(0) = −∞ also makes intuitive sense. At least for polynomials over R, the degree of a nonzero polynomial tells us how fast it grows: when π‘ is large, π (π‘) behaves roughly like π‘ deg( π ) . What about the zero polynomial? Well, whether or not π‘ is large, 0(π‘) = 0, and π‘ −∞ can sensibly be interpreted as limπ →−∞ π‘ π = 0. So it makes sense to put deg(0) = −∞. Lemma 3.1.10 Let π be an integral domain. Then: i. deg( π π) = deg( π ) + deg(π) for all π , π ∈ π [π‘]; ii. π [π‘] is an integral domain; iii. when π is a field, the units in π [π‘] are the polynomials of degree 0 (that is, the nonzero constants); iv. when π is a field, π (π‘) ∈ π [π‘] is irreducible if and only if deg( π ) > 0 and π cannot be expressed as a product of two polynomials of degree > 0. Proof At least parts (i)–(iii) were in Honours Algebra (Section 3.3). Part (iv) follows from the general definition of irreducible element of a ring. 3.2 Factorizing polynomials Every nonzero integer can be expressed as a product of primes in an essentially unique way. But the analogous statement is not true in all rings, or even all integral domains. Some rings have elements that can’t be expressed as a product of irreducibles at all. In other rings, factorizations into irreducibles exist but are not unique. (By ‘not unique’ I mean more than just changing the order of the factors or multiplying them by units.) The big theorem of this section is that, happily, every polynomial over a field can be factorized, essentially uniquely, into irreducibles. We begin with a result on division of polynomials from Section 3.3 of Honours Algebra. 28 Proposition 3.2.1 Let πΎ be a field and π , π ∈ πΎ [π‘]. Then there is exactly one pair of polynomials π, π ∈ πΎ [π‘] such that π = ππ + π and deg(π) < deg(π). We use this to prove an extremely useful fact: Proposition 3.2.2 Let πΎ be a field. Then πΎ [π‘] is a principal ideal domain. Proof First, πΎ [π‘] is an integral domain, by Lemma 3.1.10(ii). Now let πΌ P πΎ [π‘]. If πΌ = {0} then πΌ = h0i. Otherwise, put π = min{deg( π ) : 0 ≠ π ∈ πΌ} and choose π ∈ πΌ such that deg(π) = π. I claim that πΌ = hπi. To prove this, let π ∈ πΌ; we must show that π | π . By Proposition 3.2.1, π = ππ + π for some π, π ∈ πΎ [π‘] with deg(π) < π. Now π = π − ππ ∈ πΌ since π , π ∈ πΌ, so the minimality of π implies that π = 0. Hence π = ππ, as required. If you struggled with Exercise 2.1.10, that proof should give you a clue. Warning 3.2.3 Lemma 3.1.10(ii) implies that πΎ [π‘1 , . . . , π‘ π ] is an integral domain for all π ≥ 2, but it is not a principal ideal domain. For example, the ideal hπ‘1 , π‘2 i = { π (π‘1 , π‘2 ) ∈ Q[π‘ 1 , π‘2 ] : π has constant term 0} is not principal. Also, Proposition 3.2.2 really needed the hypothesis that πΎ is a field; it’s not enough for it to be a principal ideal domain. For example, Z is a principal ideal domain, but in Z[π‘], the ideal h2, π‘i = { π (π‘) ∈ Z[π‘] : the constant term of π is even} is not principal. Exercise 3.2.4 Prove that the ideals in Warning 3.2.3 are indeed not principal. Exercise 3.2.4: a non-principal ideal At the end of Chapter 2, I promised I’d give you a way of manufacturing lots of new fields. Here it is! Corollary 3.2.5 Let πΎ be a field and let 0 ≠ π ∈ πΎ [π‘]. Then π is irreducible ⇐⇒ πΎ [π‘]/h π i is a field. Proof This follows from Propositions 2.2.14 and 3.2.2. 29 To manufacture new fields using Corollary 3.2.5, we’ll need a way of knowing which polynomials are irreducible. That’s the topic of Section 3.3, but for now let’s stick to our mission: proving that every polynomial factorizes into irreducibles in an essentially unique way. To achieve this mission, we’ll need two more lemmas. Lemma 3.2.6 Let πΎ be a field and let π (π‘) ∈ πΎ [π‘] be a nonconstant polynomial. Then π (π‘) is divisible by some irreducible in πΎ [π‘]. The word nonconstant just means ‘of degree > 0’. Proof Let π be a nonconstant polynomial of smallest possible degree such that π | π . (For this to make sense, there must be at least one nonconstant polynomial dividing π , and there is: π .) I claim that π is irreducible. Proof: if π = π1 π2 then each ππ divides π , so by the minimality of deg(π), each ππ has degree 0 or deg(π). They cannot both have degree deg(π), since deg(π1 ) + deg(π2 ) = deg(π) > 0. So at least one has degree 0, i.e., is a unit. Lemma 3.2.7 Let πΎ be a field and π , π, β ∈ πΎ [π‘]. Suppose that π is irreducible and π | πβ. Then π | π or π | β. This behaviour is familiar in the integers: if a prime π divides some product ππ, then π | π or π | π. Proof Suppose that π - π. Since π is irreducible, π and π are coprime. Since πΎ [π‘] is a principal ideal domain, Proposition 2.1.12 implies that there are π, π ∈ πΎ [π‘] such that π π + ππ = 1. Multiplying both sides by β gives π π β + ππβ = β. But π | π π β and π | πβ, so π | β. Theorem 3.2.8 Let πΎ be a field and 0 ≠ π ∈ πΎ [π‘]. Then π = π π1 π2 · · · π π for some π ≥ 0, π ∈ πΎ and monic irreducibles π1 , . . . , ππ ∈ πΎ [π‘]. Moreover, π and π are uniquely determined by π , and π1 , . . . , ππ are uniquely determined up to reordering. 30 In the case π = 0, the product π1 · · · ππ should be interpreted as 1 (as in Digression 2.1.8). Monic means that the leading coefficient is 1. Proof First we prove that such a factorization exists, by induction on deg( π ). If deg( π ) = 0 then π is a constant π and we take π = 0. Now suppose that deg( π ) > 0 and assume the result for polynomials of smaller degree. By Lemma 3.2.6, there is an irreducible π dividing π , and we can assume that π is monic by dividing by a constant if necessary. Then π /π is a nonzero polynomial of smaller degree than π , so by inductive hypothesis, π /π = πβ1 · · · βπ for some π ∈ πΎ and monic irreducibles β1 , . . . , βπ . Rearranging gives π = πβ1 · · · βπ π, completing the induction. Now we prove uniqueness, again by induction on deg( π ). If deg( π ) = 0 then π is a constant π and the only possible factorization is the one with π = 0. Now suppose that deg( π ) > 0, and take two factorizations π π1 · · · ππ = π = ππ1 · · · ππ (3.7) where π, π ∈ πΎ and ππ , π π are monic irreducible. Since deg( π ) > 0, we have π, π ≥ 1. Now ππ | ππ1 · · · ππ , so by Lemma 3.2.7, ππ | π π for some π. By rearranging, we can assume that π = π. But ππ is also irreducible, so ππ = πππ for some nonzero π ∈ πΎ, and both ππ and ππ are monic, so π = 1. Hence ππ = ππ . Cancelling in (3.7) (which we can do as πΎ [π‘] is an integral domain) gives π π1 · · · ππ−1 = ππ1 · · · ππ−1 . By inductive hypothesis, π = π, π = π, and the lists of irreducibles π1 , . . . , ππ−1 and π1 , . . . , ππ−1 are the same up to reordering. This completes the induction. One way to find an irreducible factor of a polynomial π (π‘) ∈ πΎ [π‘] is to find a root (an element π ∈ πΎ such that π (π) = 0): Lemma 3.2.9 Let πΎ be a field, π (π‘) ∈ πΎ [π‘] and π ∈ πΎ. Then π (π) = 0 ⇐⇒ (π‘ − π) | π (π‘). Proof ⇒: suppose that π (π) = 0. By Proposition 3.2.1, π (π‘) = (π‘ − π)π(π‘) + π (π‘) (3.8) for some π, π ∈ πΎ [π‘] with deg(π) < 1. Then π is a constant, so putting π‘ = π in (3.8) gives π = 0. ⇐: if π (π‘) = (π‘ − π)π(π‘) for some polynomial π then π (π) = 0. 31 Definition 3.2.10 Let πΎ be a field, let 0 ≠ π ∈ πΎ [π‘], and let π ∈ πΎ be a root of π . The multiplicity of π is the unique integer π ≥ 1 such that (π‘ − π) π | π (π‘) but (π‘ − π) π+1 - π (π‘). Exercise 3.2.11 This definition assumes that there is a unique π with these properties (that is, there is one and only one such π). Prove it. Example 3.2.12 Over any field πΎ, the polynomial π‘ has a root at 0 with multiplicity 1, and the polynomial π‘ 2 has a root at 0 with multiplicity 2. This is true even when πΎ = F2 , in which case π‘ and π‘ 2 induce the same function πΎ → πΎ (Warning 3.1.2). Proposition 3.2.13 Let πΎ be a field and let 0 ≠ π ∈ πΎ [π‘]. Write π 1 , . . . , π π for the distinct roots of π in πΎ, and π 1 , . . . , π π for their multiplicities. Then π (π‘) = (π‘ − π 1 ) π1 · · · (π‘ − π π ) π π π(π‘) for some π(π‘) ∈ πΎ [π‘] that has no roots. Proof By induction on π. If π = 0, this is immediate (again interpreting an empty product as 1). Now suppose that π ≥ 1. By definition, (π‘ − π π ) π π | π (π‘), so we can put π (π‘) e ∈ πΎ [π‘]. π (π‘) = (π‘ − π π ) π π Any root of e π is a root of π , so it is one of π 1 , . . . , π π . But e π (π π ) ≠ 0: for e e if π (π π ) = 0 then (π‘ − π π ) | π (π‘) by Lemma 3.2.9, so (π‘ − π π ) π π +1 | π (π‘), a contradiction. Hence any root of e π is one of π 1 , . . . , π π−1 . These are indeed roots e of π , with multiplicities π 1 , . . . , π π−1 . So by inductive hypothesis, e π (π‘) = (π‘ − π 1 ) π1 · · · (π‘ − π π−1 ) π π−1 π(π‘) for some π(π‘) ∈ πΎ [π‘] with no roots, completing the induction. Corollary 3.2.14 Let πΎ be a field and π ∈ πΎ [π‘]. Suppose that π has distinct roots π 1 , . . . , π π ∈ πΎ with multiplicities π 1 , . . . , π π . Then π 1 + · · · + π π ≤ deg( π ). In other words, a polynomial of degree π has no more than π roots, even when you count the roots with multiplicities (e.g. count a double root twice). A field is algebraically closed if every nonconstant polynomial has at least one root. For example, C is algebraically closed (the fundamental theorem of algebra). Proposition 3.2.13 implies: Corollary 3.2.15 Let πΎ be an algebraically closed field and let 0 ≠ π ∈ πΎ [π‘]. Write π 1 , . . . , π π for the distinct roots of π in πΎ, and π 1 , . . . , π π for their multiplicities. Then π (π‘) = π(π‘ − π 1 ) π1 · · · (π‘ − π π ) π π , where π is the leading coefficient of π . 32 3.3 Irreducible polynomials Determining whether an integer is prime is generally hard, and determining whether a polynomial is irreducible is hard too. This section presents a few techniques for doing so. Let’s begin with the simplest cases. Recall Lemma 3.1.10(iv): a polynomial is irreducible if and only if it is nonconstant (has degree > 0) and cannot be expressed as a product of two nonconstant polynomials. Lemma 3.3.1 Let πΎ be a field and π ∈ πΎ [π‘]. i. If π is constant then π is not irreducible. ii. If deg( π ) = 1 then π is irreducible. iii. If deg( π ) ≥ 2 and π has a root then π is reducible. iv. If deg( π ) ∈ {2, 3} and π has no root then π is irreducible. Proof Parts (i) and (ii) follow from what we just recalled, and (iii) follows from Lemma 3.2.9. For (iv), suppose for a contradiction that π = πβ with deg(π), deg(β) ≥ 1. We have deg(π) + deg(β) ∈ {2, 3}, so without loss of generality, deg(π) = 1. Also without loss of generality, π is monic, say π(π‘) = π‘ + π; but then π (−π) = 0, a contradiction. Warning 3.3.2 The converse of (iii) is false! To show a polynomial is irreducible, it’s not enough to show it has no root (unless it has degree 2 or 3). For instance, (π‘ 2 + 1) 2 ∈ Q[π‘] has no root but is reducible. Examples 3.3.3 i. Let π be a prime. Then π (π‘) = 1 + π‘ + · · · + π‘ π−1 ∈ F π [π‘] is reducible, since π (1) = 0. ii. Let π (π‘) = π‘ 3 − 10 ∈ Q[π‘]. Then deg( π ) = 3 and π has no root in Q, so π is irreducible by part (iv) of the lemma. iii. Over C or any other algebraically closed field, the irreducibles are exactly the polynomials of degree 1. Exercise 3.3.4 If I gave you a quadratic over Q, how would you decide whether it is reducible or irreducible? From now on we focus on πΎ = Q. Any polynomial over Q can be multiplied by a nonzero rational constant to get a polynomial over Z, and that’s often a helpful move, so we’ll look at Z[π‘] too. 33 Definition 3.3.5 A polynomial over Z is primitive if its coefficients have no common divisor except for ±1. For example, 15 + 6π‘ + 10π‘ 2 is primitive but 15 + 6π‘ + 30π‘ 2 is not. Lemma 3.3.6 Let π (π‘) ∈ Q[π‘]. Then there exist a primitive polynomial πΉ (π‘) ∈ Z[π‘] and πΌ ∈ Q such that π = πΌπΉ. Í Proof Write π (π‘) = π (ππ /ππ )π‘ π , where ππ ∈ Z and 0 ≠ ππ ∈ Z. Take any common Í multiple π of the ππ s; then writing ππ = ππ π/ππ ∈ Z, we have π (π‘) = (1/π) ππ π‘ π . Now let π be the greatest common divisor of the ππ s, put ππ = ππ /π ∈ Z, and put Í πΉ (π‘) = ππ π‘ π . Then πΉ (π‘) is primitive and π (π‘) = (π/π)πΉ (π‘). If the coefficients of a polynomial π (π‘) ∈ Q[π‘] happen to all be integers, the word ‘irreducible’ could mean two things: irreducibility in the ring Q[π‘] or in the ring Z[π‘]. We say that π is irreducible over Q or Z to distinguish between the two. Suppose we have a polynomial over Z that’s irreducible over Z. In principle it could still be reducible over Q: although there’s no nontrivial way of factorizing it over Z, perhaps it can be factorized when you give yourself the freedom of non-integer coefficients. But the next result tells us that you can’t. Lemma 3.3.7 (Gauss) primitive. i. The product of two primitive polynomials over Z is ii. If a polynomial over Z is irreducible over Z, it is irreducible over Q. Proof For (i), let π and π be primitive polynomials over Z. Let π be a prime number. (We’re going to show that π doesn’t divide all the coefficients of π π.) Write π : Z → Z/πZ = F π for the canonical homomorphism, which induces a homomorphism π∗ : Z[π‘] → F π [π‘] as in Definition 3.1.6. Since π is primitive, π does not divide all the coefficients of π . Equivalently, π∗ ( π ) ≠ 0. Similarly, π∗ (π) ≠ 0. But F π [π‘] is an integral domain, so π∗ ( π π) = π∗ ( π )π∗ (π) ≠ 0, so π does not divide all the coefficients of π π. This holds for all primes π, so π π is primitive. For (ii), let π ∈ Z[π‘] be a polynomial irreducible over Z. Let π, β ∈ Q[π‘] with π = πβ. By Lemma 3.3.6, π = πΌπΊ and β = π½π» for some πΌ, π½ ∈ Q and primitive πΊ, π» ∈ Z[π‘]. Then πΌπ½ = π/π for some coprime integers π and π, giving π π = ππΊπ». (All three of these polynomials are over Z.) Now π divides every coefficient of π π , hence every coefficient of ππΊπ». Since π and π are coprime, π divides every coefficient of πΊπ». But πΊπ» is primitive by (i), so π = ±1, so π = ±ππΊπ». Since π is irreducible over Z, either πΊ or π» is constant, so π or β is constant, as required. 34 Gauss’s lemma quickly leads to a test for irreducibility. It involves taking a polynomial over Z and reducing it mod π, for some prime π. This means applying the map π∗ : Z[π‘] → F π [π‘] from the last proof. As we saw after Definition 3.1.6, Í Í if π (π‘) = ππ π‘ π then π∗ ( π )(π‘) = π(ππ )π‘ π , where π(ππ ) is the congruence class of ππ mod π. I’ll write π = π(π) and π = π∗ ( π ). That is, π is ‘ π mod π’. Proposition 3.3.8 (Mod π method) Let π (π‘) = π 0 + π 1 π‘ + · · · + π π π‘ π ∈ Z[π‘]. If there is some prime π such that π - π π and π ∈ F π [π‘] is irreducible, then π is irreducible over Q. I’ll give some examples first, then the proof. Examples 3.3.9 i. Let’s use the mod π method to show that π (π‘) = 9+14π‘ −8π‘ 3 is irreducible over Q. Take π = 7: then π (π‘) = 2 − π‘ 3 ∈ F7 [π‘], so it’s enough to show that 2 − π‘ 3 is irreducible over F7 . Since this has degree 3, it’s enough to show that π‘ 3 = 2 has no solution in F7 (by Lemma 3.3.1(iv)). And you can easily check this by computing 03 , (±1) 3 , (±2) 3 and (±3) 3 mod 7. ii. The condition in Proposition 3.3.8 that π - π π can’t be dropped. For instance, consider π (π‘) = 6π‘ 2 + π‘ and π = 2. Warning 3.3.10 Take π (π‘) as in Example 3.3.9(i), but this time take π = 3. Then π (π‘) = −π‘ + π‘ 3 ∈ F3 [π‘], which is reducible. But that doesn’t mean π is reducible! The mod π method only ever lets you show that a polynomial is irreducible over Q, not reducible. Proof of Proposition 3.3.8 Take a prime π satisfying the conditions in the Proposition. By Gauss’s lemma, it is enough to prove that π is irreducible over Z. Since π is irreducible, deg( π ) > 0, so deg( π ) > 0. Now let π = πβ in Z[π‘]. We have π = πβ and π is irreducible, so without loss of generality, π is constant. The leading coefficient of π is the product of the leading coefficients of π and β, and is not divisible by π, so the leading coefficient of π is not divisible by π. Hence deg(π) = deg(π) = 0. We finish with an irreducibility test that turns out to be surprisingly powerful. Proposition 3.3.11 (Eisenstein’s criterion) Let π (π‘) = π 0 + · · · + π π π‘ π ∈ Z[π‘], with π ≥ 1. Suppose there exists a prime π such that: • π - ππ; • π | ππ for all π ∈ {0, . . . , π − 1}; • π2 - π0. 35 Then π is irreducible over Q. To prove this, we will use the concept of the codegree codeg( π ) of a polynomial Í π (π‘) = π ππ π‘ π , which is defined to be the least π such that ππ ≠ 0 (if π ≠ 0), or as the formal symbol ∞ if π = 0. For polynomials π and π over an integral domain, codeg( π π) = codeg( π ) + codeg(π). Clearly codeg( π ) ≤ deg( π ) unless π = 0. Proof By Gauss’s lemma, it is enough to show π is irreducible over Z. Let π, β ∈ Z[π‘] with π = πβ. Continue to write π (π‘) ∈ F π [π‘] for π reduced mod π; then π = πβ. Since π 2 - π 0 = π (0) = π(0)β(0), we may assume without loss of generality that π - π(0). Hence codeg(π) = 0. Also, codeg( π ) = π, since π divides each of π 0 , . . . , π π−1 but not π π . So π = codeg( π ) = codeg(π) + codeg(β) = codeg(β) ≤ deg(β) ≤ deg(β), (3.9) giving π ≤ deg(β). But π = πβ with deg( π ) = π, so deg(β) = π and deg(π) = 0. Exercise 3.3.12 The last step in (3.9) was ‘deg(β) ≤ deg(β)’. Why is that true? And when does equality hold? Example 3.3.13 Let 5 1 2 π(π‘) = π‘ 5 − π‘ 4 + π‘ 3 + ∈ Q[π‘]. 9 3 3 Then π is irreducible over Q if and only if 9π(π‘) = 2π‘ 5 − 15π‘ 4 + 9π‘ 3 + 3 Testing for irreducibility is irreducible over Q, which it is by Eisenstein’s criterion with π = 3. Exercise 3.3.14 Use Eisenstein’s criterion to show that for every π ≥ 1, there is an irreducible polynomial over Q of degree π. I’ll give you one more example, and it’s not just any old polynomial: it’s an important one that we’ll need when we come to think about solvability by radicals. It needs a lemma. Lemma 3.3.15 Let π be a prime and 0 < π < π. Then π | ππ . 36 For example, the 7th row of Pascal’s triangle is 1, 7, 21, 35, 35, 21, 7, 1, and 7 divides all of these numbers apart from the first and last. Proof We have π!( π − π)! ππ = π!, and π divides π! but not π! or ( π − π)! (since π π is prime and 0 < π < π), so π must divide π . Example 3.3.16 Let π be a prime. The πth cyclotomic polynomial is π‘π − 1 . (3.10) π‘−1 I claim that Φ π is irreducible. We can’t apply Eisenstein to Φ π as it stands, because whichever prime we choose (whether it’s π or another one) doesn’t divide any of the coefficients. However, we saw on p. 27 that Φ π (π‘) is irreducible if and only if Φ π (π‘ − π) is irreducible, for any π ∈ Q. We’ll take π = −1. We have Φ π (π‘) = 1 + π‘ + · · · + π‘ π−1 = (π‘ + 1) π − 1 (π‘ + 1) − 1 π 1Õ π‘ π = π‘ π‘ π=1 π π π = π+ π‘ +···+ π‘ π−2 + π‘ π−1 . 2 π−1 Φ π (π‘ + 1) = So Φ π (π‘ + 1) is irreducible by Eisenstein’s criterion and Lemma 3.3.15, hence Φ π (π‘) is irreducible too. Digression 3.3.17 I defined the πth cyclotomic polynomial Φ π only when π is prime. The definition of Φπ for general π ≥ 1 is not the obvious generalization of (3.10). It’s this: Ö Φπ (π‘) = (π‘ − π), π where the product runs over all primitive πth roots of unity π. (In this context, ‘primitive’ means that π is the smallest number satisfying π π = 1; it’s a different usage from ‘primitive polynomial’.) Many surprising things are true. It’s not obvious that the coefficients of Φπ are real, but they are. Even given that they’re real, it’s not obvious that they’re rational, but they are. Even given that they’re rational, it’s not obvious that they’re integers, but they are. The degree of Φπ is π(π), the number of integers between 1 and π that are coprime with π (Euler’s function). It’s also true that the polynomial Φπ is irreducible for all π, not just primes. Some of these things are quite hard to prove, and results from Galois theory help. We probably won’t get into all of this, but you can read more here. 37 Chapter 4 Field extensions Introduction to Week 4 Roughly speaking, an ‘extension’ of a field πΎ is a field π that contains πΎ as a subfield. It’s not much of an exaggeration to say that field extensions are the central objects of Galois theory, in much the same way that vector spaces are the central objects of linear algebra. It will be a while before it becomes truly clear why field extensions are so important, but here are a couple of indications: • For any polynomial π over Q, we can take the smallest subfield π of C that contains all the complex roots of π , and that’s an extension of Q. • For any irreducible polynomial π over a field πΎ, the quotient ring π = πΎ [π‘]/h π i is a field. The constant polynomials form a subfield of π isomorphic to πΎ, so π is an extension of πΎ. It’s important to distinguish between these two types of example. The first extends Q by all the roots of π , whereas the second extends πΎ by just one root of π —as we’ll see. But let’s begin at the beginning. 4.1 Definition and examples Definition 4.1.1 A field extension consists of a field πΎ, a field π, and a homomorphism π : πΎ → π. Since homomorphisms between fields are injective (Lemma 2.2.2), πΎ is isomorphic to the subfield im(π) of π. It is usually safe to identify π(π) with π for each π ∈ πΎ—in other words, pretend that πΎ is actually a subfield of π and π(π) = π. We then say that π is an extension of πΎ and write π : πΎ. 38 It’s worth taking a minute to make sure you understand the relationship between subsets and injections. This is a fundamental point about sets, not fields. Given a set π΄ and a subset π΅ ⊆ π΄, there’s an inclusion function π : π΅ → π΄, defined by π(π) = π for all π ∈ π΅. (Remember that any function has a specified domain and codomain, so this isn’t the same thing as the identity on π΅.) This inclusion function is an injective. On the other hand, given any injective function between sets, say π : π → π΄, the image im π is a subset of π΄, and there’s a bijection π0 : π → im π given by π0 (π₯) = π(π₯) (π₯ ∈ π). Hence π is ‘isomorphic to’ (in bijection with) the subset im π of π΄. This back-and-forth process means that subsets and injections are more or less the same thing. Digression 4.1.2 If a field extension is basically the same thing as a field π together with a subfield πΎ, you might wonder why we bother with the more general Definition 4.1.1, involving an arbitrary homomorphism π. It turns out that in the long run, it makes things easier. There are two factors at play. The first is purely set-theoretic. The concept of subset isn’t actually as simple as it appears, at least when you look at what mathematicians do rather than what we claim we do. For example, (1) it’s common to define the set C as R2 , (2) everyone treats R as a subset of C, but (3) almost no one would say that R is a subset of R2 (if you wrote ‘the point π of R2 ’ on an exam, you’d be marked wrong). In truth, the common conventions are inconsistent. A good way to make everything respectable is to do everything in terms of injections rather than subsets. It would take up too much space to go into this here, but Definition 4.1.1 is one example of this approach in action. The second factor is algebraic. According to Definition 4.1.1, a field extension is simply a homomorphism between fields, so it includes examples such as the conjugation map : C → C. You might feel this example obeys the letter but not the spirit of Definition 4.1.1. But again, it turns out to be useful to include such examples. When we come to count isomorphisms between fields in a few weeks, you’ll see why. Examples 4.1.3 i. The inclusion π : Q → C is a field extension, usually just written as C : Q. Similarly, there are field extensions C : R and R : Q. ii. Let √ √ Q( 2) = {π + π 2 : π, π ∈ Q}. √ Then Q( 2) is a subring of C (easily), and in fact it’s a subfield: for if (π, π) ≠ (0, 0) then √ 1 π−π 2 √ = π + π 2 π 2 − 2π 2 39 √ (noting that the denominators are not 0 because 2√ is irrational). So we √ have an extension C : Q( 2). Also, because Q ⊆ Q( 2), we have another √ extension Q( 2) : Q. iii. By direct calculation or later theory (which will make it much easier), √ √ √ Q( 2, π) = {π + π 2 + ππ + π 2π : π, π, π, π ∈ Q} √ is also a subfield of C, so we have an extension Q( 2, π) : Q. iv. Let πΎ be a field. A rational expression over πΎ is a ratio of two polynomials π (π‘) , π(π‘) where π (π‘), π(π‘) ∈ πΎ [π‘] with π ≠ 0. Two such expressions, π1 /π1 and π2 /π2 , are regarded as equal if π1 π2 = π2 π1 in πΎ [π‘]. So formally, a rational expression is an equivalence class of pairs ( π , π) under the equivalence relation in the last sentence. The set of rational expressions over πΎ is denoted by πΎ (π‘). Rational expressions are added, subtracted and multiplied in the ways you’d expect, making πΎ (π‘) into a field. There is a homomorphism π : πΎ → πΎ (π‘) given by π(π) = π/1 (π ∈ πΎ). In other words, πΎ (π‘) contains a copy of πΎ as the constant rational expressions. So, we have a field extension πΎ (π‘) : πΎ. v. In particular, when πΎ = F π for some prime π, we have the field extension F π (π‘) : F π . Note that F π (π‘) is an infinite field of characteristic π! Fields of positive characteristic don’t have to be finite. So I’ve now fulfilled the promise I made in Warning 2.2.12. Warning 4.1.4 People sometimes say ‘rational function’ to mean ‘rational expression’. But just as for polynomials (Warnings 3.1.2 and 3.1.3), I want to emphasize that rational expressions are not functions. For instance, 1/(π‘ − 1) is a totally respectable element of πΎ (π‘). You don’t have to—and shouldn’t—worry about what happens when π‘ = 1, because π‘ is just a formal symbol (a mark on a piece of paper) rather than a variable, and 1/(π‘ − 1) is just a formal expression, not a function. If this puzzles you, I suggest going back to those warnings about polynomials, which make the same point in a simpler setting. 40 Exercise 4.1.5 Find two examples of fields πΎ such that Q ( πΎ ( √ Q( 2, π). (The symbol ( means proper subset.) For any kind of algebraic structure, there is a notion of the ‘substructure generated by’ a given subset. For example, when π is a subset of a group πΊ, you know what the subgroup generated by π is: it’s the intersection of all the subgroups containing π. Similarly, when π is a subset of a vector space π, the linear subspace generated by π (or ‘spanned by π’, as one usually says) is the intersection of all the linear subspaces containing π. We now make a similar definition for fields. Definition 4.1.6 Let πΎ be a field and π a subset of πΎ. The subfield of πΎ generated by π is the intersection of all subfields of πΎ containing π. You can check that any intersection of subfields of πΎ is a subfield of πΎ (even if it’s an uncountably infinite intersection). In fact, you already did most of the work for this in Exercise 2.1.2. So, the subfield πΉ of πΎ generated by π really is a subfield of πΎ. It contains π itself. By definition of intersection, πΉ is the smallest subfield of πΎ containing π, in the sense that any subfield of πΎ containing π contains πΉ. Exercise 4.1.7 Check the truth of all the statements in the previous paragraph. Examples 4.1.8 of πΎ. i. The subfield of πΎ generated by ∅ is the prime subfield ii. Let πΏ be the subfield of C generated by {π}. I claim that πΏ = {π + ππ : π, π ∈ Q}. To prove this, we have to show that πΏ is the smallest subfield of C containing π. First, it is a subfield of C (by an argument similar to Example 4.1.3(ii)) and it contains 0 + 1π = π. Now let πΏ 0 be any subfield of C containing π. Then πΏ 0 contains the prime subfield of C (by definition of prime subfield), which is Q. So whenever π, π ∈ Q, we have π, π, π ∈ πΏ 0 and so π + ππ ∈ πΏ 0. Hence πΏ ⊆ πΏ 0, as required. √ iii. A very similar argument shows √ that the subfield of C generated by 2 is what we have been calling Q( 2). 41 Exercise 4.1.9 What is the subfield of C generated by {7/8}? By {2 + 3π}? By R ∪ {π}? We will be very interested in chains of fields πΎ⊆πΏ⊆π in which πΎ and π are regarded as fixed and πΏ as variable. You can think of πΎ as the floor, π as the ceiling, and πΏ as varying in between. Definition 4.1.10 Let π : πΎ be a field extension and π ⊆ π. We write πΎ (π ) for the subfield of π generated by πΎ ∪ π , and call it πΎ with π adjoined. So, πΎ (π ) is the smallest subfield of π containing both πΎ and π . When π is a finite set {πΌ1 , . . . , πΌπ }, we write πΎ ({πΌ1 , . . . , πΌπ }) as πΎ (πΌ1 , . . . , πΌπ ). √ Examples 4.1.11 i. Take π : πΎ to be C√: Q and π = { 2}. Then πΎ (π ) is the smallest subfield of C containing Q ∪ { 2}. But every subfield of C contains Q: that’s what it means for Q to be√the prime subfield of C. So, πΎ (π ) is the 2. By Example 4.1.8(iii), smallest subfield of C containing √ that’s exactly √ √ what we’ve been calling Q( 2) all along. We refer to Q( 2) as ‘Q with 2 adjoined’. ii. Similarly, Q with π adjoined is Q(π) = {π + ππ : π, π ∈ Q} √ (Example 4.1.8(ii)), and Q with { 2, π} adjoined is the subfield denoted by √ Q( 2, π) in Example 4.1.3(iii). iii. Let π be a field and π ⊆ π. Write πΎ for the prime subfield of π. Then πΎ (π) is the smallest subfield of π containing πΎ and π. But every subfield of π contains πΎ, by definition of prime subfield. So πΎ (π) is the smallest subfield of π containing π; that is, it’s the subfield of π generated by π. √ We already saw this argument in (i), in the case π = C and π = { 2}. iv. Let πΎ be any field and let π be the field πΎ (π‘) of rational expressions over πΎ, which is an extension of πΎ. You might worry that there’s some ambiguity in the notation: πΎ (π‘) could either mean the field of rational expressions over πΎ (as defined in Example 4.1.3(iv)) or the subfield of πΎ (π‘) obtained by adjoining the element π‘ of πΎ (π‘) to πΎ (as in Definition 4.1.10). 42 In fact, they’re the same. In other words, the smallest subfield of πΎ (π‘) containing πΎ and π‘ is πΎ (π‘) itself. Or equivalently, the only subfield of πΎ (π‘) containing πΎ and π‘ is πΎ (π‘) itself. To see this, let πΏ be any such subfield. For Í any polynomial π (π‘) = ππ π‘ π over πΎ, we have π (π‘) ∈ πΏ, since ππ , π‘ ∈ πΏ and πΏ is closed under multiplication and addition. Hence for any polynomials π (π‘), π(π‘) over πΎ with π(π‘) ≠ 0, we have π (π‘), π(π‘) ∈ πΏ, so π (π‘)/π(π‘) ∈ πΏ as πΏ is closed under division by nonzero elements. So πΏ = πΎ (π‘). Warning 4.1.12 It is not true in general that πΎ (πΌ) = {π + ππΌ : π, π ∈ πΎ } (false!) (4.1) √ Examples like Q( 2) and Q(π) do satisfy this, but that’s only because √ 2 and π satisfy quadratic equations. Certainly the right-hand side is a subset of πΎ (πΌ), but in general it’s much smaller, and isn’t a subfield. You’ve just seen an example: the field πΎ (π‘) of rational expressions is much bigger than the set {π + ππ‘ : π, π ∈ πΎ } of polynomials of degree ≤ 1. And that set of polynomials isn’t closed under multiplication. Another example: let πΌ be the real cube root of 2. You can show that πΌ2 cannot be expressed as π + ππΌ for any π, π ∈ Q (a fact we’ll come back to in Example 4.2.9(ii)). But πΌ ∈ Q(πΌ), so πΌ2 ∈ Q(πΌ), so (4.1) fails in this case. In fact, Q(πΌ) = {π + ππΌ + ππΌ2 : π, π, π ∈ Q}. We’ll see why next week. Exercise 4.1.13 Let π : πΎ be a field extension. Show that πΎ (π ∪ π) = (πΎ (π ))(π) whenever π , π ⊆ π. (For example, πΎ (πΌ, π½) = (πΎ (πΌ))(π½) whenever πΌ, π½ ∈ π.) 4.2 Algebraic and transcendental elements A complex number πΌ is said to be ‘algebraic’ if π 0 + π 1 πΌ + · · · + π π πΌπ = 0 for some rational numbers ππ , not all zero. (You may have seen this definition with ‘integer’ instead of ‘rational number’; it makes no difference, as you can always clear the denominators.) This concept generalizes to arbitrary field extensions: 43 Definition 4.2.1 Let π : πΎ be a field extension and πΌ ∈ π. Then πΌ is algebraic over πΎ if there exists π ∈ πΎ [π‘] such that π (πΌ) = 0 but π ≠ 0, and transcendental otherwise. Exercise 4.2.2 Show that every element of πΎ is algebraic over πΎ. Examples 4.2.3 i. Let π ≥ 1. Then π 2ππ/π ∈ C is algebraic over Q, since π (π‘) = π‘ π − 1 is a nonzero polynomial such that π (π 2ππ/π ) = 0. ii. The numbers π and π are both transcendental over Q. Both statements are hard to prove (and we won’t). By Exercise 4.2.2, any complex number transcendental over Q is irrational. Proving the irrationality of π and π is already a challenge; proving they’re transcendental is even harder. iii. Although π is transcendental over Q, it is algebraic over R, since it’s an element of R. (Again, we’re using Exercise 4.2.2.) Moral: you shouldn’t say an element of a field is just ‘algebraic’ or ‘transcendental’; you should say it’s ‘algebraic/transcendental over πΎ’, specifying your πΎ. Or at least, you should do this when there’s any danger of confusion. iv. Take the field πΎ (π‘) of rational expressions over a field πΎ. Then π‘ ∈ πΎ (π‘) is transcendental over πΎ, since π (π‘) = 0 ⇐⇒ π = 0. The set of complex numbers algebraic over Q is written as Q. It’s a fact that Q is a subfield of C, but this is extremely hard to prove by elementary means. Next week I’ll show you that with a surprisingly small amount of abstract algebra, you can transform this from a very hard problem into an easy one. So that you appreciate the miracle later, I give you this unusual exercise now. Exercise 4.2.4 Attempt to prove any part of the statement that Q is a subfield of C. For example, try to show that Q is closed under addition, or multiplication, or reciprocals. I have no idea how to do any of these using only our current tools, but it’s definitely worth a few minutes of doomed effort to get a sense of the difficulties. Let π : πΎ be a field extension and πΌ ∈ π. An annihilating polynomial of πΌ is a polynomial π ∈ πΎ [π‘] such that π (πΌ) = 0. So, πΌ is algebraic if and only if it has some nonzero annihilating polynomial. It is natural to ask not only whether πΌ is annihilated by some nonzero polynomial, but which polynomials annihilate it. The situation is pleasantly simple: 44 Lemma 4.2.5 Let π : πΎ be a field extension and πΌ ∈ π. Then there is a polynomial π(π‘) ∈ πΎ [π‘] such that hπi = {annihilating polynomials of πΌ over πΎ }. (4.2) If πΌ is transcendental over πΎ then π = 0. If πΌ is algebraic over πΎ then there is a unique monic polynomial π satisfying (4.2). Proof By the universal property of polynomial rings (Lemma 3.1.5), there is a unique homomorphism π : πΎ [π‘] → π such that π (π) = π for all π ∈ πΎ and π (π‘) = πΌ. (Here we’re taking the ‘π’ of Lemma 3.1.5 to be the inclusion πΎ → π.) Then Õ Õ ππ π‘ π = π π π πΌπ for all Í ππ π‘ π ∈ πΎ [π‘], so ker π = {annihilating polynomials of πΌ over πΎ }. Now ker π is an ideal of the principal ideal domain πΎ [π‘] (using Proposition 3.2.2), so ker π = hπi for some π ∈ πΎ [π‘]. If πΌ is transcendental then ker π = {0}, so π = 0. If πΌ is algebraic then π ≠ 0. Multiplying a polynomial by a nonzero constant does not change the ideal it generates (by Exercise 2.1.11 and Lemma 3.1.10(iii)), so we can assume that π is monic. It remains to prove that π is the only monic polynomial such that hπi = ker π. But if π e is another monic polynomial such that hπ ei = ker π then π e = ππ for some nonzero constant π (again by Exercise 2.1.11 and Lemma 3.1.10(iii)), and both are monic, so π = 1 and π e = π. Definition 4.2.6 Let π : πΎ be a field extension and let πΌ ∈ π be algebraic over πΎ. The minimal polynomial of πΌ is the unique monic polynomial π satisfying (4.2). Exercise 4.2.7 What is the minimal polynomial of an element π of πΎ? This is an important definition, so we give some equivalent ways of stating it. Lemma 4.2.8 Let π : πΎ be a field extension, let πΌ ∈ π be algebraic over πΎ, and let π ∈ πΎ [π‘] be a monic polynomial. The following are equivalent: i. π is the minimal polynomial of πΌ over πΎ; 45 ii. π(πΌ) = 0, and π | π for all annihilating polynomials π of πΌ over πΎ; iii. π(πΌ) = 0, and deg(π) ≤ deg( π ) for all nonzero annihilating polynomials π of πΌ over πΎ; iv. π(πΌ) = 0 and π is irreducible over πΎ. Part (iii) says the minimal polynomial is a monic annihilating polynomial of least degree. Proof (i)⇒(ii) follows from the definition of minimal polynomial. (ii)⇒(iii) because if π | π ≠ 0 then deg(π) ≤ deg( π ). (iii)⇒(iv): assume (iii). First, π is not constant: for if π is constant then π = 1 (since π is monic); but π(πΌ) = 0, so 1 = 0 in πΎ, a contradiction. Next, suppose that π = π π for some π , π ∈ πΎ [π‘]. Then 0 = π(πΌ) = π (πΌ)π(πΌ), so without loss of generality, π (πΌ) = 0. By (iii), deg( π ) ≥ deg(π), so deg( π ) = deg(π) and deg(π) = 0. This proves (iv). (iv)⇒(i): assume (iv), and write π πΌ for the minimal polynomial of πΌ. We have π πΌ | π by definition of π πΌ and since π(πΌ) = 0. But π is irreducible and π πΌ is not constant, so π is a nonzero constant multiple of π πΌ . Since both are monic, π = π πΌ , proving (i). √ Examples 4.2.9 i. The minimal polynomial of 2 over Q is π‘ 2 − 2. There are several ways to see this. √ First method: π‘ 2 −2 is a monic annihilating polynomial of 2, and no nonzero √ polynomial of degree ≤ 1 over Q annihilates 2 since it is irrational. Then use Lemma 4.2.8(iii). Second method: π‘ 2 − 2 is an irreducible monic annihilating polynomial. It is irreducible either because π‘ 2 − 2 has degree 2 and has no rational roots (using Lemma 3.3.1(iv)), or by Eisenstein’s criterion with prime 2. Then use Lemma 4.2.8(iv). √3 ii. The minimal polynomial of 2 over Q is π‘ 3 − 2. This will follow from Lemma 4.2.8(iv) as long as π‘ 3 − 2 is irreducible, which you can show using either Lemma 3.3.1(iv) or Eisenstein. But unlike in (i), it’s not so easy to show directly that π‘ 3 −2 is the annihilating √3 polynomial of least degree. Try proving with your bare hands that 2 satisfies no quadratic equation over Q, i.e. that the equation √3 2 √3 2 = π 2+π has no solution for π, π ∈ Q. It’s not impossible, but it’s a mess. (You naturally begin by cubing both sides, but look what happens next. . . ) So the theory really gets us something here. Two traps 46 iii. Let π be a prime number, and put π = π 2ππ/π ∈ C. Then π is a root of π‘ π − 1, but that is not the minimal polynomial of π, since it is reducible: π‘ π − 1 = (π‘ − 1)π(π‘) where π(π‘) = π‘ π−1 + · · · + π‘ + 1. Since π π −1 = 0 but π−1 ≠ 0, we must have π(π) = 0. By Example 3.3.16, π is irreducible over Q. Hence π is the minimal polynomial of π over Q. 4.3 Simple extensions We can say a lot about extensions generated by a single element. Definition 4.3.1 A field extension π : πΎ is simple if there exists πΌ ∈ π such that π = πΎ (πΌ). Examples 4.3.2 √ i. √Surprisingly many algebraic extensions are simple. For 2, 3) : Q√is a simple extension (despite appearances), because instance, Q( √ √ √ in fact Q( 2, 3) = Q( 2 + 3). ii. πΎ (π‘) : πΎ is simple, where πΎ (π‘) is the field of rational expressions over πΎ. √ √ √ √ Exercise√4.3.3√Prove that Q( 2, 3) = Q( 2 + 3). Hint: begin by finding ( 2 + 3) 3 . Given a simple extension π : πΎ and an element πΌ that generates it, we can take the minimal polynomial of πΌ, which is irreducible over πΎ. But in the opposite direction, if you hand me a field πΎ and an irreducible polynomial π over πΎ, I can cook up for you a simple extension π : πΎ and an element πΌ ∈ π whose minimal polynomial is the π you gave me. This works as follows. Whenever π is an irreducible polynomial over a field πΎ, the quotient πΎ [π‘]/hπi is a field (Corollary 3.2.5). We have ring homomorphisms π πΎ → πΎ [π‘] −→ πΎ [π‘]/hπi, where the first map sends π ∈ πΎ to the constant polynomial π and π is the canonical homomorphism. Their composite is a homomorphism of fields πΎ → πΎ [π‘]/hπi. So, we have a field extension πΎ [π‘]/hπi : πΎ. And one of the elements of πΎ [π‘]/hπi is π(π‘), which I will call πΌ. 47 Lemma 4.3.4 Let π be a monic irreducible polynomial over a field πΎ. Then πΎ [π‘]/hπi : πΎ is a simple extension generated by πΌ (in the notation above), and the minimal polynomial of πΌ over πΎ is π. Proof We have ker π = hπi by definition of π, and ker π is the set of annihilating polynomials of πΌ over πΎ, so π is the minimal polynomial of πΌ over πΎ (by definition of minimal polynomial). To see that πΌ generates πΎ [π‘]/hπi over πΎ, let πΏ be a subfield of πΎ [π‘]/hπi containing πΎ and πΌ. Then π −1 πΏ is a subring of πΎ [π‘] containing πΎ and π‘ (since π(π‘) = πΌ), which forces π −1 πΏ = πΎ [π‘] and so πΏ = πΎ [π‘]/hπi. We’ll show that πΎ [π‘]/hπi is the only simple extension of πΎ by an element with minimal polynomial π. But the word ‘only’ is going to have to be interpreted in an up-to-isomorphism sense (as in ‘there’s only one group of order 2’). Here it is formally. Definition 4.3.5 Let πΎ be a field, and let π : πΎ → π and π0 : πΎ → π 0 be extensions of πΎ. A homomorphism π : π → π 0 is said to be a homomorphism over πΎ if / π π` 0 π > π0 π πΎ commutes. If π is invertible then we call π an isomorphism over πΎ. Exercise 4.3.6 In this definition, show that if π is invertible then π−1 is also a homomorphism over πΎ. How to understand simple algebraic extensions The next result not only classifies the simple extensions by an algebraic element (part (i)), but also those by a transcendental element (part (ii)). Theorem 4.3.7 (Classification of simple extensions) Let πΎ be a field. i. Let π ∈ πΎ [π‘] be a monic irreducible polynomial. Then there exist an extension π : πΎ and an algebraic element πΌ ∈ π such that π = πΎ (πΌ) and πΌ has minimal polynomial π over πΎ. Moreover, if (π, πΌ) and (π 0, πΌ0) are two such pairs, there is an isomorphism π : π → π 0 over πΎ such that π(πΌ) = πΌ0. ii. There exist an extension π : πΎ and a transcendental element πΌ ∈ π such that π = πΎ (πΌ). Moreover, if (π, πΌ) and (π 0, πΌ0) are two such pairs, there is an isomor- 48 phism π : π → π 0 over πΎ such that π(πΌ) = πΌ0. Proof The first part of (i) follows from Lemma 4.3.4. For ‘Moreover’, we may as well take π = πΎ [π‘]/hπi and πΌ as in Lemma 4.3.4. The homomorphism π: πΎ [π‘] → π0 Í π Í ππ π‘ β¦→ ππ πΌ0π (4.3) has kernel hπi, so πΎ [π‘]/hπi im π by the first isomorphism theorem, so im π is a field. Also, π (π) = π for all π ∈ πΎ. Hence im π is a subfield of π 0 containing π (π‘) = πΌ0 and πΎ. But we are assuming that π 0 = πΎ (πΌ0), which means that the only subfield of π 0 containing πΎ and πΌ0 is π 0 itself. So im π = π 0, giving πΎ [π‘]/hπi π 0. Diagram: π πΎ [π‘] e π / πΎ [π‘]/hπi O π / % 9π 0 πΎ (If you’re losing the thread, it may help to go back to the review of the universal property of quotients and the first isomorphism theorem on p. 16.) The isomorphism π : πΎ [π‘]/hπi → π 0 that we have constructed is an isomorphism over πΎ, and π (πΌ) = π (π(π‘)) = πΌ0. So π = π satisfies the conditions of the theorem. To prove the first part of (ii), we simply take π to be the field πΎ (π‘) of rational expressions over πΎ and πΌ = π‘. Then π‘ ∈ π is transcendental over πΎ (Example 4.2.3(iv)). For ‘Moreover’, take any simple extension π 0 of πΎ by a transcendental element πΌ0. Any π , π ∈ πΎ [π‘] with π ≠ 0 give rise to an element π (πΌ0)/π(πΌ0) ∈ π 0, where π(πΌ0) ≠ 0 because πΌ0 is transcendental. One can check that this gives a well-defined homomorphism π : πΎ (π‘) → π0 π (π‘) π (πΌ0) β¦→ π(π‘) π(πΌ0) ( π , π ∈ πΎ [π‘], π ≠ 0). Now π is injective (being a homomorphism of fields), so πΎ (π‘) im π, so im π is a subfield of π 0. Also, im π contains π (π‘) = πΌ0, and π (π) = π for each π ∈ πΎ. So im π is a subfield of π 0 containing πΌ0 and πΎ, which since π 0 = πΎ (πΌ0) implies that im π = π 0. So π : πΎ (π‘) → π 0 is an isomorphism, and it is an isomorphism over πΎ satisfying π (π‘) = πΌ0. Hence π = π satisfies the conditions of the theorem. 49 Conclusion: given any field πΎ (not necessarily Q!) and any monic irreducible π(π‘) ∈ πΎ [π‘], we can say the words ‘adjoin to πΎ a root πΌ of π’, and this unambiguously defines an extension πΎ (πΌ) : πΎ. (At least, unambiguously up to isomorphism over πΎ—but who could want more?) Similarly, we can unambiguously adjoin to πΎ a transcendental element. Examples 4.3.8 i. Let πΎ be any field not containing a square root of 2. Then 2 2 π‘ − 2 is irreducible √ over πΎ. So we can adjoin to πΎ a root of π‘ − 2, giving an extension πΎ ( 2) : πΎ. We√have already seen this example many times when πΎ = Q, in which case any πΎ ( 2) can be seen as a subfield of C. But the construction works for √ πΎ. For instance, 2 has no square root in F3 , so there is an extension F3 ( 2) of F3 . It can be constructed as F3 [π‘]/hπ‘ 2 − 2i. ii. The polynomial π(π‘) = 1 + π‘ + π‘ 2 is irreducible over F2 , so we may adjoin to F2 a root πΌ of π. Then F2 (πΌ) = F2 [π‘]/h1 + π‘ + π‘ 2 i. √ Exercise 4.3.9 How many elements does the field F3 ( 2) have? What about F2 (πΌ)? Warning 4.3.10 Take πΎ = Q and π(π‘) = π‘ 3 − 2, which is irreducible. Write πΌ1 , πΌ2 , πΌ3 for the roots of π in C. Then Q(πΌ1 ), Q(πΌ2 ) and Q(πΌ3 ) are all different as subsets of C. For example, one of the πΌπ is the real cube root of 2 (say πΌ1 ), which implies that Q(πΌ1 ) ⊆ R, whereas the other two are not real, so Q(πΌπ ) P R for π ≠ 1. However, Q(πΌ1 ) : Q, Q(πΌ2 ) : Q and Q(πΌ3 ) : Q are all isomorphic as abstract field extensions of Q. This follows from Theorem 4.3.7, since all the πΌπ have the same minimal polynomial, π. You’re already very familiar with this kind of situation in other branches of algebra, whether you realize it or not. For instance, in linear algebra, take three vectors v1 , v2 , v3 in R2 , none a scalar multiple of any other. Then span(v1 ), span(v2 ) and span(v3 ) are all different as subsets of R2 , but they are all isomorphic as abstract vector spaces (since they’re all 1-dimensional). A similar example could be given with a group containing several subgroups that are all isomorphic. You’ve seen that Galois theory involves aspects of group theory and ring theory. In the next chapter, you’ll see how linear algebra enters the picture too. 50 Chapter 5 Degree We’ve already seen that if you adjoin to Q a square root of 2, then each element of the resulting field can be specified using two rational numbers, π and π: √ √ Q( 2) = π + π 2 : π, π ∈ Q . Introduction to Week 5 We’ve also seen that if you adjoin to Q a cube root of 2, then it takes three rational numbers to specify each element of the resulting field: n o √3 √3 √3 2 Q( 2) = π + π 2 + π 2 : π, π, π ∈ Q √3 2) : Q is in some sense (Warning 4.1.12). This might lead us to suspect that Q( √ a ‘bigger’ extension than Q( 2) : Q. The first thing we’ll do in this chapter is to make√this intuition rigorous. We’ll √3 define the ‘degree’ of an extension and see that Q( 2) : Q and Q( 2) : Q have degrees 2 and 3, respectively. The concept of degree is incredibly useful, and not only in Galois theory. In fact, I’ll show you how it can be used to solve three problems that remained unsolved for literally millennia, since the time of the ancient Greeks. 5.1 Degrees of extensions and polynomials The concept of degree is an excellent illustration of a powerful mathematical technique: forgetting. Let π : πΎ be a field extension. What happens if we forget how to multiply together elements of π that aren’t in πΎ? We still have the field πΎ. What remains of π is its underlying additive abelian group (π, +, 0), and because we haven’t forgotten how to multiply elements of πΎ 51 with elements of π, we still have the multiplication function πΎ × π → π. So, we have a field πΎ, an abelian group π, and an action of πΎ on π. In other words, whenever π : πΎ is a field extension, π is a vector space over πΎ in a natural way. Definition 5.1.1 The degree [π : πΎ] of a field extension π : πΎ is the dimension of π as a vector space over πΎ. If π is a finite-dimensional vector space over πΎ, it’s clear what this means. If π is infinite-dimensional over πΎ, we write [π : πΎ] = ∞, where ∞ is a formal symbol which we give the properties π < ∞, π·∞=∞ (π ≥ 1), ∞·∞=∞ (where π is an integer). An extension π : πΎ is finite if [π : πΎ] < ∞. Digression 5.1.2 You know that whenever π is a finite-dimensional vector space, (i) there exists a basis of π, and (ii) there is a bijection between any two bases. This makes it possible to define the dimension of a vector space as the number of elements in a basis. In fact, both (i) and (ii) are true for every vector space, not just the finite-dimensional ones. So we can define the dimension of an arbitrary vector space as the ‘number’ of elements in a basis, where now ‘number’ means cardinal, i.e. isomorphism class of sets. We could interpret Definition 5.1.1 using this general definition of dimension. For instance, suppose we had one field extension π : πΎ such that π had a countably infinite basis over πΎ, and another, π 0 : πΎ, such that π 0 had an uncountably infinite basis over πΎ. Then [π : πΎ] and [π 0 : πΎ] would be different. However, we’ll lump all the infinite-dimensional extensions together and say that their degrees are all ∞. We’ll mostly be dealing with finite extensions anyway, and won’t need to distinguish between sizes of ∞. It’s a bit like the difference between a house that costs a million pounds and a house that costs ten million: although the difference in cost is huge, most of us would lump them together in a single category called ‘unaffordable’. Examples 5.1.3 i. Every field π contains at least one nonzero element, namely, 1. So [π : πΎ] ≥ 1 for every field extension π : πΎ. If π = πΎ then {1} is a basis, so [π : πΎ] = 1. On the other hand, if [π : πΎ] = 1 then the one-element linearly independent set {1} must be a basis, which implies that every element of π is equal to π · 1 = π for some π ∈ πΎ, and so π = πΎ. Hence [π : πΎ] = 1 ⇐⇒ π = πΎ. 52 ii. Every element of C is equal to π₯ + π¦π for a unique pair (π₯, π¦) of elements of R. That is, {1, π} is a basis of C over R. Hence [C : R] = 2. iii. Let πΎ be a field and πΎ (π‘) the field of rational expressions over πΎ. Then 1, π‘, π‘ 2 , . . . are linearly independent over πΎ, so [πΎ (π‘) : πΎ] = ∞. Warning 5.1.4 The degree [πΎ : πΎ] of πΎ over itself is 1, not 0. Degrees of extensions are never 0. See Example 5.1.3(i). Theorem 5.1.5 Let πΎ (πΌ) : πΎ be a simple extension, with πΌ algebraic over πΎ. Write π ∈ πΎ [π‘] for the minimal polynomial of πΌ and π = deg(π). Then 1, πΌ, . . . , πΌπ−1 is a basis of πΎ (πΌ) over πΎ. In particular, [πΎ (πΌ) : πΎ] = deg(π). Proof By Lemma 4.3.4 and Theorem 4.3.7(i), we might as well take πΎ (πΌ) = πΎ [π‘]/hπi and πΌ = π(π‘), where π : πΎ [π‘] → πΎ [π‘]/hπi is the canonical homomorphism. Since π is surjective, every element of πΎ (πΌ) is equal to π( π ) for some π ∈ πΎ [π‘]. By Proposition 3.2.1, there are unique π, π ∈ πΎ [π‘] such that π = ππ + π and deg(π) < π. In particular, there is a unique polynomial π ∈ πΎ [π‘] such that π − π ∈ hπi and deg(π) < π. Equivalently, there are unique π 0 , . . . , π π−1 ∈ πΎ such that π (π‘) − π 0 + π 1 π‘ + · · · + π π−1 π‘ π−1 ∈ hπi. Equivalently, there are unique π 0 , . . . , π π−1 ∈ πΎ such that π( π ) = π π 0 + π 1 π‘ + · · · + π π−1 π‘ π−1 . Equivalently (since π(π‘) = πΌ), there are unique π 0 , . . . , π π−1 ∈ πΎ such that π( π ) = π 0 + π 1 πΌ + · · · π π−1 πΌπ−1 . We have now shown that every element of πΎ (πΌ) can be expressed as a πΎ-linear combination of 1, πΌ, . . . , πΌπ−1 in a unique way. In other words, 1, πΌ, . . . , πΌπ−1 is a basis of πΎ (πΌ) over πΎ. Exercise 5.1.6 For a prime π, find [Q(π 2ππ/π ) : Q]. 53 Theorem 5.1.5 implies that when πΌ ∈ π is algebraic over πΎ, with minimal polynomial of degree π, the subset {π 0 + π 1 πΌ + · · · + π π−1 πΌπ−1 : ππ ∈ πΎ } is a subfield of π. This isn’t particularly obvious: for instance, why is it closed under taking reciprocals? But it’s true. Corollary 5.1.7 Let π : πΎ be a field extension and πΌ ∈ π. Then πΎ (πΌ) : πΎ is finite ⇐⇒ πΌ is algebraic over πΎ. Proof For ⇒, we prove the contrapositive. If πΌ is transcendental over πΎ then πΎ (πΌ) is isomorphic to πΎ (π‘) over πΎ (by Theorem 4.3.7(ii)), and [πΎ (π‘) : πΎ] = ∞ by Example 5.1.3(iii). Theorem 5.1.5 gives ⇐. For a field extension π : πΎ and πΌ ∈ π, the degree of πΌ over πΎ is [πΎ (πΌ) : πΎ]. We write it as degπΎ (πΌ). So, degπΎ (πΌ) < ∞ if and only if πΌ is algebraic over πΎ, and in that case, degπΎ (πΌ) is the degree of the minimal polynomial of πΌ over πΎ. Examples 5.1.8 i. Let πΌ ∈ C be an algebraic number over Q whose minimal polynomial is quadratic. Then by Theorem 5.1.5, Q(πΌ) = {π + ππΌ : π, π ∈ Q}. √ We’ve already seen this in many examples, such as πΌ = 2 and πΌ = π. ii. Let πΌ be the real cube root of 2. By Example 4.2.9(ii), the minimal polynomial of πΌ over Q is π‘ 3 − 2, so degQ (πΌ) = 3. It follows that Q(πΌ) ≠ {π + ππΌ : π, π ∈ Q}, since otherwise the two-element set {1, πΌ} would span the three-dimensional vector space Q(πΌ). So we have another proof that 22/3 cannot be written as a Q-linear combination of 1 and 21/3 . As observed in Example 4.2.9(ii), this is messy to prove directly. Theorem 5.1.5 is quite powerful. Here are two more corollaries of it. Corollary 5.1.9 i. Let π : πΏ : πΎ be field extensions and π½ ∈ π. Then [πΏ (π½) : πΏ] ≤ [πΎ (π½) : πΎ]. ii. Let π : πΎ be a field extension and πΌ, π½ ∈ π. Then [πΎ (πΌ, π½) : πΎ (πΌ)] ≤ [πΎ (π½) : πΎ]. Proof For (i): if [πΎ (π½) : πΎ] = ∞ then the inequality is clear. Otherwise, π½ is algebraic over πΎ (by Corollary 5.1.7), with minimal polynomial π ∈ πΎ [π‘], say. Then π is an annihilating polynomial for π½ over πΏ, so the minimal polynomial of π½ over πΏ has degree ≤ deg(π). The result follows from Theorem 5.1.5. Part (ii) follows by taking πΏ = πΎ (πΌ). 54 π π½ [πΎ (π½):πΎ] [πΏ(π½):πΏ] πΏ πΎ Figure 5.1: Visualization of Corollary 5.1.9(i) (not to be taken too seriously). Informally, I think of part (i) as in Figure 5.1. The degree of π½ over πΎ measures how far π½ is from being in πΎ. Since πΏ contains πΎ, it might be that π½ is closer to πΏ than to πΎ (i.e. [πΏ (π½) : πΏ] < [πΎ (π½) : πΎ]), and it’s certainly no further away. Exercise 5.1.10 Give an example to show that the inequality in Corollary 5.1.9(ii) can be strict. Your example can be as trivial as you like. Corollary 5.1.11 Let π : πΎ be a field extension and πΌ1 , . . . , πΌπ ∈ π. Suppose that each πΌπ is algebraic over πΎ of degree ππ . Then every element πΌ ∈ πΎ (πΌ1 , . . . , πΌπ ) can be represented as a polynomial in πΌ1 , . . . , πΌπ over πΎ: Õ ππ1 ,...,π π πΌ1π1 · · · πΌππ π πΌ= π 1 ,...,π π for some ππ1 ,...,π π ∈ πΎ, where ππ ranges over 0, . . . , ππ − 1. Proof By induction on π. When π = 0, this is trivial. Now let π ≥ 1 and suppose the result holds for π − 1. Let πΌ ∈ πΎ (πΌ1 , . . . , πΌπ ) = πΎ (πΌ1 , . . . , πΌπ−1 ) (πΌπ ). By Theorem 5.1.5 applied to the extension (πΎ (πΌ1 , . . . , πΌπ−1 ))(πΌπ ) πΎ (πΌ1 , . . . , πΌπ−1 ), noting that degπΎ (πΌ1 ,...,πΌπ−1 ) (πΌπ ) ≤ degπΎ (πΌπ ) = ππ , we have πΌ= πÕ π −1 ππ πΌππ : (5.1) π=0 for some π 0 , . . . , π ππ −1 ∈ πΎ (πΌ1 , . . . , πΌπ−1 ). By inductive hypothesis, for each π we have Õ π π−1 ππ = ππ1 ,...,π π−1 ,π πΌ1π1 · · · πΌπ−1 (5.2) π 1 ,...,π π−1 for some ππ1 ,...,π π−1 ,π ∈ πΎ, where ππ ranges over 0, . . . , ππ − 1. Substituting (5.2) into (5.1) completes the induction. 55 Example 5.1.12 Back in Example 4.1.11(ii), I claimed that √ √ √ Q( 2, π) = {π + π 2 + ππ + π 2π : π, π, π, π ∈ Q}. √ √ Corollary 5.1.11 applied to Q( 2, π) : Q proves this, since degQ ( 2) = degQ (π) = 2. Exercise 5.1.13 Let π : πΎ be a field extension and πΌ a transcendental element of π. Can every element of πΎ (πΌ) be represented as a polynomial in πΌ over πΎ? 5.2 The tower law We now know about the degrees of simple extensions—those obtained by adjoining a single element. What about extensions obtained by adjoining several elements? The following result is invaluable. Theorem 5.2.1 (Tower law) Let π : πΏ : πΎ be field extensions. i. If (πΌπ )π∈πΌ is a basis of πΏ over πΎ and (π½ π ) π ∈π½ is a basis of π over πΏ, then (πΌπ π½ π )(π, π)∈πΌ×π½ is a basis of π over πΎ. ii. π : πΎ is finite ⇐⇒ π : πΏ and πΏ : πΎ are finite. iii. [π : πΎ] = [π : πΏ] [πΏ : πΎ]. The sets πΌ and π½ here could be infinite. I’ll say that a family (ππ )π∈πΌ of elements of a field is finitely supported if the set {π ∈ πΌ : ππ ≠ 0} is finite. Proof To prove (i), we show that (πΌπ π½ π )(π, π)∈πΌ×π½ is a linearly independent spanning set of π over πΎ. For linear independence, let (ππ π )(π, π)∈πΌ×π½ be a finitely supported family of eleÍ Í Í Í ments of πΎ such that π, π ππ π πΌπ π½ π = 0. Then π ( π ππ π πΌπ ) π½ π = 0, with π ππ π πΌπ ∈ πΏ Í for each π ∈ π½. Since (π½ π ) π ∈π½ is linearly independent over πΏ, we have π ππ π πΌπ = 0 for each π ∈ π½. But (πΌπ )π∈πΌ is linearly independent over πΎ, so ππ π = 0 for each π ∈ πΌ and π ∈ π½. To show (πΌπ π½ π )(π, π)∈πΌ×π½ spans π over πΎ, let π ∈ π. Since (π½ π ) π ∈π½ spans π over Í πΏ, we have π = π π π π½ π for some finitely supported family (π π ) π ∈π½ of elements of Í πΏ. Since (πΌπ )π∈πΌ spans πΏ over πΎ, for each π ∈ π½ we have π π = π ππ π πΌπ for some Í finitely supported family (ππ π )π∈πΌ of πΎ. Hence π = π, π ππ π πΌπ π½ π , as required. Parts (ii) and (iii) follow. 56 √ √ Example 5.2.2 What is [Q( 2, 3) : Q]? The tower law gives √ √ √ √ √ √ Q( 2, 3) : Q = Q( 2, 3) : Q( 2) Q( 2) : Q √ √ √ = 2 Q( 2, 3) : Q( 2) . We have √ √ √ √ Q( 2, 3) : Q( 2) ≤ Q( 3) : Q = 2 √ √ √ √ √ by Corollary √ √ 5.1.9(ii). √ On the other hand, 3 ∉ Q( 2), so√Q(√2, 3) ≠ √ Q( 2), so [Q( 2, 3) : Q( 2)] √ > √1 by Example 5.1.3(i). So [Q( 2, 3) : Q( 2)] = 2, giving the answer: [Q( 2, 3) : Q] = 4. √ √ √ √ √ By the same argument as in Example √ 5.1.12, √ {1, 2, 3, 6} spans Q( 2, 3) over Q. But we have just shown that Q( 2, 3) has dimension √ 4√over Q. Hence this spanning set is a basis. That is, for every element πΌ ∈ Q( 2, 3), there is one and only one 4-tuple (π, π, π, π) of rational numbers such that √ √ √ πΌ = π + π 2 + π 3 + π 6. √ √ Exercise 5.2.3 In that example, I claimed that 3 ∉ Q( 2). Prove it. Corollary 5.2.4 Let π : πΏ 0 : πΏ : πΎ be field extensions. If π : πΎ is finite then [πΏ 0 : πΏ] divides [π : πΎ]. Proof Apply the tower law to π : πΏ 0 : πΏ then π : πΏ : πΎ. That result might remind you of Lagrange’s theorem on group orders. The resemblance is no coincidence, as we’ll see. Exercise 5.2.5 Show that a field extension whose degree is a prime number must be simple. That result might remind you of the fact that a group of prime order must be cyclic, and that’s no coincidence either! A second corollary of the tower law: Corollary 5.2.6 Let π : πΎ be a field extension and πΌ1 , . . . , πΌπ ∈ π. Then [πΎ (πΌ1 , . . . , πΌπ ) : πΎ] ≤ [πΎ (πΌ1 ) : πΎ] · · · [πΎ (πΌπ ) : πΎ]. Proof By the tower law and then Corollary 5.1.9(ii), [πΎ (πΌ1 , . . . , πΌπ ) : πΎ] = [πΎ (πΌ1 , . . . , πΌπ ) : πΎ (πΌ1 , . . . , πΌπ−1 )] · · · [πΎ (πΌ1 , πΌ2 ) : πΎ (πΌ1 )] [πΎ (πΌ1 ) : πΎ] ≤ [πΎ (πΌπ ) : πΎ] · · · [πΎ (πΌ2 ) : πΎ] [πΎ (πΌ1 ) : πΎ]. 57 finitely generated algebraic finite Figure 5.2: Finiteness conditions on a field extension Example 5.2.7 What is [Q(121/4 , 61/15 ) : Q]? You can check (hint, hint) that degQ (121/4 ) = 4 and degQ (61/15 ) = 15. So by Corollary 5.2.4, [Q(121/4 , 61/15 ) : Q] is divisible by 4 and 15. But also, Corollary 5.2.6 implies that [Q(121/4 , 61/15 ) : Q] ≤ 4 × 15 = 60. Since 4 and 15 are coprime, the answer is 60. Exercise 5.2.8 Generalize Example 5.2.7. In other words, what general result does the argument of Example 5.2.7 prove, not involving the particular numbers chosen there? 5.3 Algebraic extensions We defined a field extension π : πΎ to be finite if [π : πΎ] < ∞, that is, π is finite-dimensional as a vector space over πΎ. Here are two related conditions. Definition 5.3.1 A field extension π : πΎ is finitely generated if π = πΎ (π ) for some finite subset π ⊆ π. Definition 5.3.2 A field extension π : πΎ is algebraic if every element of π is algebraic over πΎ. Recall from Corollary 5.1.7 that πΌ is algebraic over πΎ if and only if πΎ (πΌ) : πΎ is finite. So for a field extension to be algebraic is also a kind of finiteness condition. Examples 5.3.3 i. For any field πΎ, the extension πΎ (π‘) : πΎ is finitely generated (take the ‘π ’ above to be {π‘}) but not finite, by Corollary 5.1.7. ii. In Section 4.2 you met the set Q of complex numbers algebraic over Q. We’ll very soon prove that it’s a subfield of C. It is algebraic over Q, by definition. But you’ll show in Workshop 3 that it is not finite over Q. Our three finiteness conditions are related as follows (Figure 5.2). 58 Proposition 5.3.4 The following conditions on a field extension π : πΎ are equivalent: i. π : πΎ is finite; ii. π : πΎ is finitely generated and algebraic; iii. π = πΎ (πΌ1 , . . . , πΌπ ) for some finite set {πΌ1 , . . . , πΌπ } of elements of π algebraic over πΎ. Proof (i)⇒(ii): suppose that π : πΎ is finite. To show that π : πΎ is finitely generated, take a basis πΌ1 , . . . , πΌπ of π over πΎ. Every subfield πΏ of π containing πΎ is a πΎ-linear subspace of π, so if πΌ1 , . . . , πΌπ ∈ πΏ then πΏ = π. This proves that the only subfield of π containing πΎ ∪ {πΌ1 , . . . , πΌπ } is π itself; that is, π = πΎ (πΌ1 , . . . , πΌπ ). So π : πΎ is finitely generated. To show that π : πΎ is algebraic, let πΌ ∈ π. Then by part (ii) of the tower law (Theorem 5.2.1), πΎ (πΌ) : πΎ is finite, so by Corollary 5.1.7, πΌ is algebraic over πΎ. (ii)⇒(iii) is immediate from the definitions. (iii)⇒(i): suppose that π = πΎ (πΌ1 , . . . , πΌπ ) for some πΌπ ∈ π algebraic over πΎ. Then [π : πΎ] ≤ [πΎ (πΌ1 ) : πΎ] · · · [πΎ (πΌπ ) : πΎ] by Corollary 5.2.6. For each π, we have [πΎ (πΌπ ) : πΎ] < ∞ since πΌπ is algebraic over πΎ (using Corollary 5.1.7 again). So [π : πΎ] < ∞. We already saw that when π = πΎ (πΌ1 , . . . , πΌπ ) with each πΌπ algebraic, every element of π is a polynomial in πΌ1 , . . . , πΌπ (Corollary 5.1.11). So for any finite extension π : πΎ, there is some finite set of elements such that everything in π can be expressed as a polynomial over πΎ in these elements. Exercise 5.3.5 Let π : πΎ be a field extension and πΎ ⊆ πΏ ⊆ π. In the proof of Proposition 5.3.4, I said that if πΏ is a subfield of π then πΏ is a πΎ-linear subspace of π. Why is that true? And is the converse also true? Give a proof or a counterexample. Corollary 5.3.6 Let πΎ (πΌ) : πΎ be a simple extension. The following are equivalent: i. πΎ (πΌ) : πΎ is finite; ii. πΎ (πΌ) : πΎ is algebraic; iii. πΌ is algebraic over πΎ. 59 Proof (i)⇒(ii) follows from (i)⇒(ii) of Proposition 5.3.4. (ii)⇒(iii) is immediate from the definitions. (iii)⇒(i) follows from (iii)⇒(i) of Proposition 5.3.4. Here’s a spectacular application of Corollary 5.3.6. Proposition 5.3.7 Q is a subfield of C. Proof By Corollary 5.3.6, Q = {πΌ ∈ C : [Q(πΌ) : Q] < ∞}. For all πΌ, π½ ∈ Q, [Q(πΌ, π½) : Q] ≤ [Q(πΌ) : Q] [Q(π½) : Q] < ∞ by Corollary 5.2.6. Hence [Q(πΌ + π½) : Q] ≤ [Q(πΌ, π½) : Q] < ∞, giving πΌ + π½ ∈ Q. Similarly, πΌ · π½ ∈ Q. For all πΌ ∈ Q, [Q(−πΌ) : Q] = [Q(πΌ) : Q] < ∞, giving −πΌ ∈ Q. Similarly, 1/πΌ ∈ Q (if πΌ ≠ 0). And clearly 0, 1 ∈ Q. If you did Exercise 4.2.4, you’ll appreciate how hard that result is to prove from first principles, and how amazing it is that the proof above is so clean and simple. 5.4 Ruler and compass constructions The ancient Greeks developed planar geometry to an extraordinary degree, discovering how to perform a very wide range of constructions using only ruler and compasses. But there were three particular constructions that they couldn’t figure out how to do using only these instruments: • Trisect the angle: given an angle π, construct the angle π/3. • Duplicate the cube: given a length, construct a new length whose cube is twice the cube of the original.√ That is, given two points distance πΏ apart, 3 construct two points distance 2πΏ apart. • Square the circle: given a circle, construct a square with the same area. That is, given two points distance πΏ apart, construct two points distance √ ππΏ apart. 60 The challenge of finding constructions lay unanswered for thousands of years. And it wasn’t for lack of attention: mathematicians kept on looking. My Galois theory lecture notes from when I was an undergraduate contain the following words: Thomas Hobbes claimed to have solved these. John Wallis disagreed. A 17th century pamphlet war ensued. Twitter users may conclude that human nature has not changed. It turns out that the reason why no one could find a way to do these constructions is that they’re impossible. We’ll prove it using field theory. In order to prove that you can’t do these things using ruler and compasses, it’s necessary to know that you can do certain other things using ruler and compasses. I’ve made a video showing how to do the various constructions we’ll need. Ruler and compass constructions Digression 5.4.1 The standard phrase is ‘ruler and compass constructions’, but it’s slightly misleading. A ruler has distance markings on it, whereas for the problems of ancient Greece, you’re supposed to use only a ‘straight edge’: a ruler without markings (and no, you’re not allowed to mark it). As Stewart explains (Section 7.1), with a marked or markable straight edge, you can solve all three problems. Also, for what it’s worth, an instrument for drawing circles is strictly speaking a pair of compasses. But like everyone else, we’ll say ‘ruler and compass’— —when we really mean ‘straight edge and compasses’— The problems as stated above are maybe not quite precise; let’s formalize them. Starting from a subset Σ of the plane, our instruments allow the following constructions: • given two distinct points π΄, π΅ of Σ, draw the (infinite) line through π΄ and π΅; • given two distinct points π΄, π΅ of Σ, draw the circle with centre π΄ passing through π΅. 61 A point in the plane is immediately constructible from Σ if it is a point of intersection between two distinct lines, or two distinct circles, or a line and a circle, of the form above. A point πΆ in the plane is constructible from Σ if there is a finite sequence πΆ1 , . . . , πΆπ = πΆ of points such that πΆπ is immediately constructible from Σ ∪ {πΆ1 , . . . , πΆπ−1 } for each π. So far I have written in the Greek spirit by saying ‘the plane’ rather than R2 . But now fix a coordinate system. For Σ ⊆ R2 , write πΎΣ = Q {πΌ ∈ R : πΌ is a coordinate of some point in Σ} , which is a subfield of R. The condition on πΌ means that (πΌ, π¦) ∈ Σ for some π¦ ∈ R or (π₯, πΌ) ∈ Σ for some π₯ ∈ R. The key to the impossibility proofs is the following definition. For subfields πΎ ⊆ π ⊆ R, let us say that π : πΎ is an iterated quadratic extension if there is some finite sequence of subfields πΎ = πΎ0 ⊆ πΎ1 ⊆ · · · ⊆ πΎπ = π such that [πΎπ : πΎπ−1 ] = 2 for all π ∈ {1, . . . , π}. Theorem 5.4.2 Let Σ ⊆ R2 and (π₯, π¦) ∈ R2 . If (π₯, π¦) is constructible from Σ then there is an iterated quadratic extension of πΎΣ containing π₯ and π¦. Before I show you the proof, I’ll state a corollary that reveals how this theorem leads to proofs of impossibility. Corollary 5.4.3 Let Σ ⊆ R2 and (π₯, π¦) ∈ R2 . If (π₯, π¦) is constructible from Σ then π₯ and π¦ are algebraic over πΎΣ , and their degrees over πΎΣ are powers of 2. Proof Take an iterated quadratic extension π of πΎΣ with π₯ ∈ π. Then [π : πΎΣ ] = 2π for some π ≥ 0, by the tower law. But then degπΎΣ (π₯) = [πΎΣ (π₯) : πΎΣ ] divides 2π by Corollary 5.2.4, and is therefore a power of 2. And similarly for π¦. We will show that if (for instance) we could trisect angles, we would be able to construct a point whose coordinates do not have degree a power of 2, giving a contradiction. To prove Theorem 5.4.2, we need a lemma. Lemma 5.4.4 Let πΎ be a subfield of R and πΌ, π½ ∈ R. Suppose that πΌ and π½ are each contained in some iterated quadratic extension of πΎ. Then there is some iterated quadratic extension of πΎ containing both πΌ and π½. 62 Proof Take subfields πΎ = πΎ0 ⊆ πΎ1 ⊆ · · · ⊆ πΎπ ⊆ R, πΎ = πΏ0 ⊆ πΏ1 ⊆ · · · ⊆ πΏπ ⊆ R with πΌ ∈ πΎπ , π½ ∈ πΏ π , and [πΎπ : πΎπ−1 ] = 2 = [πΏ π : πΏ π−1 ] for all π, π. For each π ∈ {1, . . . , π}, choose some π½ π ∈ πΏ π \ πΏ π−1 ; then πΏ π = πΏ π−1 (π½ π ). Hence πΏ π = πΎ (π½1 , . . . , π½ π ) for each π. Now consider the chain of subfields πΎ = πΎ0 ⊆ πΎ1 ⊆ · · · ⊆ πΎπ ⊆ πΎπ (π½1 ) ⊆ πΎπ (π½1 , π½2 ) ⊆ · · · ⊆ πΎπ (π½1 , . . . , π½π ). (5.3) For each π ∈ {1, . . . , π}, Corollary 5.1.9(i) gives the inequality [πΎπ (π½1 , . . . , π½ π ) : πΎπ (π½1 , . . . , π½ π−1 )] ≤ [πΎ (π½1 , . . . , π½ π ) : πΎ (π½1 , . . . , π½ π−1 )] = [πΏ π : πΏ π−1 ] = 2. So in the chain of subfields (5.3), each successive extension has degree 1 or 2. An extension of degree 1 is an equality, so by ignoring repeats, we see that πΎπ (π½1 , . . . , π½π ) is an iterated quadratic extension of πΎ. Finally, πΌ ∈ πΎπ ⊆ πΎπ (π½1 , . . . , π½π ), π½ ∈ πΏ π = πΎ (π½1 , . . . , π½π ) ⊆ πΎπ (π½1 , . . . , π½π ), so πΌ, π½ ∈ πΎπ (π½1 , . . . , π½π ), as required. Exercise 5.4.5 In the second paragraph of the proof, I claimed that πΏ π = πΏ π−1 (π½ π ). The general principle here is that if π : πΎ is a field extension of degree 2 and πΎ ∈ π \ πΎ then π = πΎ (πΎ). Prove this. Proof of Theorem 5.4.2 Suppose that (π₯, π¦) is constructible from Σ in π steps. If π = 0 then (π₯, π¦) ∈ Σ, so π₯, π¦ ∈ πΎΣ , which is trivially an iterated quadratic extension of πΎΣ . Now let π ≥ 1, and suppose inductively that each coordinate of each point of R2 constructible from Σ in < π steps lies in some iterated quadratic extension of πΎΣ . By definition, (π₯, π¦) is an intersection point of two distinct lines/circles through points constructible in < π steps. By inductive hypothesis, each coordinate of each of those points lies in some iterated quadratic extension of πΎΣ , so by Lemma 5.4.4, there is an iterated quadratic extension πΏ of πΎΣ containing all the points’ coordinates. The coefficients in the equations of the lines/circles then also lie in πΏ. We now show that deg πΏ (π₯) ∈ {1, 2}. 63 If (π₯, π¦) is the intersection point of two distinct lines, then π₯ and π¦ satisfy two linearly independent equations ππ₯ + ππ¦ + π = 0, π π₯ + π 0 π¦ + π0 = 0 0 with π, π, π, π0, π0, π0 ∈ πΏ. Solving gives π₯ ∈ πΏ. If (π₯, π¦) is an intersection point of a line and a circle, then ππ₯ + ππ¦ + π = 0, π₯ 2 + π¦ 2 + ππ₯ + ππ¦ + π = 0 with π, π, π, π, π, π ∈ πΏ. If π = 0 then π ≠ 0 and π₯ = −π/π ∈ πΏ. Otherwise, we can eliminate π¦ to give a quadratic over πΏ satisfied by π₯, so that deg πΏ (π₯) ∈ {1, 2}. If (π₯, π¦) is an intersection point of two circles, then π₯ 2 + π¦ 2 + ππ₯ + ππ¦ + π = 0, π₯ 2 + π¦ 2 + π 0π₯ + π0 π¦ + π 0 = 0 with π, π, π , π 0, π0, π 0 ∈ πΏ. Subtracting, we reduce to the case of a line and a circle, again giving deg πΏ (π₯) ∈ {1, 2}. Hence deg πΏ (π₯) ∈ {1, 2}. If deg πΏ (π₯) = 1 then π₯ ∈ πΏ, which is an iterated quadratic extension of πΎΣ . If deg πΏ (π₯) = 2, i.e. [πΏ(π₯) : πΏ] = 2, then πΏ (π₯) is an iterated quadratic extension of πΎΣ . In either case, π₯ lies in some iterated quadratic extension of πΎΣ . The same is true of π¦. Hence by Lemma 5.4.4, there is an iterated quadratic extension of πΎΣ containing π₯ and π¦. This completes the induction. Now we solve the problems of ancient Greece. In all cases, we take Σ = {(0, 0), (1, 0)}. Then πΎΣ = Q. Proposition 5.4.6 The angle cannot be trisected by ruler and compass. Proof Suppose it can be. Construct an equilateral triangle with (0, 0) and (1, 0) as two of its vertices (which can be done by ruler and compass; Figure 5.3). Trisect the angle of the triangle at (0, 0). Plot the point (π₯, π¦) where the trisector meets the circle with centre (0, 0) through (1, 0). Then π₯ = cos(π/9), so by Corollary 5.4.3, degQ (cos(π/9)) is a power of 2. Now we use the trigonometric formula cos 3π = 4(cos π) 3 − 3 cos π. Taking π = π/9 and using cos(π/3) = 1/2, we get 8π₯ 3 − 6π₯ − 1 = 0. Reduced mod 5, this cubic has no roots and is therefore irreducible (by Lemma 3.3.1(iv)). So by the mod π test, 8π‘ 3 − 6π‘ − 1 is irreducible over Q. Hence (8π‘ 3 − 6π‘ − 1)/8 is the minimal polynomial of π₯ over Q, giving degQ (π₯) = 3. Since 3 is not a power of 2, this is a contradiction. 64 (π₯, π¦) (0, 0) (1, 0) Figure 5.3: The impossibility of trisecting 60β¦ . Proposition 5.4.7 The cube cannot be duplicated by ruler and compass. Proof Suppose it can be. Since (0, 0) and (1, 0) √3 are distance 1 apart, we can construct from them two points π΄ and π΅ distance √3 2 apart. From π΄ and π΅ we can construct, using ruler and compass, the point ( 2, 0). (The video shows how to √3 ‘transport distances’ like this.) So degQ ( 2) is a power of 2, by Corollary 5.4.3. √3 But degQ ( 2) = 3 by Example 5.1.8(ii), a contradiction. Proposition 5.4.8 The circle cannot be squared by ruler and compass. This one is the most outrageously false, yet the hardest to prove. Proof Suppose it can be. Since the circle with centre (0, 0) through (1, 0) √has π, area π, we can construct by ruler and compass a square with side-length √ and from that, we √ can construct by ruler and compass the point ( π, 0). So by Corollary 5.4.3, π is algebraic over Q with degree a power of 2. Since Q is a subfield of C, it follows that π is algebraic over Q. But it is a (hard) theorem that π is transcendental over Q. Digression 5.4.9 Stewart has a nice alternative approach to all this, in his Chapter 7. He treats the plane as the complex plane, and he shows that the set of all points in C constructible from 0 and 1 is a subfield. In fact, it is the smallest subfield of C closed under taking square roots. He calls it Qpy , the ‘Pythagorean closure’ of Q. It can also be described as the set of complex numbers contained in some iterated quadratic extension of Q. There is one more famous ruler and compass problem: for which integers π is the regular π-sided polygon constructible, starting from just a pair of points in the plane? The answer has to do with Fermat primes, which are prime numbers of the form 2π’ + 1 for some π’ ≥ 1. A little exercise in number theory shows that if 2π’ + 1 is prime then π’ must itself be a power of 2. The only known Fermat primes are 0 22 + 1 = 3, 1 22 + 1 = 5, 2 22 + 1 = 17, 65 3 22 + 1 = 257, 4 22 + 1 = 65537. Whether there are any others is a longstanding open question. In any case, it can be shown that the regular π-sided polygon is constructible if and only if π = 2π π 1 · · · π π for some π, π ≥ 0 and distinct Fermat primes π 1 , . . . , π π . We will not do the proof, but it involves cyclotomic polynomials. A glimpse of the connection: let π be a prime such that the regular π-sided polygon is constructible. Consider the regular π-sided polygon inscribed in the unit circle in C, with one of its vertices at 1. Then another vertex is at π 2ππ/π , and from constructibility, one can deduce that degQ (π 2ππ/π ) is a power of 2. But we saw in Example 3.3.16 that degQ (π 2ππ/π ) = π − 1. So π − 1 is a power of 2, that is, π is a Fermat prime. Galois theory, number theory and Euclidean geometry come together! 66 Chapter 6 Splitting fields Introduction to Week 6 In Chapter 1, we met a definition of the symmetry group of a polynomial over Q. It was phrased in terms of indistinguishable tuples, it was possibly a little mysterious, and it was definitely difficult to work with (e.g. we couldn’t compute the symmetry group of 1 + π‘ + π‘ 2 + π‘ 3 + π‘ 4 ). In this chapter, we’re going to give a different but equivalent definition of the symmetry group of a polynomial. It’s a two-step process: 1. We show how every polynomial π over πΎ gives rise to an extension of πΎ, called the ‘splitting field’ of π . 2. We show how every field extension has a symmetry group. The symmetry group, or ‘Galois group’, of a polynomial is then defined to be the symmetry group of its splitting field extension. How does these two steps work? 1. When πΎ = Q, the splitting field of π is the smallest subfield of C containing all the complex roots of π . For a general field πΎ, it’s constructed by adding the roots of π one at a time, using simple extensions, until we obtain an extension of πΎ in which π splits into linear factors. 2. The symmetry group of a field extension π : πΎ is defined as the group of automorphisms of π over πΎ. This is the same idea you’ve seen many times before, for symmetry groups of other mathematical objects. Why bother? Why not define the symmetry group of π directly, as in Chapter 1? • Because this strategy works over every field πΎ, not just Q. • Because there are field extensions that do not arise from a polynomial, and their symmetry groups are sometimes important. For example, an important 67 structure in number theory, somewhat mysterious to this day, is the symmetry group of the algebraic numbers Q over Q. • Because using abstract algebra means you can cut down on explicit calculations with polynomials. (By way of analogy, you’ve seen how abstract linear algebra with vector spaces and linear maps allows you to cut down on calculations with matrices.) It also makes connections with other parts of mathematics more apparent. 6.1 Extending homomorphisms In your degree so far, you’ll have picked up the general principle that for many kinds of mathematical object (such as groups, rings, fields, vector spaces, modules, metric spaces, topological spaces, measure spaces, . . . ), it’s important to consider the appropriate notion of mapping between them (such as homomorphisms, linear maps, continuous maps, . . . ). And since Chapter 4, you know that the basic objects of Galois theory are field extensions. So it’s no surprise that sooner or later, we’ll have to think about mappings from one field extension to another. That moment is now: we’ll need what’s in this section in order to establish the basic facts about splitting fields. When we think about a field extension π : πΎ, we generally regard the field πΎ as our starting point and π as a field that extends it. (This doesn’t mean anything rigorous.) Similarly, we might start with a homomorphism π : πΎ → πΎ 0 between fields, together with extensions π of πΎ and π 0 of πΎ 0, and look for a homomorphism π → π 0 that extends π. The language is as follows. Definition 6.1.1 Let π : πΎ → π and π0 : πΎ 0 → π 0 be field extensions. Let π : πΎ → πΎ 0 be a homomorphism of fields. A homomorphism π : π → π 0 extends π if the square π Extension problems πO / πO 0 π0 π / πΎ πΎ0 π commutes (that is, π β¦ π = π0 β¦ π). Here I’ve used the definition of a field extension as a homomorphism π of fields (Definition 4.1.1). Most of the time we view πΎ as a subset of π and πΎ 0 as a subset of π 0, with π and π0 being the inclusions. In that case, for π to extend π just means that π(π) = π(π) for all π ∈ πΎ. 68 The basic questions are: given the two field extensions and the homomorphism π, is there some π that extends π? If so, how many? We’ll consider these questions later. This section simply gathers together three general results about extensions of field homomorphisms. Recall that any ring homomorphism π : π → π induces a homomorphism π∗ : π [π‘] → π[π‘] (Definition 3.1.6). Explanation of Lemma 6.1.2 Lemma 6.1.2 Let π : πΎ and π 0 : πΎ 0 be field extensions, let π : πΎ → πΎ 0 be a homomorphism, and let π : π → π 0 be a homomorphism extending π. Let πΌ ∈ π and π (π‘) ∈ πΎ [π‘]. Then π (πΌ) = 0 ⇐⇒ π∗ ( π ) π(πΌ) = 0. Í Í Proof Write π (π‘) = π ππ π‘ π , where ππ ∈ πΎ. Then (π∗ ( π ))(π‘) = π π(ππ )π‘ π ∈ πΎ 0 [π‘], so Õ Õ (π∗ ( π ))(π(πΌ)) = π(ππ )π(πΌ) π = π(ππ )π(πΌ) π = π( π (πΌ)), π π where the second equality holds because π extends π. Since π is injective (Lemma 2.2.2), the result follows. Exercise 6.1.3 Show that if π is injective then so is π∗ , and if π is an isomorphism then so is π∗ . Lemma 6.1.4 Let π : πΎ and π 0 : πΎ 0 be field extensions, let π : πΎ → πΎ 0 be an isomorphism, and let π : π → π 0 be a homomorphism extending π. Let πΌ ∈ π be algebraic over πΎ with minimal polynomial π. Then π(πΌ) is algebraic over πΎ 0 with minimal polynomial π∗ (π). Proof By Lemma 6.1.2, π∗ (π) is an annihilating polynomial of π(πΌ) over πΎ 0. Also π∗ (π) ≠ 0, since π ≠ 0 and π∗ is injective. So π(πΌ) is algebraic over πΎ 0. (Recall that an element is algebraic if it has a nonzero annihilating polynomial.) Since π∗ : πΎ [π‘] → πΎ 0 [π‘] is an isomorphism and π ∈ πΎ [π‘] is irreducible, π∗ (π) ∈ πΎ 0 [π‘] is irreducible. It is also monic. Hence π∗ (π) is a monic irreducible annihilating polynomial of π(πΌ), so it is the minimal polynomial of π(πΌ). An isomorphism between fields, rings, groups, vector spaces, etc., can be understood as simply a renaming of the elements. For example, if I tell you that the ring π is left Noetherian but not right Artinian, and that π is isomorphic to π , then you can deduce that π is left Noetherian but not right Artinian without having the slightest idea what those terms mean. Just as long as they don’t depend on the names of the elements of the ring concerned (which such definitions never do), you’re fine. 69 Proposition 6.1.5 Let π : πΎ → πΎ 0 be an isomorphism of fields, let πΎ (πΌ) : πΎ be a simple extension where πΌ has minimal polynomial π over πΎ, and let πΎ 0 (πΌ0) : πΎ 0 be a simple extension where πΌ0 has minimal polynomial π∗ (π) over πΎ 0. Then there is a unique isomorphism π : πΎ (πΌ) → πΎ 0 (πΌ0) that extends π and satisfies π(πΌ) = πΌ0. Diagram: πΎ (πΌ) O πΎ π / π 0) πΎ 0 (πΌ O / πΎ0 We often use a dotted arrow to denote a map whose existence is part of the conclusion of a theorem, rather than a hypothesis. π Proof View πΎ 0 (πΌ0) as an extension of πΎ via the composite homomorphism πΎ −→ πΎ 0 → πΎ 0 (πΌ0). Then the minimal polynomial of πΌ0 over πΎ is π. (If this isn’t intuitively clear to you, think of the isomorphism π as renaming.) Hence by the classification of simple extensions, Theorem 4.3.7, there is an isomorphism π : πΎ (πΌ) → πΎ 0 (πΌ0) over πΎ such that π(πΌ) = πΌ0. Then π extends π. It only remains to prove uniqueness. Let π e be any homomorphism πΎ (πΌ) → πΎ 0 (πΌ0) that extends π and satisfies π e(πΌ) = πΌ0. Then π e(π) = π(π) = π(π) for all 0 π ∈ πΎ and π e(πΌ) = πΌ = π(πΌ). Since every element of πΎ (πΌ) is a polynomial in πΌ with coefficients in πΎ, it follows that π e = π. 6.2 Existence and uniqueness of splitting fields Let π be a polynomial over a field πΎ. Informally, a splitting field for π is an extension of πΎ where π has all its roots, and which is no bigger than it needs to be. Warning 6.2.1 If π is irreducible, we know how to create an extension of πΎ where π has at least one root: take the simple extension πΎ [π‘]/h π i, in which the equivalence class of π‘ is a root of π (Lemma 4.3.4). But πΎ [π‘]/h π i is not usually a splitting field for π . For example, take πΎ = Q and π (π‘) = π‘ 3 − 2, as in Warning 4.3.10. Write πΌ for the real cube root of 2. (Half the counterexamples in Galois theory involve the real cube root of 2.) Then Q[π‘]/h π i is isomorphic to the subfield Q(πΌ) of R, which only contains one root of π : the other two are non-real, hence not in Q(πΌ). 70 Definition 6.2.2 Let π be a polynomial over a field π. Then π splits in π if π (π‘) = π½(π‘ − πΌ1 ) · · · (π‘ − πΌπ ) for some π ≥ 0 and π½, πΌ1 , . . . , πΌπ ∈ π. Equivalently, π splits in π if all its irreducible factors in π [π‘] are linear. Examples 6.2.3 i. A field π is algebraically closed if and only if every polynomial over π splits in π. √ ii. Let π (π‘) = π‘ 4 − 4π‘ 2 − 5. Then π splits in Q(π, 5), since π (π‘) = (π‘ 2 + 1)(π‘ 2 − 5) √ √ = (π‘ − π)(π‘ + π)(π‘ − 5)(π‘ + 5). But π does not split in Q(π), as its factorization into irreducibles in Q(π) [π‘] is π (π‘) = (π‘ − π)(π‘ + π)(π‘ 2 − 5), which contains a nonlinear factor. Moral: π may have one root or even several roots in π, but still not split in π. iii. Let π = F2 (πΌ), where πΌ is a root of π (π‘) = 1 + π‘ + π‘ 2 , as in Example 4.3.8(ii). We have π (1 + πΌ) = 1 + (1 + πΌ) + (1 + 2πΌ + πΌ2 ) = 1 + πΌ + πΌ2 = 0, so π has two distinct roots in π, giving π (π‘) = (π‘ − πΌ)(π‘ − (1 + πΌ)) in π [π‘]. Hence π splits in π. In this example, adjoining one root of π gave us a second root for free. But this doesn’t typically happen (Warning 6.2.1). Definition 6.2.4 Let π be a polynomial over a field πΎ. A splitting field of π over πΎ is an extension π of πΎ such that: i. π splits in π; ii. π = πΎ (πΌ1 , . . . , πΌπ ), where πΌ1 , . . . , πΌπ are the roots of π in π. 71 Exercise 6.2.5 Show that (ii) can equivalently be replaced by: ‘if πΏ is a subfield of π containing πΎ, and π splits in πΏ, then πΏ = π’. Examples 6.2.6 i. Let π ∈ Q[π‘]. Write πΌ1 , . . . , πΌπ for the complex roots of π . Then Q(πΌ1 , . . . , πΌπ ) (the smallest subfield of C containing πΌ1 , . . . , πΌπ ) is a splitting field of π over Q. ii. Let π (π‘) = π‘ 3 − 2 ∈ Q[π‘]. Its complex roots are πΌ, ππΌ and π2 πΌ, where πΌ is the real cube root of 2 and π = π 2ππ/3 . Hence a splitting field of π over Q is Q(πΌ, ππΌ, π2 πΌ) = Q(πΌ, π). Now degQ (πΌ) = 3 as π is irreducible, and degQ (π) = 2 as π has minimal polynomial 1 + π‘ + π‘ 2 . By an argument like that in Example 5.2.7, it follows that [Q(πΌ, π) : Q] = 6. On the other hand, [Q(πΌ) : Q] = 3. So again, the extension we get by adjoining all the roots of π is bigger than the one we get by adjoining just one root of π . iii. Take π (π‘) = 1 + π‘ + π‘ 2 ∈ F2 [π‘], as in Example 6.2.3(iii). By Theorem 5.1.5, {1, πΌ} is a basis of F2 (πΌ) over F2 , so F2 (πΌ) = {0, 1, πΌ, 1 + πΌ} = F2 ∪ {the roots of π in F2 (πΌ)}. Hence F2 (πΌ) is a splitting field of π over F2 . Exercise 6.2.7 In Example 6.2.6(ii), I said that Q(πΌ, ππΌ, π2 πΌ) = Q(πΌ, π). Why is that true? Digression 6.2.8 Splitting fields over Q are easy, as Example 6.2.6(i) shows. That’s because we have a ready-made algebraically closed field containing Q, namely, C. In fact, for any field πΎ, it’s possible to build an algebraically closed field containing πΎ. And there is a unique ‘smallest’ algebraically closed field containing πΎ, called its algebraic closure πΎ. For example, the algebraic closure of Q is Q, the subfield of C consisting of the algebraic numbers. (For a proof that Q is indeed algebraically closed, see Workshop 3.) Many texts on Galois theory include constructions of the algebraic closure of a field, but we won’t do it. 72 Our mission for the rest of this section is to show that every polynomial π has exactly one splitting field. So that’s actually two tasks: first, show that π has at least one splitting field, then, show that π has only one splitting field. The first task is easy, and in fact we prove a little bit more: Lemma 6.2.9 Let π be a polynomial over a field πΎ. Then there exists a splitting field π of π over πΎ such that [π : πΎ] ≤ deg( π )!. (So that this lemma holds for π = 0, we had better define (−∞)! = 1.) Proof If π is constant then πΎ is a splitting field of π over πΎ, and the result holds trivially. Now suppose inductively that deg( π ) ≥ 1. We may choose an irreducible factor π of π . By Theorem 4.3.7, there is an extension πΎ (πΌ) of πΎ with π(πΌ) = 0. Then (π‘ − πΌ) | π (π‘) in πΎ (πΌ) [π‘], giving a polynomial π(π‘) = π (π‘)/(π‘ − πΌ) over πΎ (πΌ). We have deg(π) = deg( π ) − 1, so by inductive hypothesis, there is a splitting field π of π over πΎ (πΌ) with [π : πΎ (πΌ)] ≤ deg(π)!. Then π is a splitting field of π over πΎ. (Check that you understand why.) Also, by the tower law, [π : πΎ] = [π : πΎ (πΌ)] [πΎ (πΌ) : πΎ] ≤ (deg( π ) − 1)! · deg(π) ≤ deg( π )!, completing the induction. Proving that every polynomial has only one splitting field is harder. As ever, ‘only one’ has to be understood up to isomorphism: after all, if you’re given a splitting field, you can always rename its elements to get an isomorphic copy that’s not literally identical to the original one. But isomorphism is all that matters. Our proof of the uniqueness of splitting fields depends on the following result, which will also be useful for other purposes as we head towards the fundamental theorem of Galois theory. Proposition 6.2.10 Let π : πΎ → πΎ 0 be an isomorphism of fields, let π ∈ πΎ [π‘], let π be a splitting field of π over πΎ, and let π 0 be a splitting field of π∗ ( π ) over πΎ 0. Then: i. there exists an isomorphism π : π → π 0 extending π; ii. there are at most [π : πΎ] such extensions π. Diagram: π πO / / πO 0 πΎ π 73 πΎ0 Proof We prove both statements by induction on deg( π ). If π is constant then both field extensions are trivial, so there is exactly one isomorphism π extending π. Now suppose that deg( π ) ≥ 1. We can choose a monic irreducible factor π of π . Then π splits in π since π does and π | π ; choose a root πΌ ∈ π of π. We have π (πΌ) = 0, so (π‘ − πΌ) | π (π‘) in πΎ (πΌ) [π‘], giving a polynomial π(π‘) = π (π‘)/(π‘ − πΌ) over πΎ (πΌ). Then π is a splitting field of π over πΎ (πΌ), and deg(π) = deg( π ) − 1. Also, π∗ (π) splits in π 0 since π∗ ( π ) does and π∗ (π) | π∗ ( π ). Write πΌ10 , . . . , πΌ0π for the distinct roots of π∗ (π) in π 0. Note that 1 ≤ π ≤ deg(π∗ (π)) = deg(π). Counting isomorphisms: the proof of Proposition 6.2.10 (6.1) Since π∗ is an isomorphism, π∗ (π) is monic and irreducible, and is therefore the minimal polynomial of πΌ0π for each π ∈ {1, . . . , π }. Hence by Proposition 6.1.5, for each π, there is a unique isomorphism π π : πΎ (πΌ) → πΎ 0 (πΌ0π ) that extends π and satisfies π π (πΌ) = πΌ0π . (See diagram below.) For each π ∈ {1, . . . , π }, we have a polynomial π π∗ (π) = π π∗ ( π ) π∗ ( π ) = π π∗ (π‘ − πΌ) π‘ − πΌ0π over πΎ 0 (πΌ0π ), and π 0 is a splitting field of π∗ ( π ) over πΎ 0, so π 0 is also a splitting field of π π∗ (π) over πΎ 0 (πΌ0π ). To prove that there is at least one isomorphism π extending π, choose any π ∈ {1, . . . , π } (as we may since π ≥ 1). By applying the inductive hypothesis to π and π π , there is an isomorphism π extending π π : / π πO πΎ (πΌ) O ππ / πO 0 πΎ 0 (πΌ0π ) / O πΎ πΎ0 π But then π also extends π, as required. To prove there are at most [π : πΎ] isomorphisms π : π → π 0 extending π, first note that any such π satisfies (π∗ ( π ))(π(πΌ)) = 0 (by Lemma 6.1.2), so π(πΌ) = πΌ0π for some π ∈ {1, . . . , π }. Every element of πΎ (πΌ) is a polynomial in πΌ over πΎ, and π(π) = π(π) ∈ πΎ 0 for all π ∈ πΎ, so π maps πΎ (πΌ) into πΎ 0 (πΌ0π ). Now π(πΎ (πΌ)) contains πΎ 0 (since π is an isomorphism) and πΌ0π , so π(πΎ (πΌ)) = πΎ 0 (πΌ0π ). Since homomorphisms of fields are injective, π restricts to an isomorphism πΎ (πΌ) → πΎ 0 (πΌ0π ) satisfying πΌ β¦→ πΌ0π . By the uniqueness part of 74 Proposition 6.1.5, this restricted isomorphism must be π π . Thus, π extends π π for a unique π ∈ {1, . . . , π }, giving (number of isos π extending π) = π Õ (number of isos π extending π π ). π=1 For each π, the number of isomorphisms π extending π π is ≤ [π : πΎ (πΌ)], by inductive hypothesis. So, using the tower law and (6.1), (number of isos π extending π) ≤ π · [π : πΎ (πΌ)] = π · [π : πΎ] ≤ [π : πΎ], deg(π) completing the induction. Exercise 6.2.11 Why does the proof of Proposition 6.2.10 not show that there are exactly [π : πΎ] isomorphisms π extending π? How could you strengthen the hypotheses in order to obtain that conclusion? (The second question is a bit harder, and we’ll see the answer next week.) This brings us to the foundational result on splitting fields. Recall that an automorphism of an object π is an isomorphism π → π. Theorem 6.2.12 Let π be a polynomial over a field πΎ. Then: i. there exists a splitting field of π over πΎ; ii. any two splitting fields of π are isomorphic over πΎ; iii. when π is a splitting field of π over πΎ, (number of automorphisms of π over πΎ) ≤ [π : πΎ] ≤ deg( π )!. Proof Part (i) is immediate from Lemma 6.2.9, and part (ii) follows from Proposition 6.2.10 by taking πΎ 0 = πΎ and π = idπΎ . The first inequality in (iii) follows from Proposition 6.2.10 by taking πΎ 0 = πΎ, π 0 = π and π = idπΎ , and the second follows from Lemma 6.2.9. Up to now we have been saying ‘a’ splitting field. Theorem 6.2.12 gives us the right to speak of the splitting field of a given polynomial π over a given field πΎ. We write it as SFπΎ ( π ). We finish with a left over lemma that will be useful later. 75 Lemma 6.2.13 i. Let π : π : πΎ be field extensions, π ∈ πΎ [π‘], and π ⊆ π. Suppose that π is the splitting field of π over πΎ. Then π(π ) is the splitting field of π over πΎ (π ). ii. Let π be a polynomial over a field πΎ, and let πΏ be a subfield of SFπΎ ( π ) containing πΎ (so that SFπΎ ( π ) : πΏ : πΎ). Then SFπΎ ( π ) is the splitting field of π over πΏ. Proof For (i), π splits in π, hence in π(π ). Writing π for the set of roots of π in π, we have π = πΎ (π) and so π(π ) = πΎ (π)(π ) = πΎ (π ∪ π ) = πΎ (π )(π); that is, π(π ) is generated over πΎ (π ) by π. This proves (i), and (ii) follows by taking π = SFπΎ ( π ) and π = πΏ. 6.3 The Galois group What gives Galois theory its special flavour is the use of groups to study fields and polynomials. Here is the central definition. Definition 6.3.1 The Galois group Gal(π : πΎ) of a field extension is the group of automorphisms of π over πΎ, with composition as the group operation. Exercise 6.3.2 Check that this really does define a group. You’ll need the result of Exercise 4.3.6, for instance. In other words, an element of Gal(π : πΎ) is an isomorphism π : π → π such that π (π) = π for all π ∈ πΎ. Examples 6.3.3 i. What is Gal(C : R)? Certainly the identity is an automorphism of C over R. So is complex conjugation π , as implicitly shown in the first proof of Lemma 1.1.3. So {id, π } ⊆ Gal(C : R). I claim that Gal(C : R) has no other elements. For let π ∈ Gal(C : R). Then (π (π)) 2 = π (π 2 ) = π (−1) = −π (1) = −1 as π is a homomorphism, so π (π) = ±π. If π (π) = π then π (π + ππ) = π (π) + π (π)π (π) = π + ππ for all π, π ∈ R (since π is an automorphism over R), giving π = id. Similarly, if π (π) = −π then π = π . So Gal(C : R) = {id, π } πΆ2 . 76 ii. Let πΌ be the real cube root of 2. For each π ∈ Gal(Q(πΌ) : Q), we have (π (πΌ)) 3 = π (πΌ3 ) = π (2) = 2 and π (πΌ) ∈ Q(πΌ) ⊆ R, so π (πΌ) = πΌ. Every element of Q(πΌ) can be expressed as a polynomial in πΌ over Q (by Theorem 5.1.5), so π = id. Hence Gal(Q(πΌ) : Q) is trivial. Exercise 6.3.4 Prove that Gal(Q(π 2ππ/3 ) : Q) = {id, π }, where π (π§) = π§. (Hint: imitate Example 6.3.3(i).) The Galois group of a polynomial is defined to be the Galois group of its splitting field extension: Definition 6.3.5 Let π be a polynomial over a field πΎ. The Galois group GalπΎ ( π ) of π over πΎ is Gal(SFπΎ ( π ) : πΎ). So the definitions fit together like this: polynomial β¦−→ field extension β¦−→ group. We will soon prove that Definition 6.3.5 is equivalent to the definition of Galois group in Chapter 1, where we went straight from polynomials to groups. Theorem 6.2.12(iii) says that | GalπΎ ( π )| ≤ [SFπΎ ( π ) : πΎ] ≤ deg( π )!. (6.2) In particular, GalπΎ ( π ) is always a finite group. Examples 6.3.6 i. GalQ (π‘ 2 + 1) = Gal(Q(π) : Q) = {id, π } πΆ2 , where π is complex conjugation on Q(π). The second equality is proved by the same argument as in Example 6.3.3(i), replacing C : R by Q(π) : Q. Calculating the Galois group with bare hands, part 1 Calculating the Galois group with bare hands, part 2 ii. Let π (π‘)√ = (π‘ 2 + 1)(π‘ 2 − 2). Then GalQ ( π ) is the group of automorphisms of Q(π, 2) over Q. Similar arguments to those in Examples √ √6.3.3 show that every π ∈ GalQ ( π ) must satisfy π (π) = ±π and π ( 2) = ± 2, and that the two choices of sign determine π completely. And one can show that all four choices are possible, so that | GalQ ( π )| = 4. There are two groups of order four, πΆ4 and πΆ2 × πΆ2 . But each element of GalQ ( π ) has order 1 or 2, so GalQ ( π ) is not πΆ4 , so GalQ ( π ) πΆ2 × πΆ2 . I’ve been sketchy with the details here, because it’s not really sensible to try to calculate Galois groups until we have a few more tools at our disposal. We start to assemble them now. 77 Figure 6.1: The Galois group πΊ = GalπΎ ( π ) permutes the roots πΌπ of π . (Image adapted from @rowvector.) In the examples so far, we’ve seen that if πΌ is a root of π then so is π (πΌ) for every π ∈ GalπΎ ( π ). This is true in general, and is best expressed in terms of group actions (Figure 6.1). In a slogan: the Galois group permutes the roots. Lemma 6.3.7 Let π be a polynomial over a field πΎ. Let π be the set of roots of π in SFπΎ ( π ). Then there is an action of GalπΎ ( π ) on π defined by GalπΎ ( π ) × π → π (π, πΌ) β¦→ π (πΌ). (6.3) Proof First, if π ∈ GalπΎ ( π ) and πΌ ∈ π then π (πΌ) ∈ π, by Lemma 6.1.2 with πΎ = πΎ 0, π = π 0 and π = id. For the function (6.3) to be an action means that (π β¦ π)(πΌ) = π(π (πΌ)) and id(πΌ) = πΌ for all π, π ∈ GalπΎ ( π ) and πΌ ∈ π, which are true by definition. The action of the Galois group An action of a group πΊ on a set π is essentially the same thing as a homomorphism from πΊ to the group Sym(π) of bijections from π to π. (If we write ππ₯ as ππ (π₯), then the homomorphism πΊ → Sym(π) is π β¦→ ππ .) We now adopt this viewpoint in the case of Galois groups. Let π be a polynomial over a field πΎ, and write πΌ1 , . . . , πΌ π for the distinct roots of π in SFπΎ ( π ). For each π ∈ {1, . . . , π }, Lemma 6.3.7 implies that π (πΌπ ) = πΌ π for a unique π. Write π as ππ (π), so that π (πΌπ ) = πΌππ (π) . Then we have a function Γ: GalπΎ ( π ) → π π π β¦→ ππ , and it is straightforward to check that Γ is a homomorphism. 78 (6.4) Exercise 6.3.8 What is the kernel of Γ, in concrete terms? If you remember the definition of Galois group in Chapter 1 (Definition 1.2.1), the mention of π π should have set your antennae tingling. There, we defined the Galois group as a certain subgroup of π π , namely, the one consisting of those permutations π for which the tuples (πΌ1 , . . . , πΌ π ), (πΌπ(1) , . . . , πΌπ(π) ) are indistinguishable. Let’s now make the definition of indistinguishability official, switch to the standard name (recall Warning 1.1.2), and generalize from Q to an arbitrary field. Definition 6.3.9 Let π : πΎ be a field extension, let π ≥ 0, and let (πΌ1 , . . . , πΌ π ) and (πΌ10 , . . . , πΌ0π ) be π-tuples of elements of π. Then (πΌ1 , . . . , πΌ π ) and (πΌ10 , . . . , πΌ0π ) are conjugate over πΎ if for all π ∈ πΎ [π‘1 , . . . , π‘ π ], π(πΌ1 , . . . , πΌ π ) = 0 ⇐⇒ π(πΌ10 , . . . , πΌ0π ) = 0. In the case π = 1, we omit the brackets and say that πΌ and πΌ0 are conjugate to mean that (πΌ) and (πΌ0) are. We show now that the two definitions of the Galois group of π are equivalent. Proposition 6.3.10 Let π be a polynomial over a field πΎ, with distinct roots πΌ1 , . . . , πΌ π in SFπΎ ( π ). Define the group homomorphism Γ : GalπΎ ( π ) → π π as in (6.4). Then Γ is injective, and its image is {π ∈ π π : (πΌ1 , . . . , πΌ π ) and (πΌπ(1) , . . . , πΌπ(π) ) are conjugate over πΎ }. (6.5) In particular, (6.5) is a subgroup of π π isomorphic to GalπΎ ( π ). Proof To show that Γ is injective, let π ∈ ker Γ. Then π (πΌπ ) = πΌπ for all π. Now SFπΎ ( π ) = πΎ (πΌ1 , . . . , πΌ π ), with each πΌπ algebraic over πΎ, so every element of SFπΎ ( π ) can be expressed as a polynomial in πΌ1 , . . . , πΌ π over πΎ (by Corollary 5.1.11). Since π : SFπΎ ( π ) → SFπΎ ( π ) fixes each element of πΎ and each πΌπ , it is the identity on all of SFπΎ ( π ). Thus, ker Γ is trivial, so Γ is injective. Now we prove that im Γ is the set (6.5). In one direction, let π ∈ im Γ. Then π = ππ for some π ∈ GalπΎ ( π ) (writing Γ(π) = ππ , as before). For every π ∈ πΎ [π‘1 , . . . , π‘ π ], π(πΌππ (1) , . . . , πΌππ (π) ) = π(π (πΌ1 ), . . . , π (πΌ π )) = π ( π(πΌ1 ), . . . , π(πΌ π )), 79 where the first equality is by definition of ππ and the second is because π is a homomorphism over πΎ. But π is an isomorphism, so it follows that π(πΌππ (1) , . . . , πΌππ (π) ) = 0 ⇐⇒ π(πΌ1 , . . . , πΌ π ) = 0. Hence π belongs to the set (6.5). In the other direction, let π be a permutation in (6.5). By Corollary 5.1.11, every element of SFπΎ ( π ) can be expressed as π(πΌ1 , . . . , πΌ π ) for some π ∈ πΎ [π‘1 , . . . , π‘ π ]. Now for π, π ∈ πΎ [π‘1 , . . . , π‘ π ], we have π(πΌ1 , . . . , πΌ π ) = π(πΌ1 , . . . , πΌ π ) ⇐⇒ π(πΌπ(1) , . . . , πΌπ(π) ) = π(πΌπ(1) , . . . , πΌπ(π) ) (by applying Definition 6.3.9 of conjugacy with π − π as the ‘π’). So there is a well-defined, injective function π : SFπΎ ( π ) → SFπΎ ( π ) satisfying π ( π(πΌ1 , . . . , πΌ π )) = π(πΌπ(1) , . . . , πΌπ(π) ) (6.6) for all π ∈ πΎ [π‘1 , . . . , π‘ π ]. Moreover, π is surjective because π is a permutation, π (π) = π for all π ∈ πΎ (by taking π = π in (6.6)), and π (πΌπ ) = πΌπ(π) for all π (by taking π = π‘π in (6.6)). You can check that π is a homomorphism of fields. Hence π ∈ GalπΎ ( π ) with Γ(π) = π, proving that π ∈ im Γ. The final sentence of the proposition follows because every injective group homomorphism πΎ : πΊ → π» induces an isomorphism between πΊ and the subgroup im πΎ of π». Exercise 6.3.11 I skipped two small bits in that proof: ‘π is surjective because π is a permutation’ (why?), and ‘You can check that π is a homomorphism of fields’. Fill the gaps. Digression 6.3.12 As you may know, an action of a group πΊ on a set π is called faithful if the corresponding homomorphism πΊ → Sym(π) is injective. A more concrete way to say that is that the only element π ∈ πΊ that fixes everything (ππ₯ = π₯ for all π₯ ∈ π) is the identity. Equivalently, if π, β ∈ πΊ and ππ₯ = βπ₯ for all π₯ then π = β. Most actions that one meets in practice are faithful; those that aren’t involve a kind of redundancy. It’s important in Galois theory to be able to move easily between fields. For example, you might start with a polynomial whose coefficients belong to one field πΎ, but later decide to consider the coefficients as belonging to some larger field πΏ. Here’s what happens to the Galois group when you do that. Corollary 6.3.13 Let πΏ : πΎ be a field extension and π ∈ πΎ [π‘]. Then Gal πΏ ( π ) embeds naturally as a subgroup of GalπΎ ( π ). 80 The phrasing here is slightly vague: it means there is an injective homomorphism Gal πΏ ( π ) → GalπΎ ( π ), and there is such an obvious choice of homomorphism that we tend to regard Gal πΏ ( π ) as being a subgroup of GalπΎ ( π ). Proof This follows from Proposition 6.3.10 together with the observation that if two π-tuples are conjugate over πΏ, they are conjugate over πΎ. Example 6.3.14 Consider the Galois group of π (π‘) = (π‘ 2 + 1)(π‘ 2 − 2) over Q, R and C. In Example 6.3.6(ii), we saw that GalQ ( π ) πΆ2 × πΆ2 . Over R, the Galois group of π is the same as that of π‘ 2 + 1, since both roots of π‘ 2 − 2 lie in R and are therefore preserved by elements of GalR ( π ). So GalR ( π ) = GalR (π‘ 2 + 1) πΆ2 . Finally, GalC ( π ) is trivial. Indeed, every polynomial π ∈ C[π‘] has trivial Galois group over C: for π splits in C, so SFC (π) = C, so SFC (π) : C is a trivial extension and so has trivial Galois group. Corollary 6.3.15 Let π be a polynomial over a field πΎ, with π distinct roots in SFπΎ ( π ). Then | GalπΎ ( π )| divides π!. Proof By Proposition 6.3.10, GalπΎ ( π ) is isomorphic to a subgroup of π π , which has π! elements. The result follows from Lagrange’s theorem. The inequalities (6.2) already gave us | GalπΎ ( π )| ≤ deg( π )!. Corollary 6.3.15 improves on this in two respects. First, it gives us | GalπΎ ( π )| ≤ π!, where π ≤ deg( π ) in all cases, and π < deg( π ) if π has repeated roots in its splitting field. A trivial example: if π (π‘) = π‘ 2 then π = 1 and deg( π ) = 2. Second, it tells us that | GalπΎ ( π )| is not only less than or equal to π!, but a factor of it. Galois theory is about the interplay between field extensions and groups. In the next chapter, we’ll see that just as every field extension giving rise to a group of automorphisms (its Galois group), every group of automorphisms gives rise to a field extension. We’ll also go deeper into the different types of field extension: normal extensions (the mirror image of normal subgroups) and separable extensions (which have to do with repeated roots). All that will lead us towards the fundamental theorem of Galois theory. 81 Chapter 7 Preparation for the fundamental theorem Introduction to Week 7 Very roughly, the fundamental theorem of Galois theory says that you can tell a lot about a field extension by looking at its Galois group. A bit more specifically, it says that the subgroups and quotients of Gal(π : πΎ), and their orders, give us information about the subfields of π containing πΎ, and their degrees. For example, one part of the fundamental theorem is that [π : πΎ] = | Gal(π : πΎ)|. The theorem doesn’t hold for all extensions, just those that are ‘nice enough’. Crucially, this includes splitting field extensions SFQ ( π ) : Q of polynomials π over Q—the starting point of classical Galois theory. Let’s dip our toes into the water by thinking about why it might be true that [π : πΎ] = | Gal(π : πΎ)|, at least for extensions that are nice enough. The simplest nontrivial extensions are the simple algebraic extensions, π = πΎ (πΌ). Write π for the minimal polynomial of πΌ over πΎ and πΌ1 , πΌ2 , . . . , πΌπ for the distinct roots of π in π. For every element π of Gal(π : πΎ), we have π(π(πΌ)) = 0 by Lemma 6.1.2, and so π(πΌ) = πΌ π for some π ∈ {1, . . . , π }. On the other hand, for each π ∈ {1, . . . , π }, there is exactly one π ∈ Gal(π : πΎ) such that π(πΌ) = πΌ π , by Proposition 6.1.5. So | Gal(π : πΎ)| = π . We know that [π : πΎ] = deg(π). So [π : πΎ] = | Gal(π : πΎ)| if and only if deg(π) is equal to π , the number of distinct roots of π in π. Certainly π ≤ deg(π). But are π and deg(π) equal? There are two reasons why √ they might not be. First, π might not split in π. 3 For√instance, if πΎ =√Q and πΌ = 2 then π(π‘) = π‘ 3 − 2, which has only one root in 3 3 Q( 2), so | Gal(Q( 2) : Q)| = 1 < 3 = deg(π). An algebraic extension is called ‘normal’ if this problem doesn’t occur, that is, if the minimal polynomial of every element does split. That’s what Section 7.1 is about. 82 Second, we might have π < deg(π) because some of the roots of π in π are repeated. If they are, the number π of distinct roots will be less then deg(π). An algebraic extension is called ‘separable’ if this problem doesn’t occur, that is, if the minimal polynomial of every element has no repeated roots in its splitting field. That’s what Section 7.2 is about. If we take any finite extension π : πΎ (not necessarily simple) that is both normal and separable, then it is indeed true that | Gal(π : πΎ)| = [π : πΎ]. And in fact, these conditions are enough to make the whole fundamental theorem work, as we’ll see next week. I hesitated before putting normality and separability into the same chapter, because you should think of them in quite different ways: • Normality has a clear conceptual meaning (as I explain in a video), and its importance was recognized by Galois himself. Despite the name, most field extensions aren’t normal; normality isn’t something to be taken for granted. • In contrast, Galois never considered separability, because it holds automatically over Q (his focus), and in fact over any field of characteristic 0, as well as any finite field. It takes some work to find an extension that isn’t separable. You can view separability as more of a technicality. There’s one more concept in this chapter: the ‘fixed field’ of a group of automorphisms (Section 7.3). Every Galois theory text I’ve seen contains at least one proof that makes you ask ‘how did anyone think of that?’ I would argue that the proof of Theorem 7.3.6 is the one and only truly ingenious argument in this course: maybe not the hardest, but the most ingenious. Your opinion may differ! 7.1 Normality Definition 7.1.1 An algebraic field extension π : πΎ is normal if for all πΌ ∈ π, the minimal polynomial of πΌ splits in π. We also say π is normal over πΎ to mean that π : πΎ is normal. Lemma 7.1.2 Let π : πΎ be an algebraic extension. Then π : πΎ is normal if and only if every irreducible polynomial over πΎ either has no roots in π or splits in π. Put another way, normality means that any polynomial over πΎ with at least one root in π has all its roots in π. Proof Suppose that π : πΎ is normal, and let π be an irreducible polynomial over πΎ. If π has a root πΌ in π then the minimal polynomial of πΌ is π /π, where π ∈ πΎ 83 What does it mean to be normal? is the leading coefficient of π . Since π : πΎ is normal, π /π splits in π, so π does too. Conversely, suppose that every irreducible polynomial over πΎ either has no roots in π or splits in π. Let πΌ ∈ π. Then the minimal polynomial of πΌ has at least one root in π (namely, πΌ), so it splits in π. √3 √3 Examples 7.1.3 i. Consider Q( 2) √ : Q. The minimal polynomial of 2 over √3 √3 3 2 3 Q is π‘ − 2, whose roots √ in C are 2 ∈ R and π 2, π 2 ∈ C \ R, where 2ππ/3 . Since Q( 3 2) ⊆ R, the minimal polynomial π‘ 3 − 2 does not split π = π√ √3 3 in Q( 2). Hence Q( 2) is not normal over Q. √3 Alternatively, using the equivalent condition in Lemma 7.1.2, Q( 2) : Q is not normal because π‘ 3 − 2 is an irreducible polynomial over Q that has a root √3 in Q( 2) but does not split there. √3 One way to think about the non-normality of Q( 2) : Q is as follows. The three roots of π‘ 3 − 2 are indistinguishable (conjugate) over Q, since they have the same minimal polynomial. But if they’re indistinguishable, it would be strange for an extension to contain some but not all of them, since that would be making a distinction between elements that are supposed to be √3 indistinguishable. In that sense, Q( 2) is ‘abnormal’. ii. Let π be a polynomial over a field πΎ. Then SFπΎ ( π ) : πΎ is always normal, as we shall see (Theorem 7.1.5). iii. Every extension of degree 2 is normal (just as, in group theory, every subgroup of index 2 is normal). You’ll be asked to show this in Workshop 4, but you also know enough to do it now. Exercise 7.1.4 What happens if you drop the word ‘irreducible’ from Lemma 7.1.2? Is it still true? Normality of field extensions is intimately related to normality of subgroups, and conjugacy in field extensions is also related to conjugacy in groups. The video ‘What does it mean to be normal?’ explains both kinds of normality and conjugacy in intuitive terms. Here’s the first of our two theorems about normal extensions. It describes which extensions arise as splitting field extensions. Theorem 7.1.5 Let π : πΎ be a field extension. Then π = SFπΎ ( π ) for some π ∈ πΎ [π‘] ⇐⇒ π : πΎ is finite and normal. 84 SF πO (π) π = πΎ (πΌ1O , . . . , πΌπ ) π πΎ (πΏ) g π / πΎ (πΌ1 , . . O . , πΌπ , π) / 7 πΎ (π) πΎ Figure 7.1: Maps used in the proof that splitting field extensions are normal. Splitting field extensions are normal Proof For ⇐, suppose that π : πΎ is finite and normal. By finiteness, there is a basis πΌ1 , . . . , πΌπ of π over πΎ, and each πΌπ is algebraic over πΎ (by Proposition 5.3.4). For each π, let ππ be the minimal polynomial of πΌπ over πΎ; then by normality, ππ splits in π. Hence π = π 1 π 2 · · · π π ∈ πΎ [π‘] splits in π. The set of roots of π in π contains {πΌ1 , . . . , πΌπ }, and π = πΎ (πΌ1 , . . . , πΌπ ), so π is generated over πΎ by the set of roots of π in π. Thus, π is a splitting field of π over πΎ. For ⇒, take π ∈ πΎ [π‘] such that π = SFπΎ ( π ). We may assume that π ≠ 0, since if π = 0 then π = πΎ, which is certainly finite and normal over πΎ. Write πΌ1 , . . . , πΌπ for the roots of π in π. Then π = πΎ (πΌ1 , . . . , πΌπ ). Each πΌπ is algebraic over πΎ (since π ≠ 0), so by Proposition 5.3.4, π : πΎ is finite. We now show that π : πΎ is normal, which is the most substantial part of the proof (Figure 7.1). Let πΏ ∈ π, with minimal polynomial π ∈ πΎ [π‘]. Certainly π splits in SF π (π), so to show that π splits in π, it is enough to show that every root π of π in SF π (π) lies in π. Since π is a monic irreducible annihilating polynomial of π over πΎ, it is the minimal polynomial of π over πΎ. Hence by Theorem 4.3.7, there is an isomorphism π : πΎ (πΏ) → πΎ (π) over πΎ. Now observe that: • π = SFπΎ (πΏ) ( π ), by Lemma 6.2.13(ii); • πΎ (πΌ1 , . . . , πΌπ , π) is a splitting field of π over πΎ (π), by Lemma 6.2.13(i) (taking the ‘π’, ‘π’ and ‘π ’ there to be SF π ( π ), π and {π}); • π ∗ ( π ) = π , since π ∈ πΎ [π‘] and π is a homomorphism over πΎ. So by Proposition 6.2.10, there is some isomorphism π : π → πΎ (πΌ1 , . . . , πΌπ , π) extending π. Then π is an isomorphism over πΎ, since π is. 85 Since πΏ is in πΎ (πΌ1 , . . . , πΌπ ), it can be expressed as a polynomial in πΌ1 , . . . , πΌπ over πΎ (by Corollary 5.1.11). Since π is a map over πΎ, it follows that π(πΏ) is a polynomial in π(πΌ1 ), . . . , π(πΌπ ) over πΎ. But π(πΏ) = π (πΏ) = π; moreover, for each π we have π (π(πΌπ )) = 0 (by Lemma 6.1.2 with π = idπΎ ) and so π(πΌπ ) ∈ {πΌ1 , . . . , πΌπ }. Hence π is a polynomial in πΌ1 , . . . , πΌπ over πΎ, giving π ∈ πΎ (πΌ1 , . . . , πΌπ ) = π, as required. Corollary 7.1.6 Let π : πΏ : πΎ be field extensions. If π : πΎ is finite and normal then so is π : πΏ. Proof Follows from Theorem 7.1.5 and Lemma 6.2.13(ii). Warning 7.1.7 does not follow that πΏ : πΎ is normal. For instance, √3 √3 It2ππ/3 ) : Q( 2) : Q. The first field is the√ splitting consider Q( 2, π 3 field of π‘ 3 − 2 over Q, and therefore normal over Q, but Q( 2) is not (Example 7.1.3(i)). Theorem 7.1.5 is the first of two theorems about normality. The second is to do with the action of the Galois group of an extension. Warning 7.1.8 By definition, the Galois group Gal(π : πΎ) of an extension π : πΎ acts on π. But if π is the splitting field of some polynomial π over πΎ then the action of Gal(π : πΎ) on π restricts to an action on the roots of π (a finite set), as we saw in Section 6.3. So there are two actions of the Galois group under consideration, one the restriction of the other. Both are important. When a group acts on a set, a basic question is: what are the orbits? For Gal(π : πΎ) acting on π, the answer is: the conjugacy classes of π over πΎ. Or at least, that’s the case when π : πΎ is finite and normal: Proposition 7.1.9 Let π : πΎ be a finite normal extension and πΌ, πΌ0 ∈ π. Then πΌ and πΌ0 are conjugate over πΎ ⇐⇒ πΌ0 = π(πΌ) for some π ∈ Gal(π : πΎ). Proof For ⇐, let π ∈ Gal(π : πΎ) with πΌ0 = π(πΌ). Then πΌ and πΌ0 are conjugate over πΎ, by Lemma 6.1.2 (with π 0 = π, πΎ 0 = πΎ and π = idπΎ ). For ⇒, suppose that πΌ and πΌ0 are conjugate over πΎ. Since π : πΎ is finite, both are algebraic over πΎ, and since they are conjugate over πΎ, they have the same minimal polynomial π ∈ πΎ [π‘]. By Theorem 4.3.7, there is an isomorphism π : πΎ (πΌ) → πΎ (πΌ0) over πΎ such that π (πΌ) = πΌ0 (see diagram below). 86 By Theorem 7.1.5, π is the splitting field of some polynomial π over πΎ. Hence π is also the splitting field of π over both πΎ (πΌ) and πΎ (πΌ0), by Lemma 6.2.13(ii). Moreover, π ∗ ( π ) = π since π is a homomorphism over πΎ and π is a polynomial over πΎ. So by Proposition 6.2.10(i), there is an automorphism π of π extending π: π /π πO O πΎ (πΌ)b π / πΎ< (πΌ0) πΎ Then π ∈ Gal(π : πΎ) with π(πΌ) = π (πΌ) = πΌ0, as required. That result was about the action of Gal(π : πΎ) on the field π, but it has a powerful corollary involving the action of the Galois group on the roots of an irreducible polynomial π , when π = SFπΎ ( π ): Corollary 7.1.10 Let π be an irreducible polynomial over a field πΎ. Then the action of GalπΎ ( π ) on the roots of π in SFπΎ ( π ) is transitive. Recall what transitive means, for an action of a group πΊ on a set π: for all ∈ π, there exists π ∈ πΊ such that ππ₯ = π₯ 0. π₯, π₯ 0 Proof Since π is irreducible, the roots of π in SFπΎ ( π ) all have the same minimal polynomial, namely, π divided by its leading coefficient. So they are all conjugate over πΎ. Since SFπΎ ( π ) : πΎ is finite and normal (by Theorem 7.1.5), the result follows from Proposition 7.1.9. Exercise 7.1.11 Show by example that Corollary 7.1.10 becomes false if you drop the word ‘irreducible’. Example 7.1.12 Let π (π‘) = 1 + π‘ + · · · + π‘ π−1 ∈ Q[π‘], where π is prime. Since (1 − π‘) π (π‘) = 1 − π‘ π , the roots of π in C are π, π2 , . . . , π π−1 , where π = π 2ππ/π . By Example 3.3.16, π is irreducible over Q. Hence by Corollary 7.1.10, for each π ∈ {1, . . . , π − 1}, there is some π ∈ GalQ ( π ) such that π(π) = ππ . This is spectacular! Until now, we’ve been unable to prove such things without a huge amount of explicit checking, which, moreover, only works on a case-by-case basis. For example, if you watched the video ‘Calculating Galois groups with bare hands, part 2’, you’ll have seen how much tedious calculation went into the single 87 case π = 5, π = 2: But the theorems we’ve proved make all this unnecessary. In fact, for each π ∈ {1, . . . , π − 1}, there’s exactly one element ππ of GalQ ( π ) such that ππ (π) = ππ . For since SFQ ( π ) = Q(π, . . . , π π−1 ) = Q(π), two elements of GalQ ( π ) that take the same value on π must be equal. Hence GalQ ( π ) = {π1 , . . . , π π−1 }. We’ll see later that GalQ ( π ) πΆ π−1 . Example 7.1.13 Let’s calculate πΊ = GalQ (π‘ 3 − 2). Since π‘ 3 − 2 has 3 distinct roots in C, it has 3 distinct roots in its splitting field. By Proposition 6.3.10, πΊ is isomorphic to a subgroup of π3 . Now πΊ acts transitively on the 3 roots, so it has at least 3 elements, so it is isomorphic to either π΄3 or π3 . Since two of the roots are non-real complex conjugates, one of the elements of πΊ is complex conjugation, which has order 2. Hence 2 divides |πΊ |, forcing πΊ π3 . We now show how a normal field extension gives rise to a normal subgroup. Whenever you meet a normal subgroup, you should immediately want to form the resulting quotient, so we do that too. Theorem 7.1.14 Let π : πΏ : πΎ be field extensions with π : πΎ finite and normal. i. πΏ : πΎ is a normal extension ⇐⇒ ππΏ = πΏ for all π ∈ Gal(π : πΎ). ii. If πΏ : πΎ is a normal extension then Gal(π : πΏ) is a normal subgroup of Gal(π : πΎ) and Gal(π : πΎ) Gal(πΏ : πΎ). Gal(π : πΏ) 88 Before the proof, here’s some context and explanation. Part (i) answers the question implicit in Warning 7.1.7: we know from Corollary 7.1.6 that π : πΏ is normal, but when is πΏ : πΎ normal? In part (i), ππΏ means {π(πΌ) : πΌ ∈ πΏ}. For ππΏ to be equal to πΏ means that π fixes πΏ as a set (in other words, permutes it within itself), not that π fixes each element of πΏ. In part (ii), it’s true for all π : πΏ : πΎ that Gal(π : πΏ) is a subset of Gal(π : πΎ), since Gal(π : πΏ) = {automorphisms π of π such that π(πΌ) = πΌ for all πΌ ∈ πΏ} ⊆ {automorphisms π of π such that π(πΌ) = πΌ for all πΌ ∈ πΎ } = Gal(π : πΎ). It’s also always true that Gal(π : πΏ) is a subgroup of Gal(π : πΎ), as you can easily check. But part (ii) tells us something much more substantial: it’s a normal subgroup when πΏ : πΎ is a normal extension. Proof of Theorem 7.1.14 For (i), first suppose that πΏ is normal over πΎ, and let π ∈ Gal(π : πΎ). For all πΌ ∈ πΏ, Proposition 7.1.9 implies that πΌ and π(πΌ) are conjugate over πΎ, so they have the same minimal polynomial, so π(πΌ) ∈ πΏ by normality. Hence ππΏ ⊆ πΏ. The same argument with π−1 in place of π gives π−1 πΏ ⊆ πΏ, and applying π to each side then gives πΏ ⊆ ππΏ. So ππΏ = πΏ. Conversely, suppose that ππΏ = πΏ for all π ∈ Gal(π : πΎ). Let πΌ ∈ πΏ with minimal polynomial π. Since π : πΎ is normal, π splits in π. Each root πΌ0 of π in π is conjugate to πΌ over πΎ, so by Proposition 7.1.9, πΌ0 = π(πΌ) for some π ∈ Gal(π : πΎ), giving πΌ0 ∈ ππΏ = πΏ. Hence π splits in πΏ and πΏ : πΎ is normal. For (ii), suppose that πΏ : πΎ is normal. To prove that Gal(π : πΏ) is a normal subgroup of Gal(π : πΎ), let π ∈ Gal(π : πΎ) and π ∈ Gal(π : πΏ). We show that π−1 ππ ∈ Gal(π : πΏ), or equivalently, π−1 ππ(πΌ) = πΌ for all πΌ ∈ πΏ, or equivalently, ππ(πΌ) = π(πΌ) for all πΌ ∈ πΏ. But by (i), π(πΌ) ∈ πΏ for all πΌ ∈ πΏ, so π (π(πΌ)) = π(πΌ) since π ∈ Gal(π : πΏ). This completes the proof that Gal(π : πΏ) P Gal(π : πΎ). Finally, we prove the statement on quotients (still supposing that πΏ : πΎ is a normal extension). Every automorphism π of π over πΎ satisfies ππΏ = πΏ (by (i)), and therefore restricts to an automorphism πˆ of πΏ. The function π: Gal(π : πΎ) → Gal(πΏ : πΎ) π β¦→ πˆ 89 is a group homomorphism, since it preserves composition. Its kernel is Gal(π : πΏ), by definition. If we can prove that π is surjective then the last part of the theorem will follow from the first isomorphism theorem. To prove that π is surjective, we must show that each automorphism π of πΏ over πΎ extends to an automorphism π of π: πO π πΏ` π / / πO > πΏ πΎ The argument is similar to the second half of the proof of Proposition 7.1.9. By Theorem 7.1.5, π is the splitting field of some π ∈ πΎ [π‘]. Then π is also the splittting field of π over πΏ. Also, π∗ ( π ) = π since π is a homomorphism over πΎ and π is a polynomial over πΎ. So by Proposition 6.2.10(i), there is an automorphism π of π extending π, as required. Example 7.1.15 Take π : πΏ : πΎ to be √3 Q 2, π : Q(π) : Q, √3 where π = π 2ππ/3 . As you will recognize by now, Q( 2, π) is the splitting field of π‘ 3 − 2 over Q, so it is a finite normal extension of Q by Theorem 7.1.5. Also, Q(π) is the splitting field of π‘ 2 +π‘ +1 over Q, so it too is a normal extension of Q. Part (i) of Theorem 7.1.14 implies that every element of GalQ (π‘ 3 −2) restricts to an automorphism of Q(π). Part (ii) implies that √3 √3 Gal Q 2, π : Q(π) P Gal Q 2, π : Q and that √3 Gal Q 2, π : Q √3 Gal(Q(π) : Q). Gal Q 2, π : Q(π) (7.1) √3 What does this say explicitly? We showed in Example 7.1.13 that Gal(Q( 2, π) : Q) π3 . That is, each element of the Galois group permutes the three roots √3 √3 √3 2, π 2, π2 2 90 of π‘ 3 − 2, and all six permutations are realized by some element of the Galois √3 group. An element√ of Gal(Q( 2, π) : Q) that √fixes π is determined by which 3 3 of the three roots 2 is mapped to, so Gal(Q( 2, π) : Q(π)) π΄3 . Finally, Gal(Q(π) : Q) πΆ2 by Example 7.1.12. So in this case, the isomorphism (7.1) states that π3 πΆ2 . π΄3 3 Exercise 7.1.16 Draw a diagram √3 showing the three roots of π‘ − 2 and the elements of π» = Gal(Q( 2, π) : Q(π)) acting on them. √3 There is a simple geometric description of the elements of Gal(Q( 2, π) : Q) that belong to the subgroup π». What is it? 7.2 Separability Theorem 6.2.12 implies that | Gal(π : πΎ)| ≤ [π : πΎ] whenever π : πΎ is a splitting field extension. Why is this an inequality, not an equality? The answer can be traced back to the proof of Proposition 6.2.10 on extension of isomorphisms. There, we had an irreducible polynomial called π∗ (π), and we wrote π for the number of distinct roots of π∗ (π) in its splitting field. Ultimately, the source of the inequality | Gal(π : πΎ)| ≤ [π : πΎ] was the fact that π ≤ deg(π∗ (π)). But is this last inequality actually an equality? That is, does an irreducible polynomial of degree π always have π distinct roots in its splitting field? Certainly it has π roots when counted with multiplicity. But there will be fewer than π distinct roots if any of the roots are repeated (have multiplicity ≥ 2, as defined in Definition 3.2.10). The question is whether this can ever happen. Exercise 7.2.1 Try to find an example of an irreducible polynomial of degree π with fewer than π distinct roots in its splitting field. Or if you can’t, see if you can prove that this is impossible over Q—that is, an irreducible over Q has no repeated roots in C. Both are quite hard, but ten minutes spent trying may help you to appreciate what’s to come. Definition 7.2.2 An irreducible polynomial over a field is separable if it has no repeated roots in its splitting field. Equivalently, an irreducible polynomial π ∈ πΎ [π‘] is separable if it splits into distinct linear factors in SFπΎ ( π ): π (π‘) = π(π‘ − πΌ1 ) · · · (π‘ − πΌπ ) 91 for some π ∈ πΎ and distinct πΌ1 , . . . , πΌπ ∈ SFπΎ ( π ). Put another way, an irreducible π is separable if and only if it has deg( π ) distinct roots in its splitting field. Example 7.2.3 π‘ 3 − 2 ∈ Q[π‘] is separable, since it has 3 distinct roots in C, hence in its splitting field. Example 7.2.4 This is an example of an irreducible polynomial that’s inseparable. It’s a little bit complicated, although in fact it’s the simplest example there is. Let π be a prime, let πΎ be the field F π (π’) of rational expressions over F π (where π’ is the variable symbol), and let π (π‘) = π‘ π − π’. By definition, π has at least one root πΌ in its splitting field. We have π (π‘ − πΌ) = π Õ π π‘ π (−πΌ) π−π = π‘ π − πΌ π = π (π‘), π π=0 where the second equality follows from Lemma 3.3.15. So π (π‘) = (π‘ − πΌ) π in SFπΎ ( π ), which means that πΌ is the only root of π in its splitting field—despite π having degree π > 1. We now show that π is irreducible over πΎ. The unique factorization of π into irreducible polynomials over SFπΎ ( π ) is π (π‘) = (π‘ − πΌ) π , so any nontrivial factorization of π in πΎ [π‘] is of the form π (π‘) = (π‘ − πΌ) π (π‘ − πΌ) π−π where 0 < π < π and both factors belong to πΎ [π‘]. The coefficient of π‘ π−1 in (π‘ − πΌ) π is −ππΌ, so −ππΌ ∈ πΎ. But π is invertible in πΎ, so πΌ ∈ πΎ, contradicting Exercise 7.2.5. Exercise 7.2.5 Show that π’ has no πth root in F π (π’); that is, there is no πΌ ∈ F π (π’) with πΌ π = π’. (Hint: consider the degree of polynomials.) Warning 7.2.6 Definition 7.2.2 is only a definition of separability for irreducible polynomials. There is a definition of separability for arbitrary polynomials, but it’s not simply Definition 7.2.2 with the word ‘irreducible’ deleted. We won’t need it, but here it is: an arbitrary polynomial is called separable if each of its irreducible factors is separable. So π‘ 2 is separable, even though it has a repeated root. In real analysis, we can test whether a root is repeated by asking whether the 92 derivative is 0 there: Over an arbitrary field, there’s no general definition of differentiation, as there is no sense of what a ‘limit’ might be. But even without limits, we can differentiate polynomials in the following sense. Íπ Definition 7.2.7 Let πΎ be a field and let π (π‘) = π=0 ππ π‘ π ∈ πΎ [π‘]. The formal derivative of π is π Õ (π· π )(π‘) = πππ π‘ π−1 ∈ πΎ [π‘]. π=1 We use π· π rather than π 0 to remind ourselves not to take the familiar properties of differentiation for granted. Nevertheless, the usual basic laws hold: Lemma 7.2.8 Let πΎ be a field. Then π· ( π + π) = π· π + π·π, π· ( π π) = π · π·π + π· π · π, π·π = 0 for all π , π ∈ πΎ [π‘] and π ∈ πΎ. Exercise 7.2.9 Check a couple of the properties in Lemma 7.2.8. The real analysis test for repetition of roots has an algebraic analogue: Lemma 7.2.10 Let π be a nonzero polynomial over a field πΎ. The following are equivalent: i. π has a repeated root in SFπΎ ( π ); ii. π and π· π have a common root in SFπΎ ( π ); iii. π and π· π have a nonconstant common factor in πΎ [π‘]. Proof (i)⇒(ii): suppose that π has a repeated root πΌ in SFπΎ ( π ). Then π (π‘) = (π‘ − πΌ) 2 π(π‘) for some π(π‘) ∈ (SFπΎ ( π )) [π‘]. Hence (π· π )(π‘) = (π‘ − πΌ) 2π(π‘) + (π‘ − πΌ) · (π·π)(π‘) , 93 so πΌ is a common root of π and π· π in SFπΎ ( π ). (ii)⇒(iii): suppose that π and π· π have a common root πΌ in SFπΎ ( π ). Then πΌ is algebraic over πΎ (since π ≠ 0), and the minimal polynomial of πΌ over πΎ is then a nonconstant common factor of π and π· π . (iii)⇒(ii): if π and π· π have a nonconstant common factor π then π splits in SFπΎ ( π ), and any root of π in SFπΎ ( π ) is a common root of π and π· π . (ii)⇒(i): suppose that π and π· π have a common root πΌ ∈ SFπΎ ( π ). Then π (π‘) = (π‘ − πΌ)π(π‘) for some π ∈ (SFπΎ ( π )) [π‘], giving (π· π )(π‘) = π(π‘) + (π‘ − πΌ) · (π·π)(π‘). But (π· π )(πΌ) = 0, so π(πΌ) = 0, so π(π‘) = (π‘ − πΌ)β(π‘) for some β ∈ (SFπΎ ( π )) [π‘]. Hence π (π‘) = (π‘ − πΌ) 2 β(π‘), and πΌ is a repeated root of π in its splitting field. The point of Lemma 7.2.10 is that condition (iii) allows us to test for repetition of roots in SFπΎ ( π ) without ever leaving πΎ [π‘], or even knowing what SFπΎ ( π ) is. Proposition 7.2.11 Let π be an irreducible polynomial over a field. Then π is inseparable if and only if π· π = 0. Proof This follows from (i) ⇐⇒ (iii) in Lemma 7.2.10. Since π is irreducible, π and π· π have a nonconstant common factor if and only if π divides π· π ; but deg(π· π ) < deg( π ), so π | π· π if and only if π· π = 0. Corollary 7.2.12 Let πΎ be a field. i. If char πΎ = 0 then every irreducible polynomial over πΎ is separable. ii. If char πΎ = π > 0 then an irreducible polynomial π ∈ πΎ [π‘] is inseparable if and only if π (π‘) = π 0 + π 1 π‘ π + · · · + ππ π‘ π π for some π 0 , . . . , ππ ∈ πΎ. In other words, the only irreducible polynomials that are inseparable are the polynomials in π‘ π in characteristic π. Inevitably, Example 7.2.4 is of this form. Í Proof Let π (π‘) = ππ π‘ π be an irreducible polynomial. Then π is inseparable if and only if π· π = 0, if and only if πππ = 0 for all π ≥ 1. If char πΎ = 0, this implies that ππ = 0 for all π ≥ 1, so π is constant, which contradicts π being irreducible. If char πΎ = π, then πππ = 0 for all π ≥ 1 is equivalent to ππ = 0 whenever π - π. Remark 7.2.13 In the final chapter we will show that every irreducible polynomial over a finite field is separable. So, it is only over infinite fields of characteristic π that you have to worry about inseparability. 94 We now build up to showing that | Gal(π : πΎ)| = [π : πΎ] whenever π : πΎ is a finite normal extension in which the minimal polynomial of every element of π is separable. First, some terminology: Definition 7.2.14 Let π : πΎ be an algebraic extension. An element of π is separable over πΎ if its minimal polynomial over πΎ is separable. The extension π : πΎ is separable if every element of π is separable over πΎ. Examples 7.2.15 i. Every algebraic extension of fields of characteristic 0 is separable, by Corollary 7.2.12. ii. Every algebraic extension of a finite field is separable, by Remark 7.2.13. iii. The splitting field of π‘ π − π’ over F π (π’) is inseparable: indeed, the element denoted by πΌ in Example 7.2.4 is inseparable over F π (π’), since its minimal polynomial is the inseparable polynomial π‘ π − π’. Exercise 7.2.16 Let π : πΏ : πΎ be field extensions. Show that if π : πΎ is algebraic then so are π : πΏ and πΏ : πΎ. Lemma 7.2.17 Let π : πΏ : πΎ be field extensions, with π : πΎ algebraic. If π : πΎ is separable then so are π : πΏ and πΏ : πΎ. Proof Both π : πΏ and πΏ : πΎ are algebraic by Exercise 7.2.16, so it does make sense to ask whether they are separable. (We only defined what it means for an algebraic extension to be separable.) That πΏ : πΎ is separable is immediate from the definition. To show that π : πΏ is separable, let πΌ ∈ π. Write π πΏ and π πΎ for the minimal polynomials of πΌ over πΏ and πΎ, respectively. Then π πΎ is an annihilating polynomial of πΌ over πΏ, so π πΏ | π πΎ in πΏ [π‘]. Since π : πΎ is separable, π πΎ splits into distinct linear factors in SFπΎ (π πΎ ). Since π πΏ | π πΎ , so does π πΏ . Hence π πΏ ∈ πΏ [π‘] is separable, so πΌ is separable over πΏ. As hinted in the introduction to this section, we will prove that | Gal(π : πΎ)| = [π : πΎ] by refining Proposition 6.2.10. Proposition 7.2.18 Let π : πΎ → πΎ 0 be an isomorphism of fields, let π ∈ πΎ [π‘], let π be a splitting field of π over πΎ, and let π 0 be a splitting field of π∗ ( π ) over πΎ 0. Suppose that the extension π 0 : πΎ 0 is separable. Then there are exactly [π : πΎ] isomorphisms π : π → π 0 extending π. Proof This is almost the same as the proof of Proposition 6.2.10, but with the inequality π ≤ deg(π∗ (π)) replaced by an equality, which holds by separability. For the inductive hypothesis to go through, we need the extension π 0 : πΎ 0 (πΌ0π ) to be separable, and this follows from the separability of π 0 : πΎ 0 by Lemma 7.2.17. 95 Theorem 7.2.19 | Gal(π : πΎ)| = [π : πΎ] for every finite normal separable extension π : πΎ. Proof By Theorem 7.1.5, π = SFπΎ ( π ) for some π ∈ πΎ [π‘]. The result follows from Proposition 7.2.18, taking π 0 = π, πΎ 0 = πΎ, and π = idπΎ . Examples 7.2.20 i. | GalπΎ ( π )| = [SFπΎ ( π ) : πΎ] for any polynomial π over a field πΎ of characteristic 0. For instance, [SFQ (π‘ 3 − 2) : Q] = 6 by a similar argument to Example 5.2.7, using that SFQ (π‘ 3 − 2) contains elements of degree 2 and 3 over Q. Hence | GalQ ( π )| = 6. But GalQ ( π ) embeds into π3 by Proposition 6.3.10, so GalQ ( π ) π3 . We already proved this in Example 7.1.15, using a different argument. ii. Consider πΎ = F π (π’) and π = SFπΎ (π‘ π − π’). With notation as in Example 7.2.4, we have π = πΎ (πΌ), so [π : πΎ] = degπΎ (πΌ) = π. On the other hand, | Gal(π : πΎ)| = 1 by Corollary 6.3.15. So Theorem 7.2.19 fails if we drop the separability hypothesis. Digression 7.2.21 With some effort, one can show that in any algebraic extension π : πΎ, the separable elements form a subfield of π. (See Stewart, Theorem 17.22.) It follows that a finite extension πΎ (πΌ1 , . . . , πΌπ ) : πΎ is separable if and only if each πΌπ is. Hence a splitting field extension SFπΎ ( π ) : πΎ is separable if and only if every root of π is separable in SFπΎ ( π ), which itself is equivalent to π being separable in the sense of Warning 7.2.6. So: SFπΎ ( π ) is separable over πΎ if and only if π is separable over πΎ. Thus, the different meanings of ‘separable’ interact nicely. Digression 7.2.22 It’s a stunning fact that every finite separable extension is simple. This is called the theorem of the primitive element. For instance, whenever πΌ1 , . . . , πΌπ are complex numbers algebraic over Q, there is some πΌ ∈ C (a ‘primitive element’) such that √ Q(πΌ √ 1 , . . . ,√πΌπ ) =√Q(πΌ). We saw one case of this in Example 4.3.2(i): Q( 2, 3) = Q( 2 + 3). The theorem of the primitive element was at the heart of most early accounts of Galois theory, and appears in many modern treatments too, but we will not prove it. 7.3 Fixed fields When a group πΊ acts on a set π, we usually want to know which elements of π are fixed by every element of πΊ. Here, we’ll ask this question for groups of 96 automorphisms of a field. The following preliminary definition will come in handy. Definition 7.3.1 Let π and π be sets, and let π» ⊆ {functions π → π }. The equalizer of π» is Eq(π») = {π₯ ∈ π : π (π₯) = π(π₯) for all π , π ∈ π»}. So the equalizer of π» is the part of π on which all the functions in π» are equal. Lemma 7.3.2 Let π and π 0 be fields, and let π» ⊆ {homomorphisms π → π 0 }. Then Eq(π») is a subfield of π. Proof We must show that 0, 1 ∈ Eq(π»), that if πΌ ∈ Eq(π») then −πΌ ∈ Eq(π») and 1/πΌ ∈ Eq(π») (when πΌ ≠ 0), and that if πΌ, π½ ∈ Eq(π») then πΌ + π½, πΌπ½ ∈ Eq(π»). I will show just the last of these, leaving the rest to you. Suppose that πΌ, π½ ∈ Eq(π»). For all π, π ∈ π», we have π(πΌπ½) = π(πΌ)π(π½) = π (πΌ)π (π½) = π (πΌπ½), so πΌπ½ ∈ Eq(π»). Write Aut(π) for the group of automorphisms of a field π. Definition 7.3.3 Let π be a field and let π» a subgroup of Aut(π). The fixed field of π» is Fix(π») = {πΌ ∈ π : π(πΌ) = πΌ for all π ∈ π»}. Lemma 7.3.4 Fix(π») is a subfield of π. Proof In fact, Fix(π») = Eq(π»), since id π is an element of π». The result follows from Lemma 7.3.2. Exercise 7.3.5 Using Lemma 7.3.4, show that every automorphism of a field is an automorphism over its prime subfield. In other words, Aut(π) = Gal(π : πΎ) whenever π is a field with prime subfield πΎ. Here’s the big, ingenious, result about fixed fields. It will play a crucial part in the proof of the fundamental theorem of Galois theory. 97 Theorem 7.3.6 Let π be a field and π» a finite subgroup of Aut(π). Then [π : Fix(π»)] ≤ |π»|. As Fix(π») gets bigger, [π : Fix(π»)] gets smaller. Theorem 7.3.6 tells us that the smaller |π»| is, the more of π must be fixed by π». For some intuition about both the statement and the proof, watch the video ‘The size of fixed fields’. Proof Write π = |π»|. It is enough to prove that any π + 1 elements πΌ0 , . . . , πΌπ of π are linearly dependent over Fix(π»). Let π = (π₯ 0 , . . . , π₯ π ) ∈ π π+1 : π₯ 0 π (πΌ0 ) + · · · + π₯ π π (πΌπ ) = 0 for all π ∈ π» . The size of fixed fields Then π is defined by π homogeneous linear equations in π + 1 variables, so it is a nontrivial π-linear subspace of π π+1 . Claim: let (π₯ 0 , . . . , π₯ π ) ∈ π and π ∈ π». Then (π(π₯ 0 ), . . . , π(π₯ π )) ∈ π. Proof: For all π ∈ π», we have π₯ 0 (π−1 β¦ π)(πΌ0 ) + · · · + π₯ π (π−1 β¦ π)(πΌπ ) = 0 since π−1 β¦ π ∈ π». Applying π to both sides gives that for all π ∈ π», π(π₯ 0 )π (πΌ0 ) + · · · + π(π₯ π )π (πΌπ ) = 0, Remarks on the proof of Theorem 7.3.6 (actually a PDF, not a video) proving the claim. Since π is nontrivial, there is a least π ≥ 0 such that π contains some nonzero vector x = (π₯ 0 , . . . , π₯ π ) with π₯π = 0 for all π > π. Then π₯ π ≠ 0, and since π is closed under scalar multiplication by π, we may assume that π₯ π = 1. We now show that π₯π ∈ Fix(π») for all π. Let π ∈ π». By the claim, (π(π₯0 ), . . . , π(π₯ π )) ∈ π. Since π is a linear subspace, (π(π₯0 ) − π₯0 , . . . , π(π₯ π ) − π₯ π ) ∈ π . But π(π₯ π ) − π₯ π = π(1) − 1 = 0 and π(π₯π ) − π₯π = 0 − 0 = 0 for all π > π, so by minimality of π, also π(π₯π ) − π₯π = 0 for all π < π. Hence π₯π ∈ Fix(π») for all π. We have shown that π contains a nonzero vector x ∈ Fix(π») π+1 . But taking Í π = id in the definition of π gives π₯π πΌπ = 0. Hence πΌ0 , . . . , πΌπ are linearly dependent over Fix(π»). Example 7.3.7 Let π : C → C be complex conjugation. Then π» = {id, π } is a subgroup of Aut(C), and Theorem 7.3.6 predicts that [C : Fix(π»)] ≤ 2. Since Fix(π») = R, this is true. 98 Exercise 7.3.8 Find another example of Theorem 7.3.6. Digression 7.3.9 In fact, Theorem 7.3.6 is an equality: [π : Fix(π»)] = |π»|. This is proved directly in many Galois theory books (e.g. Stewart, Theorem 10.5). In our approach it will be a consequence of the fundamental theorem of Galois theory (rather than a step on the way to proving it), at least under certain hypotheses on π and π». The reverse inequality, [π : Fix(π»)] ≥ |π»|, is closely related to the result called ‘linear independence of characters’. (A good reference is Lang, Algebra, 3rd edition, Theorem 4.1.) Another instance of linear independence of characters is that the functions π§ β¦→ π 2 ππππ§ (π ∈ Z) are linearly independent, a fundamental fact in the theory of Fourier series. We finish by adding a further connecting strand between the concepts of normal extension and normal subgroup, complementary to the strands in Theorem 7.1.14. It needs a lemma that fundamentally has nothing to do with fields, and is really a general result about groups acting on sets, as the proof shows. Lemma 7.3.10 Let π be a field, π» a subgroup of Aut(π), and π ∈ Aut(π). Then Fix(ππ»π−1 ) = π Fix(π»). Proof Let πΌ ∈ π. Then πΌ ∈ Fix(ππ»π−1 ) ⇐⇒ πππ−1 (πΌ) = πΌ for all π ∈ π» ⇐⇒ ππ−1 (πΌ) = π−1 (πΌ) for all π ∈ π» ⇐⇒ π−1 (πΌ) ∈ Fix(π») ⇐⇒ πΌ ∈ π Fix(π»). Proposition 7.3.11 Let π : πΎ be a finite normal extension and π» a normal subgroup of Gal(π : πΎ). Then Fix(π») is a normal extension of πΎ. Proof Since every element of π» is an automorphism over πΎ, the subfield Fix(π») of π contains πΎ. For each π ∈ Gal(π : πΎ), we have π Fix(π») = Fix(ππ»π−1 ) = Fix(π»), where the first equality holds by Lemma 7.3.10 and the second because π» is normal in Gal(π : πΎ). Hence by Theorem 7.1.14(i), Fix(π») : πΎ is a normal extension. The stage is now set for the central result of the course: the fundamental theorem of Galois theory. 99 Chapter 8 The fundamental theorem of Galois theory We’ve been building up to this moment all semester. Let’s do it! Introduction to Week 8 8.1 Introducing the Galois correspondence Let π : πΎ be a field extension, with πΎ viewed as a subfield of π, as usual. An intermediate field of π : πΎ is a subfield of π containing πΎ. Write F = {intermediate fields of π : πΎ }. For πΏ ∈ F , we draw diagrams like this: π πΏ πΎ, with the bigger fields higher up. Also write G = {subgroups of Gal(π : πΎ)}. For π» ∈ G , we draw diagrams like this: 1 π» Gal(π : πΎ), 100 where 1 denotes the trivial subgroup and the bigger groups are lower down. It will become clear soon why we’re using opposite conventions for the field and group diagrams. For πΏ ∈ F , the group Gal(π : πΏ) consists of all automorphisms π of π that fix each element of πΏ. Since πΎ ⊆ πΏ, any such π certainly fixes each element of πΎ. Hence Gal(π : πΏ) is a subgroup of Gal(π : πΎ). This process defines a function Gal(π : −) : F → G πΏ β¦→ Gal(π : πΏ). In the expression Gal(π : −), the symbol − should be seen as a blank space into which arguments can be inserted. Warning 8.1.1 The group we’re associating with πΏ is Gal(π : πΏ), not Gal(πΏ : πΎ)! Both groups matter, but only one is a subgroup of Gal(π : πΎ), which is what we’re interested in here. We showed just now that Gal(π : πΏ) is a subgroup of Gal(π : πΎ). If you wanted to show that Gal(πΏ : πΎ) is (isomorphic to) a subgroup of Gal(π : πΎ)—which it isn’t—then you’d probably do it by trying to prove that every automorphism of πΏ over πΎ extends uniquely to π. And that’s false. For instance, when πΏ = πΎ, the identity on πΏ typically has many extensions to π: they’re the elements of Gal(π : πΎ). Although Gal(πΏ : πΎ) isn’t a subgroup of Gal(π : πΎ), it is a quotient of it, at least when both extensions are finite and normal. We saw this in Theorem 7.1.14, and we’ll come back to it in Section 8.2. In the other direction, for π» ∈ G , the subfield Fix(π») of π contains πΎ. Indeed, π» ⊆ Gal(π : πΎ), and by definition, every element of Gal(π : πΎ) fixes every element of πΎ, so Fix(π») ⊇ πΎ. Hence Fix(π») is an intermediate field of π : πΎ. This process defines a function Fix : G → F π» β¦→ Fix(π»). We have now defined functions Fo Gal(π:−) Fix / G. The fundamental theorem of Galois theory tells us how these functions behave: how the concepts of Galois group and fixed field interact. The proof will bring together most of the big results we’ve proved so far, and assumes that the extension is finite, normal and separable. But first, let’s say the simple things that are true for all extensions: 101 π 1 πΏ2 Gal(π : πΏ 2 ) πΏ1 Gal(π : πΏ 1 ) πΎ Gal(π : πΎ) Figure 8.1: The function πΏ β¦→ Gal(π : πΏ) is order-reversing (Lemma 8.1.2(i)). Lemma 8.1.2 Let π : πΎ be a field extension, and define F and G as above. i. For πΏ 1 , πΏ 2 ∈ F , πΏ 1 ⊆ πΏ 2 ⇒ Gal(π : πΏ 1 ) ⊇ Gal(π : πΏ 2 ) (Figure 8.1). For π»1 , π»2 ∈ G , π»1 ⊆ π»2 ⇒ Fix(π»1 ) ⊇ Fix(π»2 ). ii. For πΏ ∈ F and π» ∈ G , πΏ ⊆ Fix(π») ⇐⇒ π» ⊆ Gal(π : πΏ). iii. For all πΏ ∈ F , πΏ ⊆ Fix(Gal(π : πΏ)). For all π» ∈ G , π» ⊆ Gal(π : Fix(π»)). Warning 8.1.3 In part (i), the functions Gal(π : −) and Fix reverse inclusions. The bigger you make πΏ, the smaller you make Gal(π : πΏ), because it gets harder for an automorphism to fix everything in πΏ. And the bigger you make π», the smaller you make Fix(π»), because it gets harder for an element of π to be fixed by everything in π». That’s why the field and group diagrams are opposite ways up. 102 Proof (i): I leave the first half as an exercise. For the second, suppose that π»1 ⊆ π»2 , and let πΌ ∈ Fix(π»2 ). Then π (πΌ) = πΌ for all π ∈ π»2 , so π (πΌ) = πΌ for all π ∈ π»1 , so πΌ ∈ Fix(π»1 ). (ii): both sides are equivalent to the statement that π (πΌ) = πΌ for all π ∈ π» and πΌ ∈ πΏ. (iii): the first statement follows from the ⇐ direction of (ii) by taking π» = Gal(π : πΏ), and the second follows from the ⇒ direction of (ii) by taking πΏ = Fix(π»). (Or, they can be proved directly.) Exercise 8.1.4 Prove the first half of Lemma 8.1.2(i). Exercise 8.1.5 Draw a diagram like Figure 8.1 for the second half of Lemma 8.1.2(i). Digression 8.1.6 If you’ve done some algebraic geometry, the formal structure of Lemma 8.1.2 might seem familiar. Given a field πΎ and a natural number π, we can form the set F of subsets of πΎ π and the set G of ideals of πΎ [π‘1 , . . . , π‘ π ], and there are functions F G defined by taking the annihilating ideal of a subset of πΎ π and the zero-set of an ideal of πΎ [π‘1 , . . . , π‘ π ]. The analogue of Lemma 8.1.2 holds. In general, a pair of ordered sets F and G equipped with functions F G satisfying the properties above is called a Galois connection. This in turn is a special case of the category-theoretic notion of adjoint functors. The functions Fo Gal(π:−) Fix / G. are called the Galois correspondence for π : πΎ. This terminology is mostly used in the case where the functions are mutually inverse, meaning that πΏ = Fix(Gal(π : πΏ)), π» = Gal(π : Fix(π»)) for all πΏ ∈ F and π» ∈ G . We saw in Lemma 8.1.2(iii) that in both cases, the left-hand side is a subset of the right-hand side. But they are not always equal: √3 Example 8.1.7 Let π : πΎ be Q( 2) : Q. Since [π : πΎ] is 3, which is a prime number, the tower law implies that there are no nontrivial intermediate fields: F = {π, πΎ }. We saw in Example 6.3.3(ii) that Gal(π : πΎ) is trivial, so G = {Gal(π : πΎ)}. Hence F has two elements and G has only one. This 103 makes it impossible for there to be mutually inverse functions between F and G . Specifically, what goes wrong is that √3 √3 Fix Gal Q 2 : Q = Fix idQ( √3 2) = Q 2 ≠ Q. Exercise 8.1.8 Let π be a prime number, let πΎ = F π (π’), and let π be the splitting field of π‘ π − π’ over πΎ, as in Examples 7.2.4 and 7.2.20(ii). Prove that Gal(π : −) and Fix are not mutually inverse. If Gal(π : −) and Fix are mutually inverse then they set up a one-to-one correspondence between the set F of intermediate fields of π : πΎ and the set G of subgroups of Gal(π : πΎ). The fundamental theorem of Galois theory tells us that this dream comes true when π : πΎ is finite, normal and separable. And it tells us more besides. 8.2 The theorem The moment has come. Theorem 8.2.1 (Fundamental theorem of Galois theory) Let π : πΎ be a finite normal separable extension. Write F = {intermediate fields of π : πΎ }, G = {subgroups of Gal(π : πΎ)}. i. The functions F o Gal(π:−) Fix / G are mutually inverse. ii. | Gal(π : πΏ)| = [π : πΏ] for all πΏ ∈ F , and [π : Fix(π»)] = |π»| for all π» ∈ G . iii. Let πΏ ∈ F . Then πΏ is a normal extension of πΎ ⇐⇒ Gal(π : πΏ) is a normal subgroup of Gal(π : πΎ), and in that case, Gal(π : πΎ) Gal(πΏ : πΎ). Gal(π : πΏ) 104 Proof First note that for each πΏ ∈ F , the extension π : πΏ is finite and normal (by Corollary 7.1.6) and separable (by Lemma 7.2.17). Also, the group Gal(π : πΎ) is finite (by Theorem 7.2.19), so every subgroup is finite too. We prove (i) and (ii) together. First let π» ∈ G . We have |π»| ≤ | Gal(π : Fix(π»))| = [π : Fix(π»)] ≤ |π»|, (8.1) where the first inequality holds because π» ⊆ Gal(π : Fix(π»)) (Lemma 8.1.2(iii)), the equality follows from Theorem 7.2.19 (since π : Fix(π») is finite, normal and separable), and the second inequality follows from Theorem 7.3.6 (since π» is finite). So equality holds throughout (8.1), giving π» = Gal(π : Fix(π»)), [π : Fix(π»)] = |π»|. Now let πΏ ∈ F . We have [π : Fix(Gal(π : πΏ))] = | Gal(π : πΏ)| = [π : πΏ], where the first equality follows from the previous paragraph by taking π» = Gal(π : πΏ), and the second follows from Theorem 7.2.19. But πΏ ⊆ Fix(Gal(π : πΏ)) by Lemma 8.1.2(iii), so πΏ = Fix(Gal(π : πΏ)) by Workshop 3, q. 1. This completes the proof of (i) and (ii). We have already proved most of (iii) as Theorem 7.1.14(ii). It only remains to show that whenever πΏ is an intermediate field such that Gal(π : πΏ) is a normal subgroup of Gal(π : πΎ), then πΏ is a normal extension of πΎ. By Proposition 7.3.11, Fix(Gal(π : πΏ)) : πΎ is normal. But by (i), Fix(Gal(π : πΏ)) = πΏ, so πΏ : πΎ is normal, as required. The fundamental theorem of Galois theory is about field extensions that are finite, normal and separable. Let’s take a moment to think about what those conditions mean. An extension π : πΎ is finite and normal if and only if π is the splitting field of some polynomial over πΎ (Theorem 7.1.5). So, the theorem can be understood as a result about splitting fields of polynomials. Not every splitting field extension is separable (Example 7.2.15(iii)). However, we know of two settings where separability is guaranteed. The first is fields of characteristic zero (Example 7.2.15(i)). The most important of these is Q, which is our focus in this chapter: we’ll consider examples in which π : πΎ is the splitting field extension of a polynomial over Q. The second is where the fields are finite (Example 7.2.15(ii)). We’ll come to finite fields in the final chapter. 105 Digression 8.2.2 Normality and separability are core requirements of Galois theory, but there are extensions of the fundamental theorem (well beyond this course) in which the finiteness condition on π : πΎ is relaxed. The first level of relaxation replaces ‘finite’ by ‘algebraic’. Then Gal(π : πΎ) is no longer a finite group, but it does acquire an interesting topology. One example is where π is the algebraic closure πΎ of πΎ, and Gal(πΎ : πΎ) is called the absolute Galois group of πΎ. It contains all splitting fields of polynomials over πΎ, so to study it is to study all polynomials over πΎ at once. Going further, we can even drop the condition that the extension is algebraic. In this realm, we need the notion of ‘transcendence degree’, which counts how many algebraically independent elements can be found in the extension. You’ll want to see some examples! Section 8.3 is devoted to a single example of the fundamental theorem, showing every aspect of the theorem in all its glory. I’ll give a couple of simpler examples in a moment, but before that, it’s helpful to review some of what we did earlier: Remark 8.2.3 When working out the details of the Galois correspondence for a polynomial π ∈ πΎ [π‘], it’s not only the fundamental theorem that’s useful. Some of our earlier results also come in handy, such as the following. i. Lemma 6.3.7 states that GalπΎ ( π ) acts on the set of roots of π in SFπΎ ( π ), and the injectivity part of Proposition 6.3.10 states that the action is faithful: if π, π ∈ GalπΎ ( π ) and π(πΌ) = π (πΌ) for every root πΌ of π in SFπΎ ( π ), then π = π. In words, an element of the Galois group is entirely determined by what it does to the roots. ii. Corollary 6.3.15 states that | GalπΎ ( π )| divides π!, where π is the number of distinct roots of π in its splitting field. iii. Let πΌ and π½ be roots of π in SFπΎ ( π ). Then there is an element of the Galois group mapping πΌ to π½ if and only if πΌ and π½ are conjugate over πΎ (have the same minimal polynomial). This follows from Proposition 7.1.9. In particular, if π is irreducible then there is always an element of the Galois group that maps πΌ to π½ (Corollary 7.1.10). iv. Let πΏ be an intermediate field of SFπΎ ( π ) : πΎ such that πΏ : πΎ is normal. Then every automorphism of πΏ over πΎ can be extended to an automorphism of SFπΎ ( π ) over πΎ. This was shown in the last paragraph of the proof of Theorem 7.1.14. Examples 8.2.4 i. Let π : πΎ be a normal separable extension of prime degree π. By the fundamental theorem, | Gal(π : πΎ)| = [π : πΎ] = π. 106 Every group of prime order is cyclic, so Gal(π : πΎ) πΆ π . By the tower law, π : πΎ has no nontrivial intermediate fields, and by Lagrange’s theorem, Gal(π : πΎ) has no nontrivial subgroups. So F = {π, πΎ } and G = {1, Gal(π : πΎ)}: π 1 πΎ Gal(π : πΎ) Both π and πΎ are normal extensions of πΎ, and both 1 and Gal(π : πΎ) are normal subgroups of Gal(π : πΎ). √ ii. Let π (π‘) = (π‘ 2 + 1)(π‘ 2 − 2) ∈ Q[π‘]. Put π = SFQ ( π ) = Q( 2, π) and πΊ = Gal(π : πΎ) = GalQ ( π ). Then π : πΎ is a finite normal separable extension, so the fundamental theorem applies. We already calculated πΊ in a sketchy way in Example 6.3.6(ii). Let’s do it again in full, using what we now know. First, √ √ √ [π : πΎ] = Q 2, π : Q 2 Q 2 : Q = 2 × 2 = 4 (much as in Example 5.2.2). √ Now consider √ how √ πΊ acts √ on the set {± 2, ±π} of roots of π . The √ conjugacy √ class of 2 is { 2, − 2}, so for each π ∈ πΊ we have π( 2) = ± 2. Similarly, π(π) = ±π for each π ∈ πΊ. The two choices of sign determine π entirely, so |πΊ | ≤ 4. But by the fundamental theorem, |πΊ | = [π : πΎ] = 4, so each of the four possibilities does in fact occur. So πΊ = {id, π+− , π−+ , π−− }, where √ √ √ √ √ √ π−+ 2 = − 2, π−− 2 = − 2, π+− 2 = 2, π+− (π) = −π, π−+ (π) = π, π−− (π) = −π. The only two groups of order 4 are πΆ4 and πΆ2 × πΆ2 , and each element of πΊ has order 1 or 2, so πΊ πΆ2 × πΆ2 . The subgroups of πΊ are 1 hπ+− i hπ−+ i πΊ 107 hπ−− i (8.2) where lines indicate inclusions. Here hπ+− i is the subgroup generated by π+− , which is {id, π+− }, and similarly for π−+ and π−− . What are the fixed fields of these subgroups? theorem √ √ The fundamental √ implies that Fix(πΊ) = Q. Also, π+− ( 2) = 2, so Q( 2) ⊆ Fix(hπ+− i). But √ √ √ Q 2, π : Q 2 = 2 = |hπ+− i| = Q 2, π : Fix(hπ+− i) √ (where the last step is by the fundamental theorem), so Q( 2) = Fix(hπ+− i). Similar arguments apply to π−+ and π−− , so the fixed fields of the groups in diagram (8.2) are Q Q √ 2 √ 2, π Q(π) Q √ 2π (8.3) Q √ Equivalently, the groups in (8.2) are√the Galois groups of Q( 2, π) over the fields in (8.3). For instance, Gal(Q( 2, π) : Q(π)) = hπ−+ i. Since the overall Galois group πΊ πΆ2 × πΆ2 is abelian, every subgroup is normal. Hence all the extensions in diagram (8.3) are normal too. Exercise 8.2.5 In this particular example, one can also see directly that all the extensions in (8.3) are normal. How? Like any big theorem, the fundamental theorem of Galois theory has some important corollaries. Here’s one. Corollary 8.2.6 Let π : πΎ be a finite normal separable extension. Then for every πΌ ∈ π \ πΎ, there is some automorphism π of π over πΎ such that π(πΌ) ≠ πΌ. Proof Theorem 8.2.1(i) implies that Fix(Gal(π : πΎ)) = πΎ. Now πΌ ∉ πΎ, so πΌ ∉ Fix(Gal(π : πΎ)), which is what had to be proved. Example 8.2.7 For any π ∈ Q[π‘] and irrational πΌ ∈ SFQ ( π ), there is some π ∈ GalQ ( π ) that does not fix πΌ. This is clear if πΌ ∉ R, as we can take π to be complex conjugation restricted to SFQ ( π ). But it is not so obvious otherwise. 108 π ππ −π π π −ππ Figure 8.2: The roots of π , and the effects on them of π, π ∈ GalQ ( π ). 8.3 A specific example Chapter 13 of Stewart’s book opens with these words: The extension that we discuss is a favourite with writers on Galois theory, because of its archetypal quality. A simpler example would be too small to illustrate the theory adequately, and anything more complicated would be unwieldy. The example is the Galois group of the splitting field of π‘ 4 − 2 over Q. We go through the same example here. My presentation of it is different from Stewart’s, so you can consult his book if anything that follows is unclear. Write π (π‘) = π‘ 4 − 2 ∈ Q[π‘], which is irreducible by Eisenstein’s criterion. Write πΊ = GalQ ( π ). Splitting field Write π for the unique real positive root of π . Then the roots of π are ±π and ±ππ (Figure 8.2). So SFQ ( π ) = Q(π, ππ) = Q(π, π). We have [Q(π, π) : Q] = [Q(π, π) : Q(π)] [Q(π) : Q] = 2 × 4 = 8, where the first factor is 2 because Q(π) ⊆ R and the second factor is 4 because π is the minimal polynomial of π over Q (being irreducible) and deg( π ) = 4. By the fundamental theorem, |πΊ | = 8. Galois group We now look for the 8 elements of the Galois group. We’ll use the principle that if π, π ∈ πΊ with π(π) = π (π) and π(π) = π (π) then π = π. To see this, note that π(πΌ) = π (πΌ) whenever πΌ is a root of π , and since the action of πΊ on the roots of π is faithful, π = π. Complex conjugation on C restricts to an automorphism π of Q(π, π) over Q, giving an element π ∈ πΊ of order 2. 109 I now claim that πΊ has an element π satisfying π(π) = ππ and π(π) = π. In that case, π will act on the roots of π as follows: π β¦→ ππ β¦→ −π β¦→ −ππ β¦→ π (Figure 8.2). This element π will have order 4. Proof of claim: since π is irreducible, πΊ acts transitively on the roots of π in SFQ ( π ), so there is some π ∈ πΊ such that π(π) = ππ. The conjugacy class of π over Q is {±π}, so π(π) = ±π. If π(π) = π then we can take π = π. If π(π) = −π then (π β¦ π )(π) = π(π) = ππ, (π β¦ π )(π) = π(−π) = −π(π) = π, so we can take π = π β¦ π . (From now on, I will usually omit the β¦ sign and write things like ππ instead. Of course, juxtaposition is also used to mean multiplication, as in ππ. But confusion shouldn’t arise: automorphisms are composed and numbers are multiplied.) Figure 8.2 suggests that πΊ is the dihedral group π· 4 , the symmetry group of the square. Warning 8.3.1 The symmetry group of a regular π-sided polygon has 2π elements: π rotations and π reflections. Some authors call it π· π and others call it π· 2π . I will call it π· π , as in the Group Theory course. If this is right, we should have π π = π −1 π . (This is one of the defining equations of the dihedral group; you saw it in Example 3.2.12 of Group Theory.) Let’s check this algebraically: π π(π) = ππ = −ππ, π −1 π (π) = π −1 (π) = −ππ, π π(π) = π (π) = −π, π −1 π (π) = π −1 (−π) = −π, so π π and π −1 π are equal on π and π, so π π = π −1 π . It follows that π ππ = π −π π for all π ∈ Z. Figure 8.3 shows the effect of 8 elements of πΊ on π, π and ππ. Since no two of them have the same effect on both π and π, they are all distinct elements of πΊ. Since |πΊ | = 8, they are the only elements of πΊ. So πΊ π· 4 . Warning 8.3.2 The ‘geometric description’ in Figure 8.3 applies only to the roots, not the whole of the splitting field Q(π, π). For example, π 2 is rotation by π on the set of roots, but it is not rotation by π on the rest of Q(π, π): it fixes each element of Q, for instance. 110 π∈πΊ π(π) π(π) π(ππ) order id π π2 3 π = π −1 π π π = π −1 π π π 2 = π 2 π π π −1 = ππ π ππ −π −ππ π −ππ −π ππ π π π π −π −π −π −π ππ −π −ππ π −ππ −π ππ π 1 4 2 4 2 2 2 2 geometric description (see Warning 8.3.2) identity rotation by π/2 rotation by π rotation by −π/2 reflection in real axis reflection in axis through 1 − π reflection in imaginary axis reflection in axis through 1 + π Figure 8.3: The Galois group of π‘ 4 − 2 over Q. Subgroups of the Galois group Since |πΊ | = 8, any nontrivial proper subgroup of πΊ has order 2 or 4. Let’s look in turn at subgroups of order 2 and 4, also determining which ones are normal. This is pure group theory, with no mention of fields. • The subgroups of order 2 are of the form hπi = {id, π} where π ∈ πΊ has order 2. So, they are hπ 2 i, hπ i, hπ πi, hπ π 2 i, hπ π −1 i. If you watched the video ‘What does it mean to be normal?’, you may be able to guess which of these subgroups are normal in πΊ, the symmetry group of the square. It should be those that can be specified without referring to particular vertices or edges of the square. So, just the first should be normal. Let’s check. We know that π π 2 = π 2 π , so π 2 commutes with both π and π, which generate πΊ. Hence π 2 is in the centre of πΊ (commutes with everything in πΊ). It follows that hπ 2 i is a normal subgroup of πΊ. On the other hand, for each π ∈ Z, the subgroup hπ ππ i is not normal, since π(π ππ ) π −1 = (ππ ) ππ−1 = (π π −1 ) ππ−1 = π ππ−2 ∉ hπ ππ i. • The subgroups of πΊ of order 4 are isomorphic to either πΆ4 or πΆ2 × πΆ2 , since these are the only groups of order 4. The only elements of πΊ of order 4 are π ±1 , so the only subgroup of πΊ isomorphic to πΆ4 is hπi = {id, π, π 2 , π 3 = π −1 }. Now consider subgroups π» of πΊ isomorphic to πΆ2 × πΆ2 . 111 Exercise 8.3.3 Show that every such π» must contain π 2 . (Hint: think geometrically.) We have π 2 ∈ π», and both other nonidentity elements of π» have order 2, so they are of the form π ππ for some π ∈ Z. The two such subgroups π» are hπ , π 2 i = {id, π , π 2 , π π 2 }, hπ π, π 2 i = {id, π π, π 2 , π π −1 }. Finally, any subgroup of index 2 of any group is normal, so all the subgroups of πΊ of order 4 are normal. Hence the subgroup structure of πΊ π· 4 is as follows, where a box around a subgroup means that it is normal in πΊ. order 1 1 hπ i hπ , π 2 i πΆ2 × πΆ2 hπ π 2 i hπ 2 i hπ πi hπi πΆ4 πΊ = hπ , πi π· 4 hπ π −1 i order 2 hπ π, π 2 i πΆ2 × πΆ2 order 4 order 8 Fixed fields We now find Fix(π») for each π» ∈ G , again considering the subgroups of orders 2 and 4 in turn. • Order 2: take Fixhπ i (officially Fix(hπ i), but let’s drop the brackets). We have π (π) = π, so π ∈ Fixhπ i, so Q(π) ⊆ Fixhπ i. But [Q(π, π) : Q(π)] = 2, and by the fundamental theorem, [Q(π, π) : Fixhπ i] = |hπ i| = 2, so Fixhπ i = Q(π). Finding fixed fields The same argument shows that for any π ∈ πΊ of order 2, if we can spot some πΌ ∈ Q(π, π) such that π(πΌ) = πΌ and [Q(π, π) : Q(πΌ)] ≤ 2, then Fixhπi = Q(πΌ). For π = π π 2 , we can take πΌ = ππ (by Figure 8.3). We have degQ (ππ) = 4 since ππ is a root of π , so [Q(ππ) : Q] = 4, or equivalently, [Q(π, π) : Q(ππ)] = 2. Hence Fixhπ π 2 i = Q(ππ). 112 Exercise 8.3.4 I took a small liberty in the sentence beginning ‘The same argument’, because it included an inequality but the previous argument didn’t. Prove the statement made in that sentence. It is maybe not so easy to spot an πΌ for π π, but the geometric description in Figure 8.3 suggests taking πΌ = π (1 − π). And indeed, one can check that π π fixes π (1 − π). One can also check that π (1 − π) is not the root of any nonzero quadratic over Q, so degQ (π (1 − π)) is ≥ 4 (since it divides 8), so [Q(π, π) : Q(π (1 − π))] ≤ 8/4 = 2. Hence Fixhπ πi = Q(π (1 − π)). Similarly, Fixhπ π −1 i = Q(π (1 + π)). Finally, π 2 (π 2 ) = (π 2 (π)) 2 = (−π) 2 = π 2 , π 2 (π) = π, so Q(π 2 , π) ⊆ Fixhπ 2 i. But [Q(π, π) : Q(π 2 , π)] = 2, so Fixhπ 2 i = Q(π 2 , π). • Order 4: for π» = hπ , π 2 i, note that π 2 is fixed by both π and π 2 , so π 2 ∈ Fix(π»), so Q(π 2 ) ⊆ Fix(π»). But π 2 ∉ Q, so [Q(π 2 ) : Q] ≥ 2, so [Q(π, π) : Q(π 2 )] ≤ 4. The fundamental theorem guarantees that [Q(π, π) : Fix(π»)] = |π»| = 4, so Fix(π») = Q(π 2 ). The same argument applies to the other two subgroups π» of order 4: if we can spot an element πΌ ∈ Q(π, π) \ Q fixed by the generators of π», then Fix(π») = Q(πΌ). This gives Fixhπi = Q(π) and Fixhπ π, π 2 i = Q(π 2π). In summary, the fixed fields of the subgroups of πΊ are as follows. degree 1 Q(π, π) Q(π) Q(π 2 ) Q(ππ) Q(π 2 , π) Q(π (1 − π)) Q(π) Q(π (1 + π)) degree 2 Q(π 2π) degree 4 degree 8 Q In the right-hand column, ‘degree’ means the degree of Q(π, π) over the subfield concerned. The fundamental theorem implies that the Galois group of Q(π, π) over each intermediate field is the corresponding subgroup of πΊ in the earlier diagram. For example, Gal(Q(π, π) : Q(π 2 , π)) = hπ 2 i. The fundamental theorem also implies that the intermediate fields that are normal over Q are the boxed ones. 113 Quotients Finally, the fundamental theorem tells us that Gal(Q(π, π) : Q) Gal(πΏ : Q) Gal(Q(π, π) : πΏ) whenever πΏ is an intermediate field normal over Q. For πΏ = Q(π 2 , π), this gives πΊ/hπ 2 i Gal(Q(π 2 , π) : Q). Normal subgroups and normal extensions (8.4) The left-hand side is the quotient of π· 4 by a subgroup isomorphic to πΆ2 . It has order 4, but it has no element of order 4: for the only elements of πΊ of order 4 are π ±1 , whose images in πΊ/hπ 2 i have order 2. Hence πΊ/hπ 2 i πΆ2 × πΆ2 . On the other hand, Q(π 2 , π) is the splitting field over Q of (π‘ 2 − 2)(π‘ 2 + 1), which by Example 8.2.4(ii) has Galois group πΆ2 × πΆ2 . This confirms the isomorphism (8.4). The other three intermediate fields normal over Q, I leave to you: Exercise 8.3.5 Choose one of Q(π 2 ), Q(π) or Q(π 2π), and do the same for it as I just did for Q(π 2 , π). As you’ve now seen, it can take quite some time to work through a particular example of the Galois correspondence. You’ll get practice at doing this in workshop questions. Beyond examples, there are at least two things we can do with the fundamental theorem of Galois theory. The first is to resolve the old question on solvability of polynomials by radicals, which we met back in Chapter 1. The second is to work out the structure of finite fields. We will carry out these two missions in the remaining two weeks. 114 Chapter 9 Solvability by radicals Introduction to Week 9 We began this course with a notorious old problem: can every polynomial be solved by radicals? Theorem 1.3.5 gave the answer and more: not only is it impossible to find a general formula that does it, but we can tell which specific polynomials can be solved by radicals. Theorem 1.3.5 states that a polynomial over Q is solvable by radicals if and only if it has the right kind of Galois group—a solvable one. In degree 5 and higher, there are polynomials that have the wrong kind of group. These polynomials are not, therefore, solvable by radicals. We’ll prove one half of this ‘if and only if’ statement: if π is solvable by radicals then GalQ ( π ) is solvable. This is the half that’s needed to show that some polynomials are not solvable by radicals. The proof of the other direction is in Chapter 18 of Stewart’s book, but we won’t do it. If you’re taking Algebraic Topology, you’ll already be familiar with the idea that groups can be used to solve problems that seem to have nothing to do with groups. You have a problem about some objects (such as topological spaces or field extensions), you associate groups with those objects (maybe their fundamental groups or their Galois groups), you translate your original problem into a problem about groups, and you solve that instead. For example, the question of whether R2 and R3 are homeomorphic is quite difficult using only general topology; but using algebraic topology, we can answer ‘no’ by noticing that the fundamental group of R2 with a point removed is not isomorphic to the fundamental group of R3 with a point removed. In much the same way, we’ll answer a difficult question about field extensions by converting it into a question about groups. For this chapter, you’ll need to remember something about solvable groups. At a minimum, you’ll need the definition, the fact that any quotient of a solvable group is solvable, and the fact that π5 is not solvable. 115 9.1 Radicals We speak of square roots, cube roots, and so on, but we also speak about roots of polynomials. To distinguish between these two related usages, we will use the word ‘radical’ for square roots etc. (Radical comes from the Latin word for root. A radish is a root, and a change or policy is radical if it gets right down to the roots of the matter.) Back in Chapter 1, I said that a complex number is called radical if ‘it can be obtained from the rationals using only the usual arithmetic operations [addition, subtraction, multiplication and division] and πth roots [for π ≥ 1]’. As an example, I said that p3 √ √2 7 1 + 2− 7 2 (9.1) r q 4 5 2 6+ 3 is radical, whichever square root, cube root, etc., we choose (p. 12). Let’s now make this definition precise. √ The first point is that the notation π π§ or π§1/π is highly dangerous: Warning 9.1.1 Let π§ be a complex number and π ≥ 2. Then there √ is no single number called π π§ or π§ 1/π . There are π elements πΌ of C √ such that πΌπ = π§. So, the notation π π§ or π§1/π makes no sense if it is intended to denote a single complex number. It is simply invalid. When π§ belongs to the set R+ of nonnegative reals, the convention is √ that π π§ or π§ 1/π denotes the unique πΌ ∈ R+ such that πΌπ = π§. There is also a widespread convention that when π§ is a negative real and π is √ odd, π π§ or π§1/π denotes the unique real πΌ such that πΌπ = π§. In these cases, there is a sensible and systematic way of choosing one of the πth roots of π§. But for a general π§ and π, there is not. Complex analysis has a lot to say about different choices of πth roots. But we don’t need to go into that. We simply treat all the πth roots of π§ on an equal footing, not attempting to pick out any of them as special. With this warning in mind, we define the radical numbers without using nota√ tion like π π§ or π§1/π . The definition of radical number Definition 9.1.2 Let Qrad be the smallest subfield of C such that for πΌ ∈ C, πΌπ ∈ Qrad for some π ≥ 1 ⇒ πΌ ∈ Qrad . A complex number is radical if it belongs to Qrad . 116 (9.2) So any rational number is radical; the sum, product, difference or quotient of radical numbers is radical; any πth root of a radical number is radical; and there are no more radical numbers than can be obtained by those rules. For the definition of Qrad to make sense, we need there to be a smallest subfield of C with the property (9.2). This will be true as long as the intersection of any family of subfields of C satisfying (9.2) is again a subfield of C satisfying (9.2): for then Qrad is the intersection of all subfields of C satisfying (9.2). Exercise 9.1.3 Check that the intersection of any family of subfields of C satisfying (9.2) is again a subfield of C satisfying (9.2). (That any intersection of subfields is a subfield is a fact we met back on p. 20; the new aspect is (9.2).) Example 9.1.4 Consider again the expression (9.1). It’s not quite as random as it looks. I chose it so that the various radicals are covered by one of the two conventionspmentioned in Warning 9.1.1: they’re all πth roots of positive reals √2 3√ 7 2 − 7, which is an odd root of a negative real. Let π§ be the except for number (9.1), choosing the radicals according to those conventions. I claim that π§ is radical, or equivalently that π§ belongs to every subfield πΎ of C satisfying (9.2). p5 2/3 ∈ πΎ First, Q ⊆ πΎ since Q is the prime subfield of C. So 2/3 ∈ πΎ, and so p5 by (9.2). Also, 6 ∈ πΎ and πΎ is a field, so 6 + 2/3 ∈ πΎ. But then by (9.2) again, the denominator of (9.1) is in πΎ. A similar argument shows that the numerator is in πΎ. Hence π§ ∈ πΎ. Definition 9.1.5 A polynomial over Q is solvable by radicals if all of its complex roots are radical. The simplest nontrivial example of a polynomial solvable by radicals is something of the form π‘ π − π, where π ∈ Q. The theorem we’re heading for is that any polynomial solvable by radicals has solvable Galois group, and if that’s true then the group GalQ (π‘ π − π) must be solvable. Let’s consider that group now. The results we prove about it will form part of the proof of the big theorem. We begin with the case π = 1. Lemma 9.1.6 For all π ≥ 1, the group GalQ (π‘ π − 1) is abelian. Proof Write π = π 2ππ/π . The complex roots of π‘ π − 1 are 1, π, . . . , ππ−1 , so SFQ (π‘ π − 1) = Q(π). Let π, π ∈ GalQ (π‘ π − 1). Since π permutes the roots of π‘ π − 1, we have π(π) = ππ for some π ∈ Z. Similarly, π (π) = π π for some π ∈ Z. Hence (π β¦ π)(π) = π(π π ) = π(π) π = ππ π , 117 and similarly (π β¦ π)(π) = ππ π . So (π β¦ π)(π) = (π β¦ π)(π). Since SFQ (π‘ π − 1) = Q(π), it follows that π β¦ π = π β¦ π. Exercise 9.1.7 In the last sentence of that proof, how exactly does it ‘follow’? Much more can be said about the Galois group of π‘ π − 1, and you’ll see a bit more at Workshop 5. But this is all we need for our purposes. Now that we’ve considered π‘ π − 1, let’s do π‘ π − π for an arbitrary π. Lemma 9.1.8 Let πΎ be a field and π ≥ 1. Suppose that π‘ π − 1 splits in πΎ. Then GalπΎ (π‘ π − π) is abelian for all π ∈ πΎ. The hypothesis that π‘ π − 1 splits in πΎ might seem so restrictive as to make this lemma useless. For instance, it doesn’t hold in Q or even R (for π > 2). Nevertheless, this turns out to be the key lemma in the whole story of solvability by radicals. Proof If π = 0 then GalπΎ (π‘ π − π) is trivial; suppose otherwise. Choose a root π of π‘ π − π in SFπΎ (π‘ π − π). For any other root π, we have (π/π) π = π/π = 1 (valid since π ≠ 0), and π‘ π − 1 splits in πΎ, so π/π ∈ πΎ. It follows that SFπΎ (π‘ π − π) = πΎ (π). Moreover, given π, π ∈ GalπΎ (π‘ π − π), we have π(π)/π ∈ πΎ (since π(π) is a root of π‘ π − π), so π(π) π(π)π (π) π(π) ·π = · π (π) = . (π β¦ π)(π) = π π π π Similarly, (πβ¦π)(π) = π(π)π (π)/π, so (π β¦π)(π) = (πβ¦π)(π). Since SFπΎ (π‘ π −π) = πΎ (π), it follows that π β¦ π = π β¦ π. Warning 9.1.9 For π ∈ Q, the Galois group of π‘ π − π over Q is not usually abelian. For instance, you saw on Assignment 4 that GalQ (π‘ 3 − 2) is the nonabelian group π3 . Exercise 9.1.10 What does the proof of Lemma 9.1.8 tell you about the eigenvectors and eigenvalues of the elements of GalπΎ (π‘ π − π)? Exercise 9.1.11 Use Lemmas 9.1.6 and 9.1.8 to show that GalQ (π‘ π −π) is solvable for all π ∈ Q. This is harder than most of these exercises, but I recommend it as a way of getting into the right frame of mind for the theory that’s coming in Section 9.2. 118 Digression 9.1.12 We’re only going to do the theory of solvability by radicals over Q. It can be done over any field, but Q has two special features. First, Q can be embedded into an algebraically closed field that we know very well, namely, C. This makes some things easier. Second, char Q = 0. For fields of characteristic π, the proof that any polynomial with a solvable Galois group is solvable by radicals has some extra complications. 9.2 Solvable polynomials have solvable groups Here we’ll prove that every polynomial over Q that is solvable by radicals has solvable Galois group. You know by now that in Galois theory, we tend not to jump straight from polynomials to groups. We go via the intermediate stage of field extensions, as in the diagram polynomial β¦−→ field extension β¦−→ group that I first drew after the definition of GalπΎ ( π ) (page 77). That is, we understand polynomials through their splitting field extensions. So it shouldn’t be a surprise that we do the same here, defining a notion of ‘solvable extension’ and showing (roughly speaking) that solvable polynomial β¦−→ solvable extension β¦−→ solvable group. Solvable polynomials have solvable groups: a map In other words, we’ll define ‘solvable extension’ in such a way that (i) if π ∈ Q[π‘] is a polynomial solvable by radicals then SFQ ( π ) : Q is a solvable extension, and (ii) if π : πΎ is a solvable extension then Gal(π : πΎ) is a solvable group. Hence if π is solvable by radicals then GalQ ( π ) is solvable—the result we’re aiming for. (I glossed over some details in that paragraph; we’ll get to those.) Definition 9.2.1 Let π : πΎ be a finite normal separable extension. Then π : πΎ is solvable (or π is solvable over πΎ) if there exist π ≥ 0 and intermediate fields πΎ = πΏ 0 ⊆ πΏ 1 ⊆ · · · ⊆ πΏπ = π such that πΏ π : πΏ π−1 is normal and Gal(πΏ π : πΏ π−1 ) is abelian for each π ∈ {1, . . . , π }. Exercise 9.2.2 Let π : π : πΎ be extensions, with π : π, π : πΎ and π : πΎ all finite, normal and separable. Show that if π : π and π : πΎ are solvable then so is π : πΎ. We will focus on subfields of C, where separability is automatic (Example 7.2.15(i)). 119 Example 9.2.3 Let π ∈ Q and π ≥ 1. Then SFQ (π‘ π − π) : Q is a finite normal separable extension, being a splitting field extension over Q. I claim that it is solvable. Proof: if π = 0 then SFQ (π‘ π − π) = Q, and Q : Q is solvable (taking π = 0 and πΏ 0 = Q in Definition 9.2.1). Now assume that π ≠ 0. Choose a complex root π of π‘ π − π and write π = π 2ππ/π . Then the complex roots of π‘ π − π are π, ππ, . . . , ππ−1 π. So SFQ (π‘ π − π) contains (ππ π)/π = ππ for all π, and so π‘ π − 1 splits in SFQ (π‘ π − π). Hence Q ⊆ SFQ (π‘ π − 1) ⊆ SFQ (π‘ π − π). Now SFQ (π‘ π − 1) : Q is normal (being a splitting field extension) and has abelian Galois group by Lemma 9.1.6. Also SFQ (π‘ π − π) : SFQ (π‘ π −1) is normal (being the splitting field extension of π‘ π − π over SFQ (π‘ π − 1), by Lemma 6.2.13(ii)), and has abelian Galois group by Lemma 9.1.8. So SFQ (π‘ π − π) : Q is a solvable extension, as claimed. The definition of solvable extension bears a striking resemblance to the definition of solvable group. Indeed: Lemma 9.2.4 Let π : πΎ be a finite normal separable extension. Then π : πΎ is solvable ⇐⇒ Gal(π : πΎ) is solvable. Proof We will only need the ⇒ direction, and that is all I prove here. For the converse, see Workshop 5. Suppose that π : πΎ is solvable. Take intermediate fields πΎ = πΏ 0 ⊆ πΏ 1 ⊆ · · · ⊆ πΏπ = π as in Definition 9.2.1. For each π ∈ {1, . . . , π }, the extension π : πΏ π−1 is finite, normal and separable (by Corollary 7.1.6 and Lemma 7.2.17), so we can apply the fundamental theorem of Galois theory to it. Since πΏ π : πΏ π−1 is a normal extension, Gal(π : πΏ π ) is a normal subgroup of Gal(π : πΏ π−1 ) and Gal(π : πΏ π−1 ) Gal(πΏ π : πΏ π−1 ). Gal(π : πΏ π ) By hypothesis, the right-hand side is abelian, so the left-hand side is too. So the sequence of subgroups Gal(π : πΎ) = Gal(π : πΏ 0 ) ⊇ Gal(π : πΏ 1 ) ⊇ · · · ⊇ Gal(π : πΏ π ) = 1 exhibits Gal(π : πΎ) as a solvable group. 120 Exercise 9.2.5 Prove the ⇐ direction of Lemma 9.2.4. It’s a very similar argument to the proof of ⇒. According to the story I’m telling, solvability by radicals of a polynomial should correspond to solvability of its splitting field extension. Thus, the subfields of C that are solvable over Q should be exactly the splitting fields SFQ ( π ) of polynomials π that are solvable by radicals. (This is indeed true, though we won’t entirely prove it.) Now if π , π ∈ Q[π‘] are both solvable by radicals then so is π π, and SFQ ( π π) is a solvable extension of Q containing both SFQ ( π ) and SFQ (π). So it should be the case that for any two subfields of C solvable over Q, there is some larger subfield, also solvable over Q, containing both. We now prove this. Lemma 9.2.6 Let πΏ and π be subfields of C such that the extensions πΏ : Q and π : Q are finite, normal and solvable. Then there is some subfield π of C such that πΏ ∪ π ⊆ π and π : Q is also finite, normal and solvable. Proof Take subfields Q = πΏ 0 ⊆ · · · ⊆ πΏ π = πΏ, Q = π0 ⊆ · · · ⊆ ππ = π such that πΏ π : πΏ π−1 is normal with abelian Galois group for each π, and similarly for π π . There is a chain of subfields Q = πΏ 0 ⊆ · · · ⊆ πΏ π = πΏ = π0 (πΏ) ⊆ · · · ⊆ ππ (πΏ) = π (πΏ) (9.3) of C, where π π (πΏ) is the subfield of C generated by π π ∪ πΏ (an instance of the notation πΎ (π ) of Definition 4.1.10). Put π = π (πΏ). Certainly πΏ ∪ π ⊆ π. We show that π : Q is finite, normal and solvable. Since πΏ : Q is finite and normal, πΏ = SFQ ( π ) for some π ∈ Q[π‘], and similarly, π = SFQ (π). Now π is the subfield of C generated by πΏ ∪ π, so it is generated by the roots of π and the roots of π, or equivalently the roots of π π. So π = SFQ ( π π), which is finite and normal over Q. To see that π : Q is solvable, we show that each successive extension in (9.3) is normal with abelian Galois group. For those to the left of πΏ, this is immediate. For those to the right, let π ∈ {1, . . . , π }. Since π π : π π−1 is finite and normal, π π = SF π π−1 (β) for some β ∈ π π−1 [π‘]. Then π π (πΏ) = SF π π−1 (πΏ) (β) by Lemma 6.2.13(i), so π π (πΏ) : π π−1 (πΏ) is normal. Its Galois group is Gal π π−1 (πΏ) (β), which by Corollary 6.3.13 is isomorphic to a subgroup of Gal π π−1 (β). But Gal π π−1 (β) is Gal(π π : π π−1 ), which is abelian, so Gal(π π (πΏ) : π π−1 (πΏ)) is abelian. 121 Digression 9.2.7 This proof, and some others in this section, can be uncluttered slightly using the notion of compositum. By definition, the compositum of subfields πΏ, π of a field πΉ is the smallest subfield containing πΏ and π. For example, the π of the proof is the compositum of πΏ and π. In our usual notation, the compositum of πΏ and π is π (πΏ) or πΏ(π). The standard notation for the compositum is πΏ π, which has the advantage of being symmetric and the disadvantage of being misleading: it is not the set {πΌπ½ : πΌ ∈ πΏ, π½ ∈ π } (although it is the smallest subfield of πΉ containing that set). The various fields used in the proof of Lemma 9.2.6, together with the fields πΏ π (π) = πΏ π π, can be drawn like this: πΏπ πΏπ π . πΏ ππ .. .. . πΏ π0 πΏ0 π π πΏ ππ .. .. . πΏπ . πΏ0 π0 Q I have chosen not to use the compositum explicitly in this course, but once you know about it, you’ll notice how often it appears implicitly in Galois theory proofs. The heart of the proof that solvable polynomials have solvable Galois groups is the following lemma (which in turn depends on Lemmas 9.1.6 and 9.1.8 on the Galois groups of π‘ π − 1 and π‘ π − π). Loosely, it says that the set of complex numbers that can be reached from Q by solvable extensions is closed under taking πth roots. Write Qsol = {πΌ ∈ C : πΌ ∈ πΏ for some subfield πΏ ⊆ C that is finite, normal and solvable over Q}. In fact, Qsol = Qrad , but we don’t know that yet. Lemma 9.2.8 Let πΌ ∈ C and π ≥ 1. If πΌπ ∈ Qsol then πΌ ∈ Qsol . 122 The proof (below) is slightly subtle. Here’s why. Let πΏ be a subfield of C that’s finite, normal and solvable over Q, and take πΌ ∈ C and π ≥ 1 such that πΌπ ∈ πΏ. To find some larger π that contains πΌ itself and is also solvable over Q, we could try putting π = SF πΏ (π‘ π − πΌπ ). Now π : Q is indeed solvable (as can be shown using Exercises 9.1.11 and 9.2.2), but the trouble is that π : Q is not in general normal. And normality is part of the definition of Qsol , ultimately because it’s an essential requirement if we want to use the fundamental theorem of Galois theory. An example should clarify. √ √ √4 2 = Example 9.2.9 Put πΌ = 2 and take π = 2. We have πΌ 2 ∈ Q( 2), and √ Q( 2) : Q is finite, normal and solvable (since its Galois group is the abelian √4 2 sol group πΆ2 ), so πΌ ∈ Q . Hence, according to Lemma 9.2.8, πΌ = 2 should be contained in some finite normal solvable extension π of Q. √ How can we find such an π? We can’t take π = SFQ(√2) (π‘ 2 − 2), since this √4 is Q( 2), which √4 is not√normal over Q. (You may have already contemplated the extensions Q( 2) : Q( 2) : Q in Workshop 4, q. 4.) To find a bigger π, still finite √ and solvable over Q but also normal, we have √ to adjoin a square root not just of 2 but also of its conjugate, − 2. This is the crucial point: the whole idea of normality is that conjugates are treated equally. (Normal behaviour means that anything you do for one element, you do for all √4 4 its conjugates.) The result is Q( 2, π) = SFQ (π‘ √4 − 2), which is indeed a finite, solvable and normal extension of Q containing 2. Proof of Lemma 9.2.8 Write π = πΌπ ∈ Qsol . Choose a subfield πΎ of C such that π ∈ πΎ and πΎ : Q is finite, normal and solvable. Step 1: enlarge πΎ to a field in which π‘ π − 1 splits. Put πΏ = SFπΎ (π‘ π − 1) ⊆ C. Since πΎ : Q is finite and normal, πΎ = SFQ ( π ) for some π ∈ Q[π‘], and then πΏ = SFQ (π‘ π − 1) π (π‘) . Hence πΏ : Q is finite and normal. It follows from Corollary 7.1.6 that πΏ : πΎ is normal. Its Galois group is GalπΎ (π‘ π − 1), which is isomorphic to a subgroup of GalQ (π‘ π − 1) (by Corollary 6.3.13), which is abelian (by Lemma 9.1.6). Hence Gal(πΏ : πΎ) is abelian. Also πΎ : Q is solvable, so πΏ : Q is solvable. In summary, πΏ is a subfield of C such that π ∈ πΏ and πΏ : Q is finite, normal and solvable, and, moreover, π‘ π − 1 splits in πΏ. We now forget about πΎ. Step 2: adjoin the πth roots of the conjugates of π. Write π ∈ Q[π‘] for the minimal polynomial of π over Q, and put π = SF πΏ (π(π‘ π )) ⊆ C. Then πΌ ∈ π, as π(πΌπ ) = π(π) = 0. We show that π : Q is finite, normal and solvable. 123 Since πΏ : Q is finite and normal, πΏ = SFQ (π) for some π ∈ Q[π‘]. Then π = SFQ (π(π‘)π(π‘ π )), so π : Q is finite and normal. It follows that π : πΏ is finite and normal too (by Corollary 7.1.6). To show that π : Q is solvable, it is enough to show that π : πΏ is solvable (by Exercise 9.2.2). Since πΏ : Q is normal and π ∈ Q[π‘] is the minimal polynomial of π ∈ πΏ, it follows by definition of normality that π splits in πΏ, say π Ö (π‘ − ππ ) π(π‘) = π=1 (ππ ∈ πΏ). Define subfields πΏ 0 ⊆ · · · ⊆ πΏ π of C by πΏ0 = πΏ πΏ 1 = SF πΏ 0 (π‘ π − π 1 ) πΏ 2 = SF πΏ 1 (π‘ π − π 2 ) .. . πΏ π = SF πΏ π −1 (π‘ π − π π ). Then πΏ π = πΏ π½ ∈ π : π½π ∈ {π 1 , . . . , ππ } . In particular, πΏ π = π. For each π ∈ {1, . . . , π }, the extension πΏ π : πΏ π−1 is finite and normal (being a splitting field extension), and its Galois group is abelian (by Lemma 9.1.8 and the fact that π‘ π − 1 splits in πΏ ⊆ πΏ π−1 ). So π : πΏ is solvable. Now we can relate the set Qrad of radical numbers, defined in terms of basic arithmetic operations, to the set Qsol , defined in terms of field extensions. Proposition 9.2.10 Qrad ⊆ Qsol . That is, every radical number is contained in some subfield of C that is a finite, normal, solvable extension of Q. As I mentioned, Qrad and Qsol are actually equal, although we won’t prove this. Proof By definition of Qrad , it is enough to show that Qsol is a subfield of C with the property that πΌπ ∈ Qsol ⇒ πΌ ∈ Qsol . We have just proved that property (Lemma 9.2.8), so it only remains to show that Qsol is a subfield of C. The argument is similar to the proof that the algebraic numbers form a field (Proposition 5.3.7). Let πΌ, π½ ∈ Qsol . Then πΌ ∈ πΏ and π½ ∈ π for some πΏ, π that are finite, normal and solvable over Q. By Lemma 9.2.6, πΌ, π½ ∈ π for some π that is finite, normal and solvable over Q. Then πΌ+ π½ ∈ π, so πΌ+ π½ ∈ Qsol , and similarly πΌ · π½ ∈ Qsol . This shows that Qsol is closed under addition and multiplication. The other parts of the proof (negatives, reciprocals, 0 and 1) are straightforward. This brings us to the main result of this chapter. Notice that it doesn’t mention field extensions: it goes straight from polynomials to groups. 124 Theorem 9.2.11 Let π ∈ Q[π‘]. If the polynomial π is solvable by radicals then the group GalQ ( π ) is solvable. Proof Suppose that π is solvable by radicals. Write πΌ1 , . . . , πΌπ ∈ C for its roots. For each π, we have πΌπ ∈ Qrad (by definition of solvability by radicals), hence πΌπ ∈ Qsol (by Proposition 9.2.10). So each of πΌ1 , . . . , πΌπ is contained in some subfield of C that is finite, normal and solvable over Q. By Lemma 9.2.6, there is some subfield π of C that is finite, normal and solvable over Q and contains all of πΌ1 , . . . , πΌπ . Thus, SFQ ( π ) = Q(πΌ1 , . . . , πΌπ ) ⊆ π. By Lemma 9.2.4, Gal(π : Q) is solvable. Now SFQ ( π ) : Q is normal, so by the fundamental theorem of Galois theory, Gal(SFQ ( π ) : Q) is a quotient of Gal(π : Q). But Gal(SFQ ( π ) : Q) = GalQ ( π ), and a quotient of a solvable group is solvable, so GalQ ( π ) is solvable. Examples 9.2.12 i. For π ∈ Q and π ≥ 1, the polynomial π‘ π − π is solvable by radicals, so the group GalQ (π‘ π − π) is solvable. You may already have proved this in Exercise 9.1.11. It also follows from Example 9.2.3 and Lemma 9.2.4. ii. Let π 1 , . . . , π π ∈ Q and π1 , . . . , π π ≥ 1. Each of the polynomials π‘ ππ −ππ is Î solvable by radicals, so their product is too. Hence GalQ π (π‘ ππ − ππ ) is a solvable group. Theorem 9.2.11 is most sensational in its contrapositive form: if GalQ ( π ) is not solvable then π is not solvable by radicals. That’s the subject of the next section. Digression 9.2.13 The converse of Theorem 9.2.11 is also true: if GalQ ( π ) is solvable then π is solvable by radicals. You can even unwind the proof to obtain an explicit formula for the solving the quartic by radicals (Stewart, Chapter 18). To prove this converse statement, we have to deduce properties of a field extension from assumptions about its Galois group. A solvable group is built up from abelian groups, and every finite abelian group is a direct sum of cyclic groups. The key step in the proof of the converse has come to be known as ‘Hilbert’s Theorem 90’ (Stewart’s Theorem 18.18), which gives information about any field extension whose Galois group is cyclic. Digression 9.2.14 The proof of Theorem 9.2.11 might not have ended quite how you expected. Given my explanations earlier in the chapter, you might justifiably have imagined we were going to show that when the polynomial π is solvable by radicals, the extension SFQ ( π ) : Q is solvable. That’s not 125 what we did. We showed that SFQ ( π ) is contained in some larger subfield π such that π : Q is solvable, then used that to prove the solvability of the group GalQ ( π ). But all is right with the world: SFQ ( π ) : Q is a solvable extension. Indeed, its Galois group GalQ ( π ) is solvable, so Lemma 9.2.4 implies that SFQ ( π ) : Q is solvable too. 9.3 An unsolvable polynomial Here we give a specific example of a polynomial over Q that is not solvable by radicals. By Theorem 9.2.11, our task is to construct a polynomial whose Galois group is not solvable. The smallest non-solvable group is π΄5 (of order 60). Our polynomial has Galois group π5 (of order 120), which is also non-solvable. Finding Galois groups is hard, and we will use a whole box of tools and tricks, from Cauchy’s theorem on groups to Rolle’s theorem on differentiable functions. First we establish a useful fact on the order of Galois groups. Lemma 9.3.1 Let π be an irreducible polynomial over a field πΎ, with SFπΎ ( π ) : πΎ separable. Then deg( π ) divides | GalπΎ ( π )|. Proof Let πΌ be a root of π in SFπΎ ( π ). By irreducibility, deg( π ) = [πΎ (πΌ) : πΎ], which divides [SFπΎ ( π ) : πΎ] by the tower law, which is equal to | GalπΎ ( π )| by Theorem 7.2.19 (using separability). Next, we need some results about the symmetric group ππ . I assume you know that ππ is generated by the ‘adjacent transpositions’ (12), (23), . . . , (π − 1 π). This may have been proved in Fundamentals of Pure Mathematics, and as the Group Theory notes say (p. 58): This is intuitively clear: suppose you have π people lined up and you want them to switch into a different order. To put them in the order you want them, it’s clearly enough to have people move up and down the line; and each time a person moves one place, they switch places with the person next to them. Here’s a different way of generating ππ . Lemma 9.3.2 For π ≥ 2, the symmetric group ππ is generated by (12) and (12 · · · π). 126 Proof We have (12 · · · π)(12)(12 · · · π) −1 = (23), either by direct calculation or the general fact that π(π 1 · · · π π )π −1 = (π(π 1 ) · · · π(π π )) for any π ∈ ππ and cycle (π 1 · · · π π ). So any subgroup π» of ππ containing (12) and (12 · · · π) also contains (23). By the same argument, π» also contains (34), . . . , (π − 1 π). But the adjacent transpositions generate ππ , so π» = ππ . Lemma 9.3.3 Let π be a prime number, and let π ∈ Q[π‘] be an irreducible polynomial of degree π with exactly π − 2 real roots. Then GalQ ( π ) π π . Proof Since char Q = 0 and π is irreducible, π is separable and therefore has π distinct roots in C. By Proposition 6.3.10, the action of GalQ ( π ) on the roots of π in C defines an isomorphism between GalQ ( π ) and a subgroup π» of π π . Since π is irreducible, π divides | GalQ ( π )| = |π»| (by Lemma 9.3.1). So by Cauchy’s theorem, π» has an element π of order π. Then π is a π-cycle, since these are the only elements of π π of order π. The complex conjugate of any root of π is also a root of π , so complex conjugation restricts to an automorphism of SFQ ( π ) over Q. Exactly two of the roots of π are non-real; complex conjugation transposes them and fixes the rest. So π» contains a transposition π. Without loss of generality, π = (12). Since π is a π-cycle, ππ (1) = 2 for some π ∈ {1, . . . , π − 1}. Since π is prime, ππ also has order π, so it is a π-cycle. Now without loss of generality, ππ = (123 · · · π). So (12), (12 · · · π) ∈ π», forcing π» = π π by Lemma 9.3.2. Hence GalQ ( π ) π π . Exercise 9.3.4 Explain why, in the last paragraph, ππ has order π. Theorem 9.3.5 Not every polynomial over Q of degree 5 is solvable by radicals. Proof We show that π (π‘) = π‘ 5 − 6π‘ + 3 satisfies the conditions of Lemma 9.3.3. Then GalQ ( π ) is π5 , which is not solvable, so by Theorem 9.2.11, π is not solvable by radicals. Evidently deg( π ) is the prime number 5, and π is irreducible by Eisenstein’s criterion with prime 3. It remains to prove that π has exactly 3 real roots. This is where we use some analysis, considering π as a function R → R (Figure 9.1). We have lim π (π₯) = −∞, π (0) > 0, π₯→−∞ π (1) < 0, lim π (π₯) = ∞, π₯→∞ 127 Figure 9.1: The function π₯ β¦→ π₯ 5 − 6π₯ + 3. and π is continuous on R, so by the intermediate value theorem, π haspat least 3 real roots. On the other hand, π 0 (π₯) = 5π₯ 4 − 6 has only 2 real roots (± 4 6/5), so by Rolle’s theorem, π has at most 3 real roots. Hence π has exactly 3 real roots, as required. Exercise 9.3.6 Prove that for every π ≥ 5, there is some polynomial of degree π that is not solvable by radicals. Digression 9.3.7 We now know that some polynomials π over Q are not solvable by radicals, which means that not all their complex roots are radical. Could it be that some of the roots are radical and others are not? Yes: simply take a polynomial π that is not solvable by radicals and put π (π‘) = π‘π(π‘). Then the roots of π are 0 (which is radical) together with the roots of π (which are not all radical). But what if π is irreducible? In that case, either all the roots of π are radical or none of them are. This follows from the fact that the extension Qrad : Q is normal, which we will not prove. Digression 9.3.8 There are many similarities between the theory of constructibility of points by ruler and compass and the theory of solvability of polynomials by radicals. In both cases, the challenge is to construct some things (points in the plane or roots of polynomials) using only certain tools (ruler and compass or a machine for taking πth roots). In both cases, there were difficult questions of constructibility that remained open for a very long time, and in both cases, they were solved by Galois theory. The solutions have something in common too. For the geometry problem, we used iterated quadratic extensions, and for the polynomial problem, we used solvable extensions, which could reasonably be called iterated abelian extensions. For the geometry problem, we showed that the coordinates of any 128 point constructible by ruler and compass satisfy a certain condition on their degree over Q (Corollary 5.4.3); for the polynomial problem, we showed that any polynomial solvable by radicals satisfies a certain condition on its Galois group. There are other similarities: compare Lemmas 5.4.4 and 9.2.6, for example, and maybe you can find more similarities still. We have now used the fundamental theorem of Galois theory to solve a major problem about Q. What else can we do with it? The fundamental theorem is about separable extensions. Our two main sources of separable extensions are: • fields of characteristic 0 such as Q (Example 7.2.15(i)), which we’ve explored extensively already; • finite fields (Example 7.2.15(ii)), which we’ve barely touched. In the next and final chapter, we’ll use the fundamental theorem and other results we’ve proved to explore the world of finite fields. In contrast to the complicated world of finite groups, finite fields are almost shockingly simple. 129 Chapter 10 Finite fields This chapter is dessert. Through this semester, we’ve developed a lot of sophisticated theory for general fields. All of it works for finite fields, but becomes much simpler there. It’s a miniature world in which life is sweet. For example: Introduction to Week 10 • If we want to apply the fundamental theorem of Galois theory to a field extension π : πΎ, we first have to ask whether it is finite, and whether it is normal, and whether it is separable. When π and πΎ are finite, all three conditions are automatic (Lemma 10.4.2). • There are many fields of different kinds, and to classify them all would be a near-impossible task. But for finite fields, the classificiation is very simple, as we’ll see. We know exactly what finite fields there are and how many elements they have. • The Galois correspondence for arbitrary field extensions can also be complicated. But again, it’s simple for finite fields. Their Galois groups are very easy (they’re all cyclic), we know what their subgroups are, and it’s easy to describe all the subfields of any finite field. So although the world of finite fields is not trivial, there’s a lot about it that’s surprisingly straightforward. Two aspects of finite fields may seem counterintuitive. First, they always have positive characteristic, which means they satisfy some equation like 1 + · · · + 1 = 0. Second (and relatedly), πth powers and πth roots behave strangely in fields of characteristic π—at least, ‘strange’ if, like most of us, characteristic 0 is what you’re most familiar with. But the behaviour of πth roots and powers is fundamental to all the nice properties of finite fields. 130 10.1 πth roots in characteristic π Recall from Lemma 2.2.11 that every finite field has positive characteristic, which, by Lemma 2.2.5, must be a prime number π. Square roots usually √come in pairs: how many times in your life have you written a ± sign before a ? But in characteristic 2, plus and minus are the same, so the two square roots become one. We’ll see that this pattern persists. Proposition 10.1.1 Let π be a prime number and π a ring of characteristic π. i. The function π → π π β¦→ π π π: is a homomorphism. ii. If π is a field then π is injective. iii. If π is a finite field then π is an automorphism of π . Proof For (i), certainly π preserves multiplication and 1. To show that π preserves addition, let π, π ∈ π : then by Lemma 3.3.15 and the hypothesis that char π = π, π π (π + π ) = (π + π ) = π Õ π π π π π−π = π π + π π = π (π) + π (π ). π π=0 Now (ii) follows since every homomorphism between fields is injective, and (iii) since every injection from a finite set to itself is bijective. The homomorphism π : π β¦→ π π is called the Frobenius map, or, in the case of finite fields, the Frobenius automorphism. That π is a homomorphism is a shocker. Writing ‘(π₯ + π¦) π = π₯ π + π¦ π ’ is a classic mistake in school-level algebra. But here, it’s true! Examples 10.1.2 i. The Frobenius automorphism of F π = Z/hπi is not very interesting. When πΊ is a finite group of order π, Lagrange’s theorem implies that π π = 1 for all π ∈ πΊ. Applying this to the multiplicative group F×π = F π \ {0} gives π π−1 = 1 whenever 0 ≠ π ∈ F π . It follows that π π = π for all π ∈ F π . That is, π is the identity. Everything is its own πth root! ii. The proof that the Frobenius map preserves addition may seem familiar. We essentially did it in Example 7.2.4, in the case of the ring (F π (π’)) [π‘] of polynomials over the field F π (π’) of rational expressions over F π . 131 iii. We don’t have many examples of finite fields yet (but we will soon). Apart from those of the form F π , the simplest is F2 (πΌ), where πΌ is a root of the irreducible polynomial 1 + π‘ + π‘ 2 over F2 (Example 4.3.8(ii)). By Theorem 5.1.5, F2 (πΌ) = {π + ππΌ : π, π ∈ F2 } = {0, 1, πΌ, 1 + πΌ}. Since 1 + πΌ + πΌ2 = 0 and F2 (πΌ) has characteristic 2, πΌ2 = 1 + πΌ, (1 + πΌ) 2 = πΌ. So the Frobenius automorphism of F2 (πΌ) interchanges πΌ and 1 + πΌ. Like all automorphisms, it fixes 0 and 1. Exercise 10.1.3 Write out the addition and multiplication tables of F2 (πΌ). Corollary 10.1.4 Let π be a prime number. i. In a field of characteristic π, every element has at most one πth root. ii. In a finite field of characteristic π, every element has exactly one πth root. Proof Part (i) says that the Frobenius map is injective, and part (ii) says that it is bijective, as Proposition 10.1.1 states. Examples 10.1.5 i. In a field of characteristic 2, every element has at most one square root. ii. In C, there are π different πth roots of unity. But in a field of characteristic π, there is only one: 1 itself. iii. Let πΎ be a field of characteristic π and π ∈ πΎ. Corollary 10.1.4(i) says that π has at most one πth root. It may have none. For instance, Exercise 7.2.5 asked you to show that in F π (π’), the element π’ has no πth root. iv. In the 4-element field F2 (πΌ) of Example 10.1.2(iii), the only square root of πΌ is 1 + πΌ, and the only square root of 1 + πΌ is πΌ. Each is the square root of the other. Exercise 10.1.6√Work out the values of the Frobenius automorphism on the field F3 ( 2), which you first met in Exercise 4.3.9. 132 10.2 Classification of finite fields If you try to write down a formula for the number of groups or rings with a given number of elements, you’ll find that it’s hard and the results are quite strange. For instance, more than 99% of the first 50 billion groups have order 1024. But fields turn out to be much, much easier. We’ll obtain a complete classification of finite fields in the next two pages. The order of a finite field π is its cardinality, or number of elements, |π |. Warning 10.2.1 Order and degree mean different things. For instance, if the order of a field is 9, then its degree over its prime subfield F3 is 2. Lemma 10.2.2 Let π be a finite field. Then char π is a prime number π, and |π | = π π where π = [π : F π ] ≥ 1. In particular, the order of a finite field is a prime power. Proof By Lemmas 2.2.5 and 2.2.11, char π is a prime number π. By Lemma 2.2.10, π has prime subfield F π . Since π is finite, 1 ≤ [π : F π ] < ∞; write π = [π : F π ]. As a vector space over F π , then, π is π-dimensional and so isomorphic to Fππ . But |Fππ | = |F π | π = π π , so |π | = π π . Example 10.2.3 There is no field of order 6, since 6 is not a prime power. Lemma 10.2.2 prompts two questions: given a prime power π π , is there some field of order π π ? And if so, how many are there? The answer to the first question is yes: Lemma 10.2.4 Let π be a prime number and π ≥ 1. Then the splitting field of π π‘ π − π‘ over F π has order π π . π Proof Put π (π‘) = π‘ π − π‘ ∈ F π [π‘] and π = SFF π ( π ). Then π· π = −1 (since π ≥ 1), so by (i)⇒(ii) of Lemma 7.2.10, π has no repeated roots in π. Hence π has at least π π elements. Write π for the Frobenius map of π. The set πΏ of roots of π in π is the set of fixed points of π π = π β¦ · · · β¦ π, which is Eq{π π , id π }. Since π is a homomorphism, πΏ is a subfield of π (by Lemma 7.3.2), and contains all the roots of π in π. Hence by definition of splitting field, πΏ = π; that is, every element of π is a root of π . Since deg( π ) = π π , this implies that π has at most π π elements. As for the second question, there is exactly one field of each prime power order. To show this, we need a lemma. 133 Lemma 10.2.5 Let π be a finite field of order π. Then πΌ π = πΌ for all πΌ ∈ π. The proof uses the same argument as in Example 10.1.2(i). Proof The multiplicative group π × has order π−1, so Lagrange’s theorem implies that πΌ π−1 = 1 for all πΌ ∈ π × = π \ {0}. Hence πΌ π = πΌ whenever 0 ≠ πΌ ∈ π, and clearly the equation holds for πΌ = 0 too. Exercise 10.2.6 Verify that π½4 = π½ for all π½ in the 4-element field F2 (πΌ) of Examples 10.1.2(iii) and 10.1.5(iv), Lemma 10.2.7 Every finite field of order π is a splitting field of π‘ π − π‘ over F π . Proof Let π be a field of order π. By Lemma 10.2.2, π = π π for some prime π and π ≥ 1, and char π = π. Hence π has prime subfield F π . By Lemma 10.2.5, π every element of π is a root of π (π‘) = π‘ π − π‘. So π has |π | = π π = deg( π ) distinct roots in π, and therefore splits in π. The set of roots of π in π generates π, since it is equal to π. Hence π is a splitting field of π . Together, these results completely classify the finite fields. Theorem 10.2.8 (Classification of finite fields) i. Every finite field has π order π for some prime π and integer π ≥ 1. ii. For each prime π and integer π ≥ 1, there is exactly one field of order π π , up to isomorphism. It has characteristic π and is a splitting field for π π‘ π − π‘ over F π . Proof This is immediate from the results above together with the uniqueness of splitting fields (Theorem 6.2.12(ii)). When π > 1 is a prime power, we write Fπ for the one and only field of order π. Warning 10.2.9 Fπ is not Z/hπi unless π is a prime. It can’t be, because Z/hπi is not a field (Example 2.2.15). To my knowledge, there is no description of Fπ simpler than the splitting field description. We now know exactly how many finite fields there are of each order. But in algebra, it’s important to think not just about the objects (such as vector spaces, groups, modules, rings, fields, . . . ), but also the maps (homomorphisms) between objects. So now that we’ve counted the finite fields, it’s natural to try to count the homomorphisms between finite fields. Field homomorphisms are injective, so this boils down to counting subfields and automorphisms. Galois theory is very well equipped to do that! We’ll come to this in the final section. But first, we consider another way in which finite fields are very simple. 134 10.3 Multiplicative structure The multiplicative group πΎ × of a finite field πΎ is as easy as can be: Proposition 10.3.1 For an arbitrary field πΎ, every finite subgroup of πΎ × is cyclic. In particular, if πΎ is finite then πΎ × is cyclic. The multiplicative group of a finite field is cyclic Proof This was Theorem 5.1.13 and Corollary 5.1.14 of Group Theory. Example 10.3.2 In examples earlier in the course, we frequently used the πth root of unity π = π 2ππ/π ∈ C, which has the property that every other πth root of unity is a power of π. Can we find an analogue of π in an arbitrary field πΎ? It’s not obvious how to generalize the formula π 2ππ/π , since the exponential is a concept from complex analysis. But Proposition 10.3.1 solves our problem. For π ≥ 1, put ππ (πΎ) = {πΌ ∈ πΎ : πΌπ = 1}. Then ππ (πΎ) is a subgroup of πΎ × , and is finite since its elements are roots of π‘ π − 1. So by Proposition 10.3.1, ππ (πΎ) is cyclic. Let π be a generator of ππ (πΎ). Then every πth root of unity in πΎ is a power of π, which is what we were aiming for. Note, however, that ππ (πΎ) may have fewer than π elements, or equivalently, the order of π may be less than π. For instance, if char πΎ = π then π π (πΎ) is trivial and π = 1, by Example 10.1.5(ii). Exercise 10.3.3 Let πΎ be a field and let π» be a finite subgroup of πΎ × of order π. Prove that π» ⊆ ππ (πΎ). Example 10.3.4 The group F×π is cyclic, for any prime π. This means that there is some π ∈ {1, . . . , π − 1} such that π, π2 , . . . runs through all elements of {1, . . . , π − 1} when taken mod π. In number theory, such an π is called a primitive root mod π (another usage of the word ‘primitive’). For instance, you can check that 3 is a primitive root mod 7, but 2 is not, since 23 ≡ 1 (mod 7). Corollary 10.3.5 Every extension of one finite field over another is simple. Proof Let π : πΎ be an extension with π finite. By Proposition 10.3.1, the group π × is generated by some element πΌ ∈ π × . Then π = πΎ (πΌ). This is yet another pleasant aspect of finite fields. 135 Exercise 10.3.6 In the proof of Corollary 10.3.5, once we know that the group π × is generated by πΌ, how does it follow that π = πΎ (πΌ)? Digression 10.3.7 In Digression 7.2.22, I mentioned the theorem of the primitive element: every finite separable extension π : πΎ is simple. One of the standard proofs involves splitting into two cases, according to whether π is finite or infinite. We’ve just done the finite case. Corollary 10.3.8 For every prime number π and integer π ≥ 1, there exists an irreducible polynomial over F π of degree π. Proof The field F π π has prime subfield F π . By Corollary 10.3.5, the extension F π π : F π is simple, say F π π = F π (πΌ). The minimal polynomial of πΌ over F π is irreducible of degree [F π (πΌ) : F π ] = [F π π : F π ] = π. This is not obvious. For example, can you find an irreducible polynomial of degree 100 over F31 ? 10.4 Galois groups for finite fields We now work out the Galois correspondence for any extension of one finite field over another. Warning 10.4.1 The term ‘finite field extension’ means an extension π : πΎ that’s finite in the sense defined on p. 52: π is finitedimensional as a vector space over πΎ. It doesn’t mean that π and πΎ are finite fields. But the safest policy is to avoid this term entirely. The three hypotheses of the fundamental theorem of Galois theory are always satisfied when both fields in the extension are finite: Lemma 10.4.2 Let π : πΎ be a field extension. i. If πΎ is finite then π : πΎ is separable. ii. If π is also finite then π : πΎ is finite and normal. Proof For (i), we show that every irreducible polynomial π over πΎ is separable. Write char πΎ = π > 0, and suppose for a contradiction that π is inseparable. By Corollary 7.2.12, π (π‘) = π 0 + π 1 π‘ π + · · · + ππ π‘ π π 136 for some π 0 , . . . , ππ ∈ πΎ. For each π, there is a (unique) πth root ππ of ππ in πΎ, by Corollary 10.1.4(ii). Then π π π π (π‘) = π 0 + π 1 π‘ π + · · · + ππ π‘ π π . But by Proposition 10.1.1(i), the function π β¦→ π π is a homomorphism πΎ [π‘] → πΎ [π‘], so π (π‘) = (π 0 + π 1 π‘ + · · · + ππ π‘ π ) π . This contradicts π being irreducible. For (ii), suppose that π is finite. Write char π = π > 0. By Theorem 10.2.8, π is a splitting field over F π , so by Lemma 6.2.13(ii), it is also a splitting field over πΎ. Hence π : πΎ is finite and normal, by Theorem 7.1.5. Part (i) fulfils the promise made in Remark 7.2.13 and Example 7.2.15(ii), and the lemma as a whole lets us use the fundamental theorem freely in the world of finite fields. We now work out the Galois correspondence for the extension F π π : F π of an arbitrary finite field over its prime subfield. Proposition 10.4.3 Let π be a prime and π ≥ 1. Then Gal(F π π : F π ) is cyclic of order π, generated by the Frobenius automorphism of F π π . By Workshop 4, q. 7, Gal(F π π : F π ) is the group of all automorphisms of F π π . Proof Write π for the Frobenius automorphism of F π π ; then π ∈ Gal(F π π : F π ). π First we calculate the order of π. By Lemma 10.2.5, πΌ π = πΌ for all πΌ ∈ F π π , or π equivalently, π π = id. If π is a positive integer such that π π = id then πΌ π = πΌ for π all πΌ ∈ F π π , so the polynomial π‘ π − π‘ has π π roots in F π π , so π π ≤ π π , so π ≤ π. Hence π has order π. On the other hand, [F π π : F π ] = π, so by the fundamental theorem of Galois theory, | Gal(F π π : F π )| = π. The result follows. Exercise 10.4.4 What is the fixed field of hπi ⊆ Gal(F π π : F π )? In Fundamentals of Pure Mathematics or Group Theory, you presumably saw that the cyclic group of order π has exactly one subgroup of order π for each divisor π of π. (And by Lagrange’s theorem, there are no subgroups of other orders.) Exercise 10.4.5 Refresh your memory by proving this fact about subgroups of cyclic groups. In the case at hand, Gal(F π π : F π ) = hπi πΆπ , and when π | π, the unique subgroup of order π is hπ π/π i. 137 Proposition 10.4.6 Let π be a prime and π ≥ 1. Then F π π has exactly one subfield of order π π for each divisor π of π, and no others. It is π πΌ ∈ F ππ : πΌ π = πΌ . Proof The subfields of F π π are the intermediate fields of F π π : F π , which by the fundamental theorem of Galois theory are precisely the fixed fields Fix(π») of subgroups π» of Gal(F π π : F π ). Any such π» is of the form hπ π/π i with π | π, and π/π Fixhπ π/π i = πΌ ∈ F π π : πΌ π = πΌ . The tower law and the fundamental theorem give [Fixhπ π/π i : F π ] = [F π π : F π ] [F π π : Fixhπ π/π i] = π |hπ π/π i| = π , π so | Fixhπ π/π i| = π π/π . As π runs through the divisors of π, the quotient π/π also runs through the divisors of π, so putting π = π/π gives the result. Warning 10.4.7 The subfields of F π π are of the form F π π where π divides π, not π ≤ π. For instance, F8 has no subfield isomorphic to F4 (that is, no 4-element subfield), since 8 = 23 , 4 = 22 , and 2 - 3. Let π be a divisor of π. By Proposition 10.4.6, F π π has exactly one subfield isomorphic to F π π . We can therefore speak of the extension F π π : F π π without ambiguity. Since F π π = Fixhπ π i and hπ π i πΆπ/π , it follows from the fundamental theorem that Gal(F π π : F π π ) πΆπ/π . (10.1) So in working out the Galois correspondence for F π π : F π , we have accidentally derived the Galois group of a completely arbitrary extension of finite fields. In the Galois correspondence for F π π : F π , all the extensions and subgroups involved are normal, either by Lemma 10.4.2 or because cyclic groups are abelian. For π | π, the isomorphism Gal(F π π : F π ) Gal(F π π : F π ) Gal(F π π : F π π ) supplied by the fundamental theorem amounts to πΆπ πΆπ . πΆπ/π Alternatively, substituting π = π/π, this is πΆπ /πΆ π πΆπ/π . 138 Example 10.4.8 Consider the Galois correspondence for F π12 : F π , where π is any prime. Writing π for the Frobenius automorphism of F π12 , the subgroups of πΊ = Gal(F π12 : F π ) are hπ 12 i πΆ1 1 order 1 hπ 6 i πΆ2 hπ 4 i order 2 order 3 πΆ3 hπ 3 i πΆ4 hπ 2 i order 4 order 6 πΆ6 πΊ = hπi πΆ12 order 12 Their fixed fields are degree 1 F π12 degree 2 F π6 degree 3 F π4 F π3 degree 4 degree 6 F π2 degree 12 Fπ Here, ‘degree’ means the degree of F π12 over the subfield, and (for instance) the subfield of F π12 called F π4 is 4 πΌ ∈ F π12 : πΌ π = πΌ F π4 . The Galois group Gal(F π12 : F π4 ) is hπ 4 i πΆ3 , and similarly for the other subfields. Exercise 10.4.9 What do the diagrams of Example 10.4.8 look like for π 8 in place of π 12 ? What about π 432 ? (Be systematic!) Ordered sets In Workshop 5, you’ll be asked to work through the Galois correspondence for an arbitrary extension F π π : F π π of finite fields, but there’s not much more to do: almost all the work is contained in the case π = 1 that we have just done. ∗ ∗ 139 ∗