# MATH 538 Class Notes 1 Week One Ed Belk ```MATH 538 Class Notes
Ed Belk
Winter, 2015
1
1.1
Week One
Lecture One
Fundamental Results
Recall that a ring element p ∈ R is said to be prime if (p) = pR is a prime ideal. In an integral domain, every prime
element is also irreducible, meaning its only factorizations in R are as the product of a unit and another element. That
is,√if p ∈ R is prime and p = ab, then one of a, b is a unit. The converse statement is not, in general, true: for example in
Z[ −5], 3 is irreducible, but not prime, since
√
√
√
√
3|(2 + −5)(2 − −5) but 3 - (2 + −5), 3 - (2 − −5).
A ring is said to be a unique factorization domain, usually abbreviated UFD, if every element can be uniquely factored
as a product of irreducible elements, up to associates. For example, in Z we have
−6 = (−1)(2)(3)
= (−1)(−2)(−3)
Any two factorizations differ by at most a unit in each factor.
Definition: A principal ideal domain, usually abbreviated PID, is an integral domain in which every ideal is principal,
i.e. is generated by a single element.
We may restate the fundamental theorem of arithmetic as follows: every nonzero ideal of Z is the product of prime ideals
uniquely. For example, (6) = (2)(3). R is said to be a Euclidean domain if it is an integral domain which may be
endowed with at least one Euclidean function; that is, a map f : R \ {0} → N ∪ {0} such that, for any a, b ∈ R, b 6= 0, one
can write b = aq + r with r = 0 or f (r) &lt; f (b).
Every√ Euclidean domain is a PID, and every PID is a UFD, but the converse inclusions do not hold. For example,
Z[ 1+ 2−19 ] is a PID which is not Euclidean, and K[X, Y ] (with K a field) is a UFD which is not principal, as is Z[X].
Consider, for example, the ideal a = (X, 2).
One of our goals as number theorists is to find a description of GQ := Gal(Q/Q). Recall that R is the ring of all elements
which are algebraic over R, and that Gal(L/K) for a field extension L/K is the group of automorphisms of L fixing K.
For example, R = C and
Gal(C/R) = {id,&macr;&middot;} ∼
= Z/2Z,
where &macr;&middot; is the complex conjugation automorphism. The Galois group of the infinite extension Fp /Fp is the topological
group Ẑ = lim← Z/nZ, the inverse limit taken over the direct system {Z/nZ : n ∈ N} with the canonical quotient maps
m mod n 7→ m mod n0 for n0 |n.
Cauchy’s theorem: If a ∈ Z with (a, p) = 1, then ap−1 ≡ 1 mod p.
Wilson’s theorem: If p ∈ Z is prime, then (p − 1)! ≡ 1 mod p.
Lemma 1.1.1 If p &gt; 2 is prime, then p can be written as the sum of two squares if and only if p ≡ 1 mod 4.
1
Proof : Necessity is clear by simply noting that any square is congruent to 0 or 1 modulo 4. For sufficiency, we invoke
Wilson’s theorem: then p−1
2 is even, and so modulo 4 we have
−1 ≡ (p − 1)! = 1 &middot; 2 &middot; &middot; &middot;
p−1
2
p+1
2
&middot; &middot; &middot; (p − 1) ≡ (1)(2) &middot; &middot; &middot;
p−1
2
(−1)(−2) &middot; &middot; &middot; − p−1
2
2
≡ (−1)(p−1)/2 (1 &middot; 2 &middot; &middot; &middot; ( p−1
2 )) .
Thus −1 is a square modulo p and we may write a2 ≡ −1 mod p. Thus p|(a2 + 1) = (a + i)(a − i). If p is prime in Z[i],
then p|(a + i) or p|(a − i). If the former case, then
a + i = p(α + iβ) = pα + i(pβ) =⇒ pβ = 1,
which is not the case or (p) = (1) and p is not prime; the case p|(a − i) is similarly disposed of. Thus pZ[i] is not prime.
Define a map
N : Z[i] −→ N ∪ {0}
a + bi 7→ a2 + b2 .
This map is multiplicative and is called the norm function of Z[i]. This defines a Euclidean function on Z[i]. Note
moreover that N (a + bi) = 1 if and only if a + bi = &plusmn;1 or &plusmn;i. Since uv = 1 implies
N (uv) = N (u)N (v) = N (1) = 1
we have that the units of Z[i] are exactly {&plusmn;1, &plusmn;i}.
Writing p = αβ ∈ Z[i] where α and β are not units, we have N (p) = N (α)N (β) = p2 , and since by assumption α and β
are not units we must have N (α) = N (β) = p. With α = x + iy, we have N (α) = p = x2 + y 2 , and we are done.
If K/Q is an algebraic number field, then [K : Q] &lt; ∞.
Definition: Let R be a ring and let A ⊆ R be a subring. Then a ∈ R is said to be integral over A if there is some monic
f (X) ∈ A[X] such that f (a) = 0.
Example: Since i ∈ C is a root of the monic polynomial X 2 + 1 ∈ R[X], we have that i is integral over R.
If a ∈ Q is integral over Z, we say that a is a rational integer – indeed, every rational integer is an element of Z. Any
nth root of unity ζ is integral over Z, while 2ζn is algebraic, but not integral, over Z.
We now prove the claim that N is a Euclidean function. Write x = a+bi, x̄ = a−bi. Then N (x) = xx̄ = a2 +b2 . We extend
the domain of N to Q[i] = Q(i) = QuotZ[i] in the natural way. Let x, y ∈ Z[i] with x 6= 0 and write xy = t + γ ∈ Q(i),
where N (γ) &lt; 1. This can always be done, since xy is at most √12 from the nearest lattice point in the complex plane.
Then
y = xt + γx = xt + γ̃ with N (γ̃) = N (γ)N (x) &lt; N (x),
and we are done.
2
1.2
Lecture Two
Recall our discussion of the Euclidean domain Z[i]. To understand Z[i] as a UFD, we must first understand the units, as
well as the prime elements.
Lemma 1.2.1 The group of units in Z[i], denoted Z[i]&times; , is the group {&plusmn;1, &plusmn;i}, i.e. the fourth roots of unity.
Proof : See Lecture One.
&times;
Definition: Two elements α, β ∈ R are said to be associated, denoted α ∼ β, if there exists a unit u ∈ R with α = βu.
Note that ∼ is an equivalence relation.
Theorem 1.2.2 The set of prime elements of Z[i] consists in all associated elements of the disjoint union of:
1. The prime π = 1 + i,
2. Elements of the form π = a + bi with N (p) a rational prime and a &gt; |b| &gt; 0, and
3. Rational primes p ∈ Z such that p ≡ 3 mod 4.
Recall from last lecture that if N (π) = p ≡ 3 mod 4 with p a rational prime, then π is not prime in Z[i].
Proof : Cases (1) and (2) are necessarily prime since their norms are prime, and π = ab ⇒ N (π) = p = N (a)N (b), so
one of a, b ∈ Z[i]&times; . In case (3): suppose π = p ≡ 3 mod 4 is a rational prime, and that p = αβ ∈ Z[i], α, β ∈
/ Z[i]&times; . Then
2
2
2
N (p) = p = N (α)N (β) =⇒ N (α) = N (β) = p. Thus with α = x + iy we have p = x + y , contradicting lemma
1.1.1.
Now, let π ∈ Z[i] be any prime element and write N (π) = p1 p2 &middot; &middot; &middot; pn = ππ̄, pi all prime and not necessarily distinct. In
Z[i] we have that π|p1 p2 &middot; &middot; &middot; pn and so since π is prime we must have thta π|pj for some j. Then
N (π)|N (p) = p2 =⇒ N (π) = p or p2 .
In the first case we must have (1) or (2), and in the second case π is associated to p since πp ∈ Z[i] has norm 1, and we
have (3). The prime in this case cannot be 2 as then it is not prime, and similarly p 6≡ 1 mod 4.
Remark: This theorem completely characterizes the factorization of rational primes in Z[i]: we have 2 = (1 + i)(1 − i) =
−i(1 + i)2 , for p ≡ 1 mod 4 we have p = (a + bi)(a − bi) with a2 + b2 = 1, and for p ≡ 3 mod 4 we have p = p.
Remark: There are two kinds of fields whose behaviour are very similar and are usually studied together in algebraic
number theory, namely
1. Global fields of positive characteristic, i.e. finite extensions of Fq [T ] with q = pn , and
2. Global fields of characteristic zero, or algebraic number fields, i.e. finite extensions of Q.
Recall from last lecture the notion of algebraic integers.
Definition: If A ⊆ B is an extension of rings, then B is said to be an integral extension of A if every element of B is
integral over A.
Theorem 1.2.3 Let A ⊆ B be an extension of rings and let x ∈ B. The following are equivalent:
1. x is integral over A.
2. The A-module A[x] is finitely generated.
3. x belongs to a subring R ⊆ B such that A ⊆ R and R is finitely generated as an A-module.
4. There is a subring R ⊆ B such that R is a finitely generated A-module and x stabilizes R, i.e. xR ⊆ R.
5. There is a faithful A[x]-module R which is finitely generated as an A-module.
Recall that an A-module M is faithful if the annihilator ideal annA M = {a ∈ A : am = 0 for all m ∈ M } is trivial,
annA M = {0}.
Proof : We prove only equivalence of the first three statements.
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(1) =⇒ (2) By assumption x is the root of a monic polynomial f (T ) =
−an−1 xn−1 − &middot; &middot; &middot; − a1 x − a0 .
P
ai T i ∈ A[T ] of degree n (say), so xn =
(2) =⇒ (3) Take R = A[x].
(3) =⇒ (1) Suppose x ∈ R ⊆ B where R is a subring of B that is finitely generated over A. Let α1 , α2 , . . . , αn generate
R as an A-module, so that each xαi is a linear combination of the αj , say
xαi =
n
X
cij αj , with cij ∈ A.
j=1
Let α = (α1 , α2 , . . . , αn )t and a = In the identity matrix, with C = (cij ). We then have the matrix equation
(xI − C)α = 0,
so with M = xI − C we have that det M = 0. The determinant of M is a monic polynomial in A[T ] which therefore has
x as a root, and the claim is proven.
Corollary 1: If A ⊆ B is an extension of rings, then the set of elements of B that are integral over A forms a subring of
B, denoted Ā.
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2
Week Two
2.1
Lecture Three
One of the most important open questions in algebraic number theory is the determination of the absolute Galois group
GQ = Gal(Q/Q). To contrast, finite extensions of Q are comparatively well-understood.
If we complete Q with respect to the usual absolute value | &middot; | we obtain R, and Q ,→ R densely. Alternatively, we can
define the p-adic absolute value on Q and complete Q with respect to | &middot; |p . It is a theorem due to Ostrowski that these
are the only nontrivial absolute values on Q, and the completion of Q with respect to | &middot; |p is denoted Qp , and again Q
injects into Qp densely. We call these Qp the p-adic local fields.
If we consider Qp and its Galois group Gp = Gal(Qp /Q), we get some “pieces of the puzzle” that is GQ . Denote by Gab
Q the
ab
abelianization of GQ , that is, GQ /[GQ , GQ ]. The fixed field corresponding to Gab
is
denoted
Q
,
the
maximal
abelian
Q
extension of Q. We now know Gal(Qab /Q), and it is also known that Gp ⊆ GQ . We analogously define the extensions
Qp ⊂ Qab
p ⊂ Qp .
Now, let G be any group. A representation of G is a group homomorphism ρ : G → Aut(V ) = GL(V ) = GLd (F ),
where V is some (finite-dimensional) vector space over a field F of dimension d. By understanding representations of G,
we may understand quotients of G.
Recall that in our last lecture we saw that Z[i] is a Euclidean domain, hence a PID, hence a UFD. The first false proof
of Fermat’s last theorem (Lamé, 1847) arose from a misunderstanding of the distiction between the latter two types of
rings.
We also saw algebraic number fields, which are a special case of global fields, namely, finite extensions of Q (of characteristic zero) and finite extensions of Fp (X), the field of rational functions over Fp (of positive characteristic). By the
correspondence Z/pZ ↔ Fp , we may understand extensions k 0 of Fp [T ] by the corresponding extensions k of Z.
Now, let K/Q be any finite extension. Let OK denote the ring of integers of K, that is, the set of all elements of
K which are integral over Z. We saw last time that this does indeed form a ring. We can ask what happens to prime
elements p ∈ Z when we pass into OK – this study is sometimes called reciprocity.
Recall that if R is a commutative ring and p ⊂ R is a prime ideal, then Sp = R \ p is a multiplicatively closed set. Denote
by Rp = Sp−1 R the localization of R at p; recall that this consists of all ordered pairs (r, s) where r ∈ R, s ∈ Sp , with
the identification
(r, s) ∼ (r0 , s0 ) ⇐⇒ rs0 − r0 s is a zero divisor of R.
There is a natural homomorphism R → Rp , which is an inclusion if R is a domain. The operations are defined:
• (r, s) + (r0 , s0 ) = (rs0 + r0 s, ss0 )
• (r, s) &middot; (r0 , s0 ) = (rr0 , ss0 )
It is left as an exercise to verify that this endows Rp with the structure of a ring.
The localization satisfies a universal property: if f : R → R̃ is a ring homomorphism such that the image of every element
of Sp under f is a unit in R̃, then f uniquely factors through Rp , i.e. there exists a unique f˜ : Rp → R̃ such that the
following diagramme commutes:
Rp
R
f˜
f
R̃
Now, if q ⊂ R is a prime ideal such that Sp ∩ q 6= ∅, then s ∈ Sp ∩ q is a unit in Rp and so (q) = qRp = Rp . Thus the only
prime ideals which remain prime are those q with q ∩ Sp = ∅.
Definition: A commutative ring R is called local if one of the following equivalent conditions holds:
• R has a unique maximal ideal.
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• The set R \ R&times; of nonunits in R forms an ideal.
Facts:
1. If p ⊂ R is prime, then Rp is local with maximal ideal pRp .
2. If R is a domain, then Rp is a domain.
Given such R and p, we can naturally contsruct two rings: one is the quotient R/p, in which the prime ideals are those of R
which contain p, and the other is the localization Rp in which the prime ideals are those of R which are contained in p. We
may also talk of the fraction field Quot(R/p) and the residue field Rp /pRp . These fields are naturally isomorphic:
Sp−1 R
Rp
= −1 = Sp−1 (R/p).
pRp
Sp p
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2.2
Lecture Four
Recall that a Pythagorean triple (x, y, z) is a triple of integers such that x2 + y 2 = z 2 . If in addition x, y, and z
have no common factor, we call the triple primitive. For example, (3, 4, 5) and (5, 12, 13) are primitive Pythagorean
triples.
Definition: An integer n ≥ 1 is said to be congruent if there exists a right triangle with rational side lengths whose
area is n. For example, n = 6 = 3&middot;4
2 is congruent.
The Congruent number problem, which has been unsolved since roughly the tenth century, is: Given an integer n ≥ 1,
determine whether or not it is congruent. Note that n is congruent if and only if there exst x, y, z ∈ Q with x2 + y 2 = z 2
and n = 12 xy.
Observations: It suffices to consider the case when n is squarefree. Moreover, there exist infinitely many primitive
Pythagorean triples.
Fermat’s conjecture: The only integer solutions to the equation xn + y n = z n for n ≥ 3 have xyz = 0.
This was proven in 1996-2000 by Wiles and Taylor, and is now widely referred to as Fermat’s last theorem. The case
n = 2 reduces to Pythagoras’s equation, which may be factored over Z[i]:
x2 + y 2 = z 2 ⇐⇒ (x + iy)(x − iy) = z 2 .
If (x, y, z) is a Pythagorean triple, then we claim that z is odd. Consequently, exactly one of x and y is odd. Indeed,
since the squares modulo 4 are 0 and 1, then if z is even we must have z 2 ≡ 0 or 2 mod 4. If z 2 ≡ 0 mod 4 then
x2 ≡ y 2 ≡ 0 mod 4, and so in particular both x and y are even, and so x, y, z have a common factor of 2, which is not the
case. Thus z 2 ≡ 2 mod 4, which has no solution, and we have exhausted all possibilities. Thus z is odd and x and y have
opposite parity – we write x + y odd.
Now, we consider x2 + y 2 = z 2 as an identity in Z[i] We know R = Z[i] is a Euclidean domain and so a fortiori a
UFD, so every element may be expressed uniquely as the product of prime elements. Let x + iy be a primitive solution
of x2 + y 2 = z 2 so that (x + iy)(x − iy) = z 2 , with (x, y, z) = 1. We claim that we may write x + iy = uα2 , where
u ∈ {&plusmn;1, &plusmn;i}. Note that to prove this, it suffices to show that any Gaussian prime π which divides x + iy divides it an
even number of times.
Indeed, since (x + iy)(x − iy) is a square, it suffices to show that if π|(x + iy), then π - (x − iy). But this is clear, since if
π divides both then π|(2x) and π|z. Since z is odd we know (2, z) = 1 and so (2x, z) = 1, a contradiction since π divides
both. Hence π - (x − iy), and we are done.
What we have shown is that we may determine all primitive Pythagorean triples by working in Z[i]. Hence the set of all
primitive Pythagorean triles is in bijection with the set of all ordered pairs &plusmn;(m2 − n2 , 2mn), where m, n ∈ Z with m + n
odd and (m, n) = 1, as proven in the homework.
Proposition 2.2.1 Fermat’s conjecture is true for n = 4.
Proof : We need to show that if x4 + y 4 = z 4 with x, y, z ∈ Z, then xyz = 0. Suppose not, so that there exists a primitive
such triple (x, y, z) with |z| minimal; such a triple may always be found by the well-ordering axiom. By our work above
we may write
x2 = m2 − n2 , y 2 = 2mn, w = z 2 = m2 + n2 ,
with m, n ∈ Z, (m, n) = 1, m + n odd. Thus (x2 , y 2 , w) is a primitive Pythagorean triple, and so there exists a right
triangle of area 12 x2 y 2 = 12 (m2 − n2 )(2mn) = mn(m + n)(m − n), which is a square. Thus we have a right triangle whose
side lengths are rational and whose area is a square; that is, 1 is a congruent number. This is not the case, as Fermat
proved using the method of descent. This implies that Fermat’s conjecture is true for n = 4; the details are left as an
exercise.
More generally: given xn + y n = z n , we can always reduce to the case where n is an odd prime, xp + y p = z p . In this case
the corresponding ring over which we work is Z[ζp ], where ζp is a primitive pth root of unity. That is, a root of X p − 1
such that {ζp , ζp2 , . . . , ζpp } is the set of all roots.
We have that Z[ζp ] ⊆ Q[ζp ] and that Q[ζp ] = Q(ζp ) is an algebraic number field. If x, y, z satisfy xp − y p = z p with
xyz 6= 0, then we may factor
xp − y p = (x − y)(x − ζp y) &middot; &middot; &middot; (x − ζpp−1 y) = z p .
7
If z is prime, then we have two distinct prime factorizations, a contradiction unless Z[ζp ] is not a UFD, which turns out
to be the case in general. We might well ask: for which primes p is Z[ζp ] a UFD?
Definition: Let A ⊆ B be an extension of rings. The integral closure of A in B, denoted Ā, is the set of all x ∈ B
which are integral over A. We showed before that Ā is a ring.
Definition: A ring A is said to be integrally closed in B if Ā = A. If A is a domain, to say without reference to B ⊇ A
that A is integrally closed is to say that it is closed in its field of fractions Quot A.
For example, Z and Fp [T ] are integrally closed (in their respective fraction fields, Q and Fp (T )).
Facts: If A ⊆ B is an extension of rings and B is integral over A, then:
• For any multiplicatively closed set S ⊂ A, we have that S −1 A ⊆ S −1 B is also an integral extension.
• If 0 −→ M −→ N −→ P −→ 0 is an exact sequence of R-modules, then
0 −→ S −1 M −→ S −1 N −→ S −1 P −→ 0
is also exact. Categorically speaking: the functor M 7→ S −1 M is an exact functor R−mod → S −1 R−mod.
• If A is a domain, then S −1 A is also a domain. Moreover, if A is integrally closed, then so is S −1 A.
8
3
Week Three
3.1
Lecture Five
Recall our discussion on Fermat’s conjecture and congruent numbers.
Proposition 3.1.1 A number k ≥ 1 is congruent if and only if there exists a rational number a such that a2 − k and
a2 + k are both squares of rational numbers.
Proof : (Necessity) Suppose k is congruent so that we have x, y, z ∈ Q such that x2 + y 2 = z 2 and k = 21 xy. Then
2
2
2
x + y &plusmn; 2xy = z &plusmn; 4k ⇐⇒
Now, with a =
z
2
x&plusmn;y
2
2
=
z 2
2
&plusmn; k.
∈ Q, one has
2
a −k =
x−y
2
2
2
,a + k =
x+y
2
2
,
and the result follows.
(Sufficiency) Let a ∈ Q such that a2 + k, a2 − k ∈ Q are squares. Write
p
p
p
p
p
√
x = a2 + k + a2 − k, y = a2 + k − a2 − k, z = x2 + y 2 = 4a2 = 2a.
Then 12 xy = k, and we are done.
4
4
2
Proposition 3.1.2 If there exist x, y, z ∈ Z with x − y = z , then 1 is a congruent number.
Proof : Note that x4 − y 4 = z 2 if and only if x4 = y 4 + z 2 . Recall that if this equation has a solution, then there exist
m, n ∈ N such that x2 = m2 + n2 , y 2 = m2 − n2 , (m, n) = 1. Hence
m2
y2
m2
x2
=
+
1,
=
− 1,
n2
n2
n2
n2
and so
x 2
n
=
m
n
+ 1,
y 2
n
=
m 2
n
− 1, and so by proposition 3.1.1 we know that 1 is congruent.
By the method of descent, Fermat proved that no square integer is congruent, and so in particular that 1 is not congruent.
Consequently, Fermat’s conjecture is true for n = 4. It is currently conjectured that if n ≡ 5, 6, or 7 mod 8, then n is
congruent. Passing to the language of elliptic curves: a number n is congruent if and only if
En : y 2 = x3 − n2 x = x(x + n)(x − n)
has a solution (x, y) over Q with y 6= 0. This ends our discussion on congruent numbers.
We aim now to study the behaviour of primes p ∈ Z in OK , the integral closure of Z in K = K/Q, a finite extension.
That is,
OK = {λ ∈ K : f (λ) = 0 for some monic f (T ) ∈ Z[T ]}.
Given an integral extension of rings A ⊆ B, let p ⊂ A be a prime ideal.
Definition: A prime ideal a ⊂ B is said to lie above p if a ∩ A = p. The ideal p is called the contraction of a, and pB
is called the extension of p in B. If B/A is integral, then:
• An ideal q ⊂ B lying over p ⊂ A is prime whenever p is prime in A.
• An extension of a prime ideal is not necessarily prime.
• For arbitrary B/A, the contraction of any prime q ⊂ B is always prime.
9
For a prime ideal p ⊂ A in an integral extension A ⊆ B, there are several possibilities for the ideal P = pB ⊂ B. The
main distinction is this: it is possible that the ideal P is itself prime in B; otherwise, more than one prime ideal q lies
above p. In this case we may uniquely factor
P = qn1 1 qn2 2 &middot; &middot; &middot; qnmm , each qi ⊂ B prime.
In case ni &gt; 1 for some i = 1, 2, . . . , m, we say that p is ramified. In fact, we know exactly which primes ramify, but we
must develop some terminology first.
Definition: Let L/K be an arbitrary finite extension of fields, so that in particular L has finite dimension as a vector
space over K. The trace and norm of x ∈ L are defined respectively as the trace and norm of the K-linear endomorphism
Tx : L → L given by Tx (λ) = xλ, with respective notation TrL/K (x) and NL/K (x). Equivalently: if the characteristic
polynomial of Tx is
fx (λ) = det(λa − Tx ) = λn − a1 λn−1 + &middot; &middot; &middot; + (−1)n an ∈ K[λ],
then TrL/K (x) = a1 and NL/K (x) = an .
Exercise: If x, y ∈ L, then Tx+y = Tx + Ty and Txy = Tx Ty . In particular, we have homomorphisms in the respective
categories
TrL/K : L −→ K, NL/K : L&times; −→ L&times; .
Proposition 3.1.3 Let L/K be a separable extension of degree n and let σ : L → K range over all K-embeddings of L
into an algebraic closure K of K. Then
Y
fx (λ) =
(λ − σx),
σ
and consequently
TrL/K (x) =
X
σx and NL/K (x) =
σ
Y
σx.
σ
Proof : (sketch) Let px (λ) be the minimal polynomial of x over K and let fx (λ) be the characteristic polynomial of
Tx .
[to be continued]
10
4
4.1
Week Four
Lecture Six
Recall: Last time we introduced the notion of the norm and trace maps defined for a finite extension of fields L/K
(which, unless otherwise stated, shall always be assumed separable). These maps are respectively homomorphisms of
multiplicative and additive groups. We also stated, but did not prove, proposition 3.1.3. Before continuing with the proof,
we see an example of the phenomenon.
√
√
Example:
Suppose√L = K( α),
is not a square. Fix an embedding σ : L ,→ K fixing K; say, σ(x + y α) =
√ where α ∈ K
√
x + y α√or σ(x + y α) = x − y α. Since X 2 − α ∈ K[X] and σ fixes K, we know that σ maps the set of roots of X 2 − α
(i.e. {&plusmn; α}) to itself. In particular,
(
√
√
√
x+y α
or
σ(x + y α) = x + y α =
√
x − y α.
√
√
Thus {1, α}forms a basis for L over K.So,
Tx for x = a + b α in this basis has
if a, b ∈ K, then the linear operator
√
√
√
bα
a bα
a
matrix Tx =
. Hence
, since Tx (1) = x =
, Tx ( α) = x α = bα + b α =
a
b a
b
TrL/K (x) = a + a = 2a and NL/K (x) = (a)(a) − (bα)(b) = a2 − αb2 .
Proof : (of proposition 3.1.3) Let x ∈ L; we deal with two cases separately.
Case one: L = K(x). Then {1, x, x2 , . . . , xn−1 } forms a basis for L as a vector space over K. Let px (T ) be the minimal
polynomial of x over K, so that px (T ) ∈ K[T ] and deg px = n. Since σ(x) must be a root of px in K for every σ, we see
that there are exactly n such choices. Moreover, the value of σ(x) completely determines σ since it determines its value
on our basis. By Vieta’s formulas, the result follows.
Case two: The general case. Suppose K ⊂ K(x) ⊂ L is a tower of field extensions, with [K(x) : K = m and [L : K(x)] = d,
write n = md, and choose an embedding σ : L ,→ K fixing K. Then σ|K(x) is determined by the image of x in K, of which
there are exactly n options. Again {1, x, x2 , . . . , xm−1 } forms a basis of K(x) over K, and fix a basis {α1 , α2 , . . . , αd } of
L over K(x). Then with α0 = 1, the set
B = {αi xj : 0 ≤ i ≤ d, 0 ≤ j ≤ m − 1} ⊂ L
forms a basis of L over K; furthermore, in this basis (using the ordering
for Tx is block-diagonal:

A1 0 &middot; &middot; &middot;
 0 A2 &middot; &middot; &middot;

[Tx ]B = A =  .
..
..
 ..
.
.
0
&middot;&middot;&middot;
0
with each Ai an m &times; m matrix. In particular, if px (T ) =
m
X
1, x, . . . , xm−1 , α1 , α1 x, . . . , αd xm−1 ), the matrix

0
0
..
.


,

ci T i ∈ K[T ] is the minimal polynomial of x over K,
i=0
then

0
1

0

Ai =  .
 ..

0
0
0 0
0 0
1 0
.. ..
. .
0 0
0 0
&middot;&middot;&middot;
&middot;&middot;&middot;
&middot;&middot;&middot;
..
.
0
0
0
..
.
&middot;&middot;&middot;
&middot;&middot;&middot;
0
1
−c0
−c1
−c2
..
.





,


−cm−2 
−cm−1
which has characteristic polynomial px for every i. Thus if fx (T ) denotes the characteristic polynomial of A, then
fx (T ) = px (T )d , and so TrL/K (x) = −c1 d and NL/K (x) = (−1)m cdm .
Finally, the embeddings L ,→ K can be partitioned into equivalence classes by the relation σ ∼ τ if and only if σ|K(x) =
τ |K(x) . Each equivalence class has d elements, and there are exactly m such classes, which completes the proof.
11
Corollary: Let K ⊂ L ⊂ M be a tower of (not necessarily separable) finite field extensions. Then
TrM/K = TrL/K ◦ TrM/L and NB/K = NL/K ◦ NM/L .
Proof : We use the proposition, grouping embeddings σ : M ,→ K according to their restriction to L. The details are left
as an exercise.
Now, let {α1 , α2 , . . . , αn } be a basis of L over K. The discriminant of this basis is defined to be
d(α1 , . . . , αn ) = det(σi (αj ))2 ,
where σi run over all embeddings L ,→ K.
Consider the bilinear form (x, y) 7→ TrL/K (xy) on L as a K-vector space; we will prove in the next lecture that this form is
non-degenerate, and thus identifies L with L∗ as a K-vector space. Using this form, it is possible to write (exercise)
d(α1 , . . . , αn ) = det(TrL/K (αi αj )).
12
4.2
Lecture Seven
Recall: Given a basis {α1 , . . . , αn } of a (separable) extension L as a vector space over K, we defined the discriminant
of the basis to be det(σi (αj ))2 , as σi run over all embeddings of L into K. We also discussed the K-bilinear form
(&middot; , &middot;) : L &times; L → K defined (x, y) = TrL/K (xy). Today, we shall prove that this form is non-degenerate: if x ∈ L is such
that (x, y) = 0 for every y, then x = 0.
nb. The statement that (&middot; , &middot;) is non-degenerate is equivalent (for finite-dimensional vector spaces) to the statement that the
matrix (σi (αj )) associated to the form is nonsingular; that is, (&middot; , &middot;) is non-degenerate if and only if det(σi (αj )) 6= 0.
Suppose L = K(θ) with {1, θ, θ2 , . . . , θn−1 } a basis of L over K. We shall compute d(1, θ, . . . , θn−1 }. Let θi = σi (θ); then
the matrix Θ = (σi (θj−1 )) = (σi (θ)j−1 ) = (θij−1 ) is a Vandermonde matrix:

1
1


Θ = 1
.
 ..
θ1
θ2
θ3
..
.
θ12
θ22
θ32
..
.
&middot;&middot;&middot;
&middot;&middot;&middot;
&middot;&middot;&middot;
..
.
1
θn
θn2
&middot;&middot;&middot;

θ1n−1
θ2n−1 

θ3n−1 
.
.. 
. 
n−1
θn
Therefore we have
d(1, θ, . . . , θn−1 ) = det Θ =
Y
(θi − θj )2 6= 0,
1≤i&lt;j≤n
since θi 6= θj for any i 6= j. This is because every σi is distinct, and is determined by its value on θ. It follows that the
form (&middot; , &middot;) is non-degenerate in this basis.
Fact: Let (&middot; , &middot;) be a quadratic form on a finite-dimensional vector space V over K and let {α1 , . . . , αn }, {β1 , . . . , βn } be
2 bases of V with change-of-basis matrix P . Then
det((βi , βj )) = (det P )2 det((αi , αj )).
Note that det P is necessarily nonzero, and thus in particular (as an element of K &times; ) is invertible.
We now return to the general, ring-theoretic case. Let A be an integrally closed domain with field of fractions K = QuotA.
Let L/K be a finite (separable) extension, and B the integral closure of A in L.
Let x ∈ B; we claim TrL/K (x) ∈ A and NL/K (x) ∈ A. Indeed, we know TrL/K (x) ∈ B ∩ K and NL/K (x) ∈ B ∩ K, and
since A is integrally closed we know that B ∩ K = A.
Proposition 4.2.1 With A, B, K, L as above, we have that x ∈ B &times; if and only if NL/K (x) ∈ A&times; .
Proof : (Necessity) Suppose NL/K (x) ∈ A&times; . We have
NL/K (x) =
Y
σ(x) = x
σ
Y
σ(x).
σ6=id
Thus if y ∈ A is such that yNL/K (x) = 1, we have

1 = yNL/K (x) = y x

Y
σ(x) = x y
σ6=id
and so x has inverse y
Q
σ6=id


Y
σ(x) ,
σ6=id
σ(x) and is therefore a unit.
(Sufficiency) Suppose x ∈ B &times; with inverse y. Then 1 = NL/K (xy) = NL/K (x)NL/K (y) ∈ B ∩ K = A; so NL/K (x) ∈ A is
a unit, and we are done.
Definition: A set {ω1 , ω2 , . . . , ωn } ⊂ B is called an integral basis of B over A if every element x ∈ B may be uniquely
written
x = a1 ω1 + &middot; &middot; &middot; + an ωn , with ai ∈ A.
13
Remark: (proposition 2.10, Neukirch) If A is a PID, then B is finitely-generated as an A-module, and is therefore a free
A-module.
Example: If A = Z, then an integral basis always exists.
√
√
Example: If A = Z[ −5] (so K = Q( −5)), then A is not a PID. If L is some finite extension of K with ring of integers
OL = B, then there exists an integral basis of B over Z, but not an integral basis of B over A.
How do we deal with the fact that we may not always have an integral basis?
Lemma 4.2.2 Let {α1 , α2 , . . . , αn } be a basis of L over K, with αi ∈ B, and let d = d(α1 , . . . , αn ). Then
dB ⊆ Aα1 + &middot; &middot; &middot; + Aαn .
Proof : Let α ∈ B and write α = a1 α1 + &middot; &middot; &middot; + an αn , with ai ∈ K, and consider TrL/K (ααi ). We have by linearity of the
trace that
n
X
TrL/K (ααi ) =
aj TrL/K (αi αj ).
j=1
Thus with
~t = (TrL/K (αα1 ), . . . , TrL/K (ααn ))t ∈ An ,
T = (TrL/K (αi αj )),
~a = (a1 , . . . , an )t ∈ K n ,
we have the matrix equation ~t = T~a. By Cramer’s rule, the entries of ~a are therefore given as quotients of the form
Ti
~
ai = det
det T , where Ti is the matrix formed by replacing the ith column of T with t. By the identity det T = d, and the
fact that det Ti ∈ A for every i, it follows that every ai has dai ∈ A, and therefore
dα ∈ Aα1 + &middot; &middot; &middot; + Aαn ,
as claimed.
Now, let K/Q be a finite extension with ring of integers OK (i.e. the integral closure of Z in K). Since Z is a PID, we
know that any finitely-generated Z-submodule of OK is free over Z, and so in particular we may take an integral basis
2
of OK . Given any other integral basis, the change-of-basis matrix (in Zn must have a unit determinant, and is therefore
&plusmn;1. By our identity above, this implies that the two bases have the same discriminant, and so we are justified in defining
the discriminant dK = dOK of K to be the discriminant of any integral basis of OK .
At this point we skip some technical lemmas from section 2 of Neukirch, and move onto section 3: Ideals.
Lemma 4.2.3 Let a and a0 be nonzero, finitely-generated OK -submodules of K. Then a, a0 admit Z-bases, and so we may
define da and da0 . If a ⊆ a0 , then da = [a0 : a]2 da0 .
Proof : Exercise. (Hint: the index of a submodule of Zn is the determinant of its change-of-basis matrix.)
√
√
Example: Let K = Q( −5) so that OK = Z[ −5]. Then OK is not a unique factorization domain, as we have two
inequivalent prime factorizations
√
√
6 = 2 &middot; 3 = (1 + −5)(1 − −5).
Exercise: Let I2 = (2, 1 +
(6) = 6OK = I22 I3 I30 .
√
−5), I3 = (3, 2 +
√
−5), I30 = (3, 2 −
14
√
−5). Show that these ideals are prime, and that
5
Week Five
5.1
Lecture Eight
There is a new instructor for the remainder of the course.
Instructor: Dragos Ghioca
Email: [email protected]
Office hours: Tuesday 11.45-13.15, Thursday 10.15-11.45 in MATX 1223
Some goals for the remainder of the course:
1. The prime factorization of ideals in OK .
2. Finiteness of the class number.
3. Dirichlet’s unit theorem.
4. Splitting of prime ideals in extensions of number fields.
Remarks:
1. If K is a number field with ring of integers OK , and a ⊂ OK is an ideal (we introduce the notation a / OK ), then
we may write
a = p1 p2 &middot; &middot; &middot; pn ,
where each pi / OK is prime; furthermore this factorization is unique up to reordering of the factors.
√ Note that this
does not imply unique prime factorization of the elements of OK themselves: as seen before, in Z[ −5], we have
√
√
6 = 2 &middot; 3 = (1 + −5)(1 − −5).
√
√
√
For the ideals, however, we may write e.g. (1 + −5) = (1 + −5, 2)(1 + −5, 3).
2. For every a / OK , there exist α, β ∈ OK such that a = (α, β). Furthermore, there exists N ∈ N such that aN is
principal.
&times;
3. Let OK
denote the group of units of OK . Dirichlet’s unit theorem tells us the structure of this group: namely, that
it is the direct sum of a finite group of roots of unity, with a free abelian group of finite rank.
4. Let Spec OK denote the set of prime ideals of OK and let p ∈ SpecOK . Let L/K be a finite extension and consider
the ideal pOL / OL . By prime factorization of ideals, we may write
pOL =
s
Y
qei i ,
i=1
where qi ∈ SpecOL are all distinct. We then have an injection of residue fields OK /p ,→ OL /qi , and moreover the
index [OL /qi : OK /p] = fi is finite for every i.
Theorem 5.1.1 With the foregoing notation, we have
s
X
ei fi = n.
i=1
For example, if K = Q, L = Q(i) so that OK = Z, OL = Z[i], we have (2) = 2OL = (1 + i)2 , and similarly 5OL =
(2 + i)(2 − i) = (1 + 2i)(1 − 2i). Note that (1 − 2i) = (−i(2 + i)) = (2 + i), so the factorization is unique, and that
(2 + i) 6= (2 − i) because the generators are not associates.
Proposition 5.1.2 Recall that a ring R is said to be Noetherian if every ideal is finitely-generated. Then OK is a
Noetherian domain, is integrally closed, and has the property that every nonzero prime ideal is maximal.
15
Recall also the Noetherianity property, also known as the ascending chain condition, which is satisfied for a ring
R if every ascending chain of ideals of R is eventually stationary. That is, if
a1 ⊆ a2 ⊆ a3 ⊆ &middot; &middot; &middot;
is an ascending chain of ideals of R, then there exists N such that aN = aN +1 = aN +2 = &middot; &middot; &middot; .
As an example of a ring which does not satisfy the ascending chain condition, let F be a field and let R = F [X1 , X2 , . . .]
be the polynomial ring over F in countably many indeterminates. Then
(X1 ) ⊂ (X1 , X2 ) ⊂ (X1 , X2 , X3 ) ⊂ &middot; &middot; &middot;
is a strictly ascending chain of ideals which consequently is never stationary. Thus, R does not satisfy the ascending chain
condition.
Proposition 5.1.3 Every Noetherian ring satisfies the ascending chain condition.
Proof : Let R be a Noetherian ring and let
a1 ⊆ a2 ⊆ a3 ⊆ &middot; &middot; &middot;
be an ascending chain of ideals. Define a =
∞
[
an ; we claim that a is an ideal of R. Indeed, given x, y ∈ a we must have
n=1
x ∈ an1 , y ∈ an2 for some n1 , n2 ∈ N, so in particular with n = max{n1 , n2 } we have that x, y ∈ an and so x + y ∈ an .
Similarly, given x ∈ a (so x ∈ an , say) and r ∈ R, we have that rx ∈ an and hence rx ∈ a, so a is indeed an ideal.
Since R is Noetherian, a is finitely-generated and so we may write a = (α1 , α2 , . . . , αn ). Since each αi ∈ a, we have that
αi ∈ aNi for Ni ∈ N, i = 1, . . . , n. In particular, we have that {α1 , . . . , αn } ∈ aN , where N = max{N1 , N2 , . . . , Nn }, and
so (α1 , . . . , αn ) = a ⊆ aN . It then follows that a = aN = aN +1 = aN +2 = &middot; &middot; &middot; , and we see that the chain is eventually
stationary.
Proof : (of proposition 5.1.2) For simplicity, we assume the extension K/Q is Galois, although the general case is also
true.
We know that OK is a free, finitely-generated Z-module, because Z is a PID, so we may write
OK = Zα1 ⊕ Zα2 ⊕ &middot; &middot; &middot; ⊕ Zαr .
Consequently, since OK has finite rank r over Z, any a / OK (as a Z-submodule) is torsion-free and also has finite rank at
most r over Z. Moreover, if a 6= (0), then its rank over Z is exactly r.
Indeed, any a 6= (0) contains some a ∈ Z \ {0}, for given any α ∈ a \ {0}, we have
0 6= NK/Q (α) ∈ a ∩ Z,
since NK/Q (α) =
Q
σ
σ(α) = α
Q
σ6=id
σ(α) ∈ a. Then a contains the ideal (a) = aOK , and so
Zaα1 ⊕ Zaα2 ⊕ &middot; &middot; &middot; ⊕ Zaαr ⊆ a,
and we have that the rank of a as a Z-module is at least r, from which the result follows. Therefore, we may write
a = Zβ1 ⊕ Zβ2 ⊕ &middot; &middot; &middot; ⊕ Zβr
for some βi . In particular, a = (β1 , . . . , βr ) is finitely-generated, so OK is Noetherian, as claimed.
We claim furthermore that OK is integrally closed. Indeed, if α ∈ K \ OK is integral over OK , then α is integral over Z
by the transitivity of integral extensions. By definition this means that α ∈ OK , a contradiction.
Finally, we aim to show that every nonzero prime ideal p / OK is maximal. From our work above, we know that p ∩ Z
is a nonzero ideal of Z and so we may write p ∩ Z = (a) = aZ; we claim that a is prime in Z. If not, then there exist
x, y ∈ Z such that xy ∈ (a) but x ∈
/ (a), y ∈
/ (a). In particular, this means that xy ∈ p while x ∈
/ p, y ∈
the primality of p.
16
nb. We have implicitly used the fact that, in Z, one has a|b if and only if aZ ⊇ bZ.
Thus we may write p ∩ Z = (p) with p ∈ Z prime. Now, consider the ring OK /p; we claim that this is a field. Indeed, the
inclusion Z ,→ OK induces an inclusion of rings Z/pZ ,→ OK /p, by the universal property of quotients, and so OK /p is a
ring containing Fp ∼
= Z/pZ; we claim that it is in fact an integral extension of Fp .
Let α ∈ OK and write
αn + an−1 αn−1 + &middot; &middot; &middot; + a1 α + a0 = 0,
where ai ∈ Z. Taking this equation modulo p, we obtain the equation (in OK /p)
ᾱn + ān−1 ᾱn−1 + &middot; &middot; &middot; + ā1 ᾱ + ā0 = 0.
Now, every āi ∈ Fp and thus ᾱ is integral over Fp , from which the claim follows.
Lemma 5.1.4 Let F be a field and let R/F be an integral extension; then R is a field.
Proof : Let x ∈ R \ {0} and write
xn + an−1 xn−1 + &middot; &middot; &middot; + a1 x + a0 = 0,
with every ai ∈ F. We may assume without loss of generality that a0 6= 0 and so a fortiori that a0 ∈ F &times; . Then
x(xn−1 + an−1 xn−2 + &middot; &middot; &middot; + a2 x + a1 ) = −a0 ,
and so
n−1
x(−a−1
+ an−1 xn−2 + &middot; &middot; &middot; + a2 x + a1 ) = 1,
0 )(x
whence x is a unit, as claimed.
Corollary: If p / OK is a nonzero prime ideal, then p is maximal.
Definition: A ring R is said to be a Dedekind domain if the following conditions hold:
• R is a Noetherian domain.
• R is integrally closed.
• Every nonzero prime ideal of R in maximal.
Our work today has shown that OK is always a Dedekind domain.
17
5.2
Lecture Nine
Proposition 5.2.1 Let R be a ring and let p ∈ SpecR. If a, b / R are two ideals such that ab ⊆ p, then a ⊆ p or b ⊆ p.
Proof : Suppose not; then there exist x ∈ a \ p, y ∈ b \ p such that xy ∈ ab ⊆ p, that is, elements x, y ∈ R with x ∈
/ p, y ∈
/ p,
and xy ∈ p, contradicting the primality of p.
Let a, b / OK . We define the sum and product ideals respectively as:
a + b = {a + b : a ∈ a, b ∈ b}
r
X
ab = {
ai bi : ai ∈ a, bi ∈ b, r ∈ N}.
and
i=1
It is clear that both sets are ideals, as is the intersection a ∩ b.
√
√
√
Example: Suppose R = Z[ −5], a = (1 + −5, 2), b = (1 + −5, 3). Then
√
√
√
√
ab = ((1 + −5)2 , 3(1 + −5), 2(1 + −5), 6) = (1 + −5),
a consequence of the fact that the product of ideals (a1 , . . . , ar ) and (b1 , . . . , bs ) is the ideal ({ai bj : 1 ≤ i ≤ r, 1 ≤ j ≤
s}).
√
√
The non-unique factorization 6 = 2 &middot; 3 = (1 + −5)(1 − −5) implies an equality of ideals
√
√
(6) = (2)(3) = (1 + −5)(1 − −5);
√
√
√
√
from √
our work above
√ we know that (1 + −5) = ab = (1 + −5, 2)(1 + −5, 3), and similarly one can check (1 − −5) =
(1 − −5, 2)(1 − −5, 3). Hence
√
√
√
√
(6) = (1 + −5, 2)(1 − −5, 2)(1 + −5, 3)(1 − −5, 3) = (2)(3);
√
it can be checked that the four ideals in the intermediate factorization√are prime: indeed, if p = (1 + −5, 2), then to
show that p is prime it suffices to show that it is maximal. Given a + b −5 ∈ OK , we have
√
√
a + b −5 = a − b + b(1 + −5) ≡ a − b mod p.
Since 2 ∈ p, we have that a − b ≡ 0 or 1 mod p, and since p is not principal (as proven in the homework) we know it is a
proper ideal, and therefore OK /p ∼
= F2 . Thus p is maximal, and in particular prime.
Definition: A fractional ideal of K is a finitely-generated OK -submodule of K.
For example, any a / OK is a fractional ideal. If K = Q, then 12 Z is a fractional ideal of K which is not an ideal of OK = Z.
A non-example is Z[ 12 ], which is an OK -submodule which is not finitely-generated.
Proposition 5.2.2 If a is a fractional ideal of K, then there exists α ∈ OK \ {0} such that αa / OK .
Proof : Suppose a = OK β1 + OK β2 + &middot; &middot; &middot; + OK βs for some βi ∈ K = QuotOK . Then for each i = 1, 2, . . . , s, there exists
αi ∈ OK such that αi βi ∈ OK , and so taking α = α1 α2 &middot; &middot; &middot; αs , we see that αa ⊆ OK . As an OK -submodule of OK , it is
also an ideal.
Corollary: Any fractional ideal may be written
1
α a,
where α ∈ OK and a / OK .
Definition: Let a be a nonzero ideal of OK . The inverse ideal of a is defined
a−1 = {x ∈ K : xa ⊆ OK }.
It is clear from the definition that OK ⊆ a−1 .
√
√
For example, if K = Q( −5) and a = (2, 1 + −5), then
√
1+ −5
2
∈ a−1 \ OK , as is
3
√
.
1+ −5
Proposition 5.2.3 For any nonzero a / OK , we have that a−1 is a fractional ideal.
18
Proof : It is clear from the definition that a−1 is an OK -submodule of K, and so it suffices to show that it is finitelygenerated. Let α ∈ a \ {0}, so that for every x ∈ a−1 we have that αx ∈ OK ; in particular, we have that αa−1 / OK , and
since OK is Noetherian we know that αa−1 is finitely-generated, say
αa−1 = OK β1 + &middot; &middot; &middot; + OK βs .
Then we have a−1 = OK βα1 + &middot; &middot; &middot; + OK βαs , and we are done.
Theorem 5.2.4 Every nonzero proper ideal of OK may be written as the product of prime ideals. Moreover, this representation is unique up to reordering of the factors.
To prove theorem 5.2.4, first we must develop some preliminary results.
Lemma 5.2.5 Every nonzero proper ideal of OK contains a product of prime ideals.
Note that, in the special case K = Q, lemma 5.2.5 reduces to the statement that every nonzero, non-unit integer divides
a product of primes.
Proof : Let S be the set of all ideals of OK which do not contain the product of prime ideals; we aim to show that S is
empty. If not, then since OK is Noetherian, the ascending chain condition implies (exercise) that any nonempty collection
of ideals of OK contains a maximal element; in our case, this means that there exists a ∈ &sect; such that a ⊂ b =⇒ b ∈
/ S.
By assumption, a is not prime and so there exist α, β ∈ OK \ a such that αβ ∈ a. Define
b1 = a + (α),
b2 = a + (β),
so that a ⊂ b1 and a ⊂ b2 . By the maximality of a this means that b1 , b2 ∈
/ S and so both contain products of prime
ideals, say
p1 p2 &middot; &middot; &middot; pr ⊆ b1 ,
q1 q2 &middot; &middot; &middot; qs ⊆ b2 ,
with each pi , qj a prime ideal of OK . Then we have
p1 &middot; &middot; &middot; pr q1 &middot; &middot; &middot; qs ⊆ b1 b2 = (a + (α))(a + (β)) ⊆ a + (αβ) = a,
so that a contains the product p1 &middot; &middot; &middot; pr q1 &middot; &middot; &middot; qs of prime ideals, a contradiction because a ∈ S. It follows that S = ∅, and
the claim is proven.
Lemma 5.2.6 Let p / OK be a nonzero prime ideal. Then OK ⊂ p−1 .
Proof : We know that OK ⊆ p−1 , and so it suffices to find an element of p−1 which is not an algebraic integer.
Let α ∈ p \ {0}. By lemma 5.2.5 we know that (α) = αOK contains a product of prime ideals p1 p2 &middot; &middot; &middot; pr ⊆ (α); without
loss of generality, we may assume that r is minimal in this regard. Since p1 p2 &middot; &middot; &middot; pr ⊆ p, we have by repeatedly applying
proposition 5.2.1 that some pi lies in p, and by relabelling if necessary we may assume that i = 1.
Since OK is a Dedekind domain we know that p1 is maximal, and since p1 ⊆ p ⊂ OK we deduce that p = p1 . By the
β
∈
/ OK .
minimality of r we know that p2 &middot; &middot; &middot; pr 6⊆ (α) and so there exists some β ∈ p2 &middot; &middot; &middot; pr such that β ∈
/ (α) and so α
We claim that
β
α
∈ p−1 ; equivalently, that
β
αp
⊆ OK . But this is clear, as
βp ⊆ p1 p2 &middot; &middot; &middot; pr ⊆ (α) =⇒
and so
β
α
β
p ⊆ OK ,
α
∈ p−1 \ OK , and we are done.
−1
Corollary 1: For any nonzero a / OK , we have p
a 6⊆ a for any nonzero p ∈ SpecOK .
Proof : By lemma 5.2.6 we know that there exists γ ∈ p−1 \ OK , and so if p−1 a ⊆ a we know in particular that γa ⊆ a.
If a = (α1 , α2 , . . . , αr ), then for i = 1, 2, . . . , r we may write
γαi =
r
X
j=1
19
aij αj
for some aij ∈ OK . With A = (aij ), α
~ = (α1 , . . . , αr )t , and I the identity matrix, we have
(γI − A)~
α = ~0,
and so γ is a root of the characteristic polynomial of A, which is monic with coefficients in OK . Thus γ is integral over
OK , and since OK is integrally closed we deduce that γ ∈ OK , contradicting our original assumption. Thus p−1 a 6⊆ a, as
claimed.
Corollary 2: Let p ∈ SpecOK be nonzero. Then p−1 p = OK .
Proof : By definition we have that p−1 p ⊆ OK ; a fortiori we have that p−1 p / OK , since it is an OK -submodule of OK .
Since 1 ∈ p−1 we have that p ⊆ p−1 p ⊆ OK , and since p is maximal it suffices to show that p 6= p−1 p, which is immediate
from corollary 1.
20
6
6.1
Week Six
Lecture Ten
Recall: Theorem 5.2.4.
Proof : (Uniqueness) Suppose that a / OK may be factored
a = p1 p2 &middot; &middot; &middot; pr = q1 q2 &middot; &middot; &middot; qs ,
with each pi , qj prime ideals of OK which are not necessarily distinct. Then since p1 ⊇ q1 q2 &middot; &middot; &middot; qs , we have by proposition
5.2.1 that some qj lies in p1 ; without loss of generality, we may assume q1 ⊆ p1 , and since q1 is prime it is maximal and
thus p1 = q1 . Multiplying both sides of our equation by p−1
1 , we obtain
p2 p3 &middot; &middot; &middot; pr = q2 q3 &middot; &middot; &middot; qs ,
and since the number of factors on either side is finite, it follows that r = s and that (by relabelling if necessary) pi = qi
for every i.
(Existence) Let S be the set of all nonzero ideals of OK which do not admit a prime factorization, and suppose that S 6= ∅.
Since OK is Noetherian, S contains a maximal element a (say) which by assumption is not prime. Since the maximal
ideals of OK are exactly the prime ideals we know by Krull’s theorem that a ⊂ p for some prime ideal p. Then
b = p−1 a ⊆ p−1 p = OK
is an ideal of OK which strictly contains a; indeed, 1 ∈ p−1 so b ⊇ a, and if b = a then p−1 a ⊆ a, contradicting corollary
1 of lemma 5.2.6. Thus in particular since a ⊂ b we have that b ∈
/ S and so we can write
b = p1 p2 &middot; &middot; &middot; pk
with pi (not necessarily distinct) prime ideals of OK . But then
a = pb = pp1 &middot; &middot; &middot; pk ,
so a ∈
/ S, a contradiction. Thus S = ∅ and we are done.
Definition: If a, b / OK are nonzero ideals, then b is said to divide a, denoted b|a, if there exists some c / OK such that
a = bc.
Proposition 6.1.1 We have that a|b if and only if a ⊇ b.
Proof : If a|b, then b = ac and so b ⊆ a. Conversely, if a ⊇ b, then writing each in their unique prime factorizations
b = pf11 pf22 &middot; &middot; &middot; pfrr ,
a = pe11 pe22 &middot; &middot; &middot; perr ,
with ei , fi ≥ 0, we have that
a ⊇ b =⇒ pe11 pe22 &middot; &middot; &middot; perr ⊇ pf11 pf22 &middot; &middot; &middot; pfrr ,
if and only if ei ≤ fi for every i. Indeed, if e1 &gt; f1 (say), then we obtain
p01 p2e2 &middot; &middot; &middot; perr ⊇ p1f1 −e1 pf22 &middot; &middot; &middot; pfrr .
The left-hand side is an integral ideal, whence we obtain a contradiction: if we factor the right-hand side as qg11 qg22 &middot; &middot; &middot; qgss ,
then we have
qg11 qg22 &middot; &middot; &middot; qgss = p1f1 −e1 pf22 &middot; &middot; &middot; pfrr |pe11 ,
hence by uniqueness
pf11 pf22 &middot; &middot; &middot; pfrr = pe11 qg11 &middot; &middot; &middot; qgss .
The left-hand side now contains pf11 and not p1f1 +1 , while the right-hand side contains pe11 ⊇ p1f1 +1 , a contradiction. It
follows that a|b, and we are done.
21
Definition: Suppose a, b / OK have prime factorizations
a=
r
Y
pei i ,
r
Y
b=
i=1
pfi i ,
i=1
with each ei , fi ≥ 0. The greatest common divisor or gcd of a and b, denoted gcd(a, b) or (a, b), is defined
(a, b) =
r
Y
min{ei ,fi }
pi
.
i=1
Similarly the least common multiple or lcm of a and b, denoted lcm(a, b) or [a, b], is defined
(a, b) =
r
Y
max{ei ,fi }
pi
.
i=1
Proposition 6.1.2 We have that (a, b) = (1) = OK if and only if a + b = (1).
Proof : (Necessity) Suppose (a, b) = 1 so that there is no p ∈ SpecOK such that p|a and p|b. So if a + b ⊂ OK , then there
is some prime p / OK such that (a + b) ⊆ p and therefore a ⊆ p and b ⊆ p, a contradiction. Thus a + b = OK .
(Sufficiency) Suppose a + b = OK , so that if p ∈ SpecOK has p|a, p|b, then p|(a + b) = (1) and so (1) ⊆ p, another
Proposition 6.1.3 One has (a, b) = a + b and [a, b] = a ∩ b.
Proof : Let d = (a, b) / OK so that we may write a = da0 , b = db0 with a0 , b0 / OK and (a0 , b0 ) = (1) = a + b. Then
a + b = da0 + db0 = d(a0 + b0 ) = dOK = d.
The proof of the second assertion is similar, and is left as an exercise.
Corollary: If (a, b) = (1), then [a, b] = ab.
Theorem 6.1.4 (The Chinese remainder theorem for rings) Let a1 , a2 , . . . , an / R be relatively prime two-sided ideals of
a unital ring R in the sense that (ai , aj ) = R for all i 6= j. Then
R/
n
\
ai ∼
=
i=1
n
M
R/ai .
i=1
In particular: if R = OK and a1 , a2 , . . . , an / OK satisfy ai + aj = OK for all i 6= j, then
OK /
n
Y
ai ∼
=
n
M
OK /ai .
i=1
i=1
Proof : We prove the special case when R = OK by induction on n. The case n = 1 is clear, so suppose the statement is
true for any collection of n ideals. Then given a1 , a2 , . . . , an+1 / OK , we have by the inductive hypothesis that
OK /
n
Y
ai ∼
=
i=1
n
M
OK /ai ,
i=1
and so it suffices to prove that
OK /
n+1
Y
ai ∼
= (OK /an+1 ) ⊕
i=1
OK /
n
Y
!
ai
.
i=1
We have a canonical homomorphism
OK −→ (OK /an+1 ) ⊕
OK /
n
Y
!
ai
given by x 7→
i=1
x + an+1 , x +
n
Y
i=1
22
!
ai
Qn
Qn+1
which has kernel an+1 ∩ i=1 ai = i=1 ai , and so by the first isomorphism theorem it suffices to show that our map is
surjective. But this is clear: since
!
n
n
Y
Y
an+1 ,
ai = an+1 +
ai = OK ,
i=1
i=1
Qn
we know in particular that there exists a ∈ an+1 , b ∈ i=1 ai such that a + b = 1. Thus, given any x, y ∈ OK , we have
with m = ay + bx that
m ≡ bx mod an+1 ≡ ax + bx mod an+1 ≡ x mod an+1 ,
m ≡ ay mod
n
Y
ai ≡ ay + by mod
i=1
n
Y
ai ≡ y mod
i=1
n
Y
ai .
i=1
Thus the image of m is exactly the pair (of cosets) (x, y), so surjectivity is proven, and we are done.
Proposition 6.1.5 Let a / OK ; there exist α, β ∈ OK such that a = (α, β).
Proof : If a = (0) then we are done. Otherwise, let α ∈ a \ {0} so that (α) ⊆ a. If a = (α) then we are done; otherwise,
since a|(α) we have
a = pe11 pe22 &middot; &middot; &middot; perr ,
(α) = pf11 pf22 &middot; &middot; &middot; pfrr qg11 &middot; &middot; &middot; qgss ,
where each ei ≤ fi and qi 6= pj for any i, j. Let β ∈ OK be the solution to the system of congruences
x ≡ 1 mod q1
x ≡ 1 mod q2
..
.
x ≡ 1 mod qs
x ≡ z1 mod pe11
x ≡ z2 mod pe22
..
.
x ≡ zr mod perr ,
where each zi ∈ piei \ pei i +1 ; note that this set is nonempty by unique factorization of ideals. Then by construction
qj - (β), pei i |β, pei i +1 - (β), and it follows that a = (α) + (β) = (α, β), as claimed.
23
6.2
Lecture Eleven
Proposition 6.2.1 The nonzero fractional ideals form a group under multiplication, where (1) = OK is the identity
element.
Proof : If b is a nonzero fractional ideal, then there exists α ∈ OK \ {0} such that (α)b / OK , so by unique factorization
of ideals we may write
(α)b = pe11 pe22 &middot; &middot; &middot; perr , hence ((α)b)−1 = p1−e1 p2−e2 &middot; &middot; &middot; pr−er ,
1
r
&middot; &middot; &middot; p−e
is a well-defined fractional ideal. The other group axioms are trivial to verify.
and thus b−1 = (α)p−e
r
1
Let JK denote the group of nonzero fractional ideals and let PK / JK denote the subgroup of principal ideals.
Definition: The ideal class group (or simply class group) of a number field K is the quotient ClK := JK /PK .
Note that we have an exact sequence
1
∗
OK
K∗
JK
ClK
1
where the map K ∗ −→ JK is given by α 7→ (α).
√
√
In Z[ −5], the ideal a = (2, 1 + −5) has
√
√
√
a(1 − −5) = (2(1 − −5), 6) = (2)(1 − −5, 3),
√
and so a ≡ (3, 1 − −5) mod PK . We saw in the homework that a2 = (2), so a2 = 1 in ClK .
Definition: Let V be an n-dimensional real vector space. A lattice Γ in V is a finitely-generated subgroup which is the
Z-span of an R-linearly independent set {v1 , v2 , . . . , vm }. If m = n, we call Γ a complete lattice.
Example: Suppose V = R2 and Γ1 = Z(1, 0)t = {(n, 0)t : n ∈ Z}; then Γ1 is a lattice which is not complete. A complete
lattice is given by
0
a
1
√
√
Γ2 = Z
+Z
=
: a, b ∈ Z ,
0
5
b 5
while a non-example is given by
Γ2 = Z
√ 1
2
,
+Z
0
0
whose generators are not R-linearly independent (although they are Z-linearly independent).
Proposition 6.2.2 A finitely-generated subgroup Γ of V ∼
= Rn is a lattice if and only if it is discrete, that is, if it has
no accumulation points.
Proof : Suppose Γ is a lattice and let γ ∈ Γ so that we can write γ = c1 v1 + &middot; &middot; &middot; + cm vm for some ci ∈ Z. Let
S = {a1 v1 + &middot; &middot; &middot; + am vm : ai ∈ [ci , ci + 1)} = γ + F,
where F = {b1 v1 + &middot; &middot; &middot; + bm vm : bi ∈ [0, 1)} is the standard fundamental parallelipiped of Γ. If γ 0 ∈ S ∩ Γ, then we
have c0i ∈ Z and ai ∈ [ci , ci + 1) so that
γ 0 = a1 v1 + &middot; &middot; &middot; + am vm = c01 v1 + &middot; &middot; &middot; + c0m vm ,
if and only if
(a1 − c01 )v1 + &middot; &middot; &middot; + (am − c0m )vm = 0,
and since v1 , . . . , vm are assumed R-linearly independent, we must have that ai = c0i for every i. Since Z ∩ [ci , ci + 1) = {ci }
we deduce that ai = ci and therefore that γ 0 = γ, so S ∩ Γ = {γ}, and since F is clearly open we have that S ∩ Γ is open
and therefore that γ is an isolated point of Γ, hence Γ is discrete.
Conversely, suppose that Γ is a discrete subgroup of V which is not a lattice. Then there exist c1 , . . . , cm ∈ R not all zero
such that
c1 v1 + &middot; &middot; &middot; + cm vm = 0,
24
where Rk ⊇ Γ = Zv1 + &middot; &middot; &middot; + Zvm ; without loss of generality, we may assume that not every ci is rational. For every N &gt; 0
there exists some ai ∈ Z such that |ci − aNi | &lt; N1 , and so taking such a1 , . . . , am we define
γN = a1 v1 + &middot; &middot; &middot; + am vm .
By the triangle inequality,
||γN || = N || γNN || = N || aN1 v1 + &middot; &middot; &middot; +
≤N
m
X
am
N vm ||
= N ||( aN1 − c1 )v1 + &middot; &middot; &middot; + ( aNm − cm )vm ||
| aNi − ci |||vi || &lt;
m
X
||vi || =: R,
i=1
i=1
some finite quantity. Thus there exist infinitely many elements of Γ ∩ BR (0), and so Γ must contain an accumulation
point, a contradiction since Γ is discrete. We deduce that Γ is indeed a lattice.
Consider the set F introduced in this proof: let Γ be a complete lattice in V ∼
= Rn , and define the fundamental
parallelipiped (or fundamental domain, or fundamental cube) of Γ = Zv1 + &middot; &middot; &middot; + Zvn to be
F = {a1 v1 + &middot; &middot; &middot; + an vn : 0 ≤ ai &lt; 1}.
The volume of Γ, denoted vol(Γ), is defined to be the volume of F. If e1 , . . . , en is an orthonormal basis of V with respect
n
X
aij ej , one has
to the ordinary Euclidean inner product, then writing
j=1
vol(Γ) = vol(F) = | det A|,
where A = (aij ).
Theorem 6.2.3 (Minkowski’s lattice point theorem) Let Γ ⊆ V ∼
= Rn be a complete lattice and let X ⊆ V be centrally
symmetric (i.e. x ∈ X if and only if −x ∈ X) and convex (i.e. x, y ∈ X implies that tx + (1 − t)y ∈ X for any
t ∈ [0, 1]). If vol(X) &gt; 2n vol(Γ), then Γ ∩ X contains a nozero element.
Note that 0 ∈ X for any such nonempty X, since x ∈ X =⇒ −x ∈ X by central symmetry and x, −x ∈ X =⇒
x + (−x) = 0 ∈ X by convexity. We record the equalities for subsets S of Rn
vol(cS) = cn vol(S),
vol(γ + S) = vol(S),
where γ ∈ V and c ∈ R.
Proof : (of theorem 6.2.3) Suppose not. If γ1 6= γ2 ∈ Γ, then
( 21 X + γ1 ) ∩ ( 21 X + γ2 ) = ∅.
Indeed, if z ∈ ( 21 X + γ1 ) ∩ ( 12 X + γ2 ) then we have x1 , x2 ∈ X such that
z = 12 x1 + γ1 = 12 x2 + γ2 ⇐⇒
1
2 (x1
− x2 ) = γ2 − γ1 ∈ Γ ∩ X = {0},
which is not the case because γ1 6= γ2 . Now, let F be a fundamental domain of Γ; for γ ∈ Γ one has
F ∩ ( 21 X + γ) = (γ + (−γ + F)) ∩ (γ + 21 X) = γ + ( 12 X ∩ (−γ + F)).
For γ 6= γ 0 we know that (( 21 X + γ) ∩ F) ∩ (( 12 X + γ 0 ) ∩ F) = ∅, and since ( 12 X + γ) ∩ F ⊆ F, we have
X
X
vol(F) ≥
vol(( 21 X + γ) ∩ F) =
vol( 21 X ∩ (−γ + F)) = vol( 12 X),
γ∈Γ
γ∈Γ
since Rn is covered by the sets −γ + F as γ varies over Γ. Thus
vol(F) ≥ vol( 21 X) =
1
2n vol(F)
&gt; vol(F),
a contradiction, and we deduce that Γ ∩ X contains a nonzero element, as claimed.
25
7
7.1
Week Seven
Lecture Twelve
Let V be a real vector space of dimension n; if V ≡ Rn or Cn we have the standard inner product
hv1 , v2 i = a1 b1 + &middot; &middot; &middot; + an bn ,
where v1 = (a1 , . . . , an )t , v2 = (b1 , . . . , bn )t ∈ V . Suppose Γ is a complete lattice in V ; write Φ for the fundamental domain
of Γ so that
a
V =
(γ + Φ).
γ∈Γ
We saw last time that vol (Γ) := vol(Φ) = | det A|, where A = (aij ) is the change-of-basis matrix for the basis {vi },
where
n
X
vi =
aij ej ,
j=1
and {ej } is the standard orthonormal basis, hei , ej i = δij .
Proposition 7.1.1 With the notation used above, we have that vol (Φ) = | det(hvi , vj i)|1/2 .
Proof : We have that
hvi , vj i =
X
n
aik ek ,
k=1
n
X
ajk ek
=
k=1
n
X
aik ajk ,
k=1
hence (hvi , vj i) = AAt , from which it follows that det(hvi , vj i) = (det A)2 and the claim is now immediate.
For a number field K, we have an n-dimensional real vector space KR (where n is the dimension of K over Q) which is
constructed as follows. Consider the n embeddings σi : K ,→ Q which fix Q. Such a σi is called a real embedding if
σi (K) ⊆ R and a complex embedding if σi (K) 6⊆ R.
Fact: The complex embeddings σi come in conjugate pairs. That is, if σi is a complex embedding, then there exists some
σj such that σj = σi , where σi (x) = σi (x).
Note that the fact is trivially true for real embeddings, since in this case σi = σi . Let r denote the number of real
embeddings and 2s the number of complex embeddings, so that we have n = r + 2s. We will write τ1 , . . . , τr for the real
embeddings and γ1 , γ1 , . . . , γs , γs for the complex embeddings, when necessary.
Now, we define a map j : K → Cn via j(x) = (σ1 (x), . . . , σn (x)) with the implied ordering; more precisely,
j(x) = (τ1 (x), . . . , τr (x), γ1 (x), γ1 (x), . . . , γs (x), γs (x)).
Clearly j is an injective linear map. Let KR denote the set of all (zσ ) = (zσ1 , . . . , zσn ) ∈ Cn such that zσ = zσ . We may
write explicitly
KR = {(x1 , . . . , xr , z1 , z1 , . . . , zs , zs ) ∈ Cn : xi ∈ R, zi ∈ C}.
It is an illustrative exercise to check that j(K) ⊂ KR ⊂ Cn and that dimR KR = n.
√
Example: In the case K = Q( 3 2), we have that KR = {(x, z, z) : x ∈ R, z ∈ C}.
Proposition 7.1.2 Let a / OK be nonzero; then j(a) is a complete lattice in KR of volume [OK : a] =
Proof : In the case a = OK we write
OK = Zβ1 ⊕ &middot; &middot; &middot; ⊕ Zβn .
Then by proposition 7.1.1 we have that vol (j(OK )) = | det(hj(βi ), j(βl )i)|1/2 . We note that
hj(βi ), j(βl )i =
n
X
m=1
26
σm (βi )σm (βl ),
p
|dK |.
and so
(hj(βi ), j(βl )i) =
X
n
σm (βi )σm (βl )
= (σl (βi ))(σl (βi ))∗ ,
m=1
∗
where A denotes the conjugate transpose of A = (ail ), in which the (i, l)th entry is ali . Thus with B = (σl (βi )), we have
that
(hj(βi ), j(βl )i) = BB ∗ .
We have det At = det A and det A = det A, and since (det B)2 = dK we have that
vol(j(OK )) = | det(BB ∗ )|1/2 = (det2 B)1/2 = | det B| =
p
dK ,
from which the claim now follows.
For the general case, we write
a = Zα1 ⊕ &middot; &middot; &middot; ⊕ Zαn .
As before, we have vol (j(a)) = | det(hj(αi ), j(αl )i)|1/2 , and we may write
αi =
n
X
aik βk ,
k=1
with βk the basis elements from before and each aik ∈ Z. We have
X
X
(hj(αi ), j(αl )i) =
aik j(βk ),
alk j(βk )
= A(hj(βi ), j(βl )i)At ,
k=1
k=1
where A = (aij ). Thus
vol(j(a)) = | det A|vol(j(OK )) = [OK : a]vol(j(OK )),
and we are done.
∼
n
There is a linear isomorphism f : KR −→ R of real vector spaces given by
f (x1 , . . . , xr , z1 , z1 , . . . , zs , zs ) = (x1 , . . . , xr , R(z1 ), I(z1 ), . . . , R(zs ), I(zs )),
√
where R(z)
and I(z) denote the real and imaginary parts of z ∈ C, respectively. For instance, if K = Q( 3 2) and
√
x = 1 + 3 2, we have
√
√ √
√
√
√
√
3
3
3
3
3
3
f ◦ j(x) = (1 + 2, 1 − 22 , 3&middot;2 2 ) ∈ R3 ,
j(x) = (1 + 2, 1 + ζ 2, 1 + ζ 2 2) ∈ KR ,
where ζ =
√
−1+ −3
2
is a primitive third root of unity.
Proposition 7.1.3 If X ⊆ KR , then volKR (X) = 2s volRn (f (X)).
Proof : (sketch) Let {ei } be the orthonormal basis of KR given by
e1 = (1, 0, 0, . . . , 0)
e2 = (0, 1, 0, . . . , 0)
..
.
er = (0, . . . , 0, 1, 0, . . . , 0)
er+1 = (0, . . . , 0, 0, √12 , √12 , 0, . . . , 0)
er+2 = (0, . . . , 0, 0, √i2 , − √i2 , 0, . . . , 0)
..
.
en−1 = (0, 0, . . . , 0, √12 , √12 )
en = (0, 0, . . . , 0, √i2 , − √i2 )
27
We have
f (e1 ) = (1, 0, 0, . . . , 0)
f (e2 ) = (0, 1, 0, . . . , 0)
..
.
f (er ) = (0, . . . , 0, 1, 0, . . . , 0)
f (er+1 ) = (0, . . . , 0, 0, √12 , 0, 0, . . . , 0)
f (er+2 ) = (0, . . . , 0, 0, 0, √12 , 0, . . . , 0)
..
.
f (en−1 ) = (0, 0, . . . , 0, √12 , 0)
f (en ) = (0, 0, . . . , 0, 0, √12 )
Hence the volume of X in KR with respect to the basis {ei } equals the volume of f (X) in Rn with respect to the basis
2s
{f (ei )}, which is seen to be precisely √12 volKR (f (X)) by considering the change-of-basis matrix.
28
7.2
Lecture Thirteen
Our goal remains to show that the class number (that is, the cardinality of the ideal class group ClK ) is finite for every
algebraic number field K. Our strategy is to show that in every ideal class of ClK , there exists an ideal a of bounded
index [OK : a] = N (a) &lt; ∞; in fact we will prove today an explicit bound for the class number. Then, given n ∈ N, we
will show that there are only finitely many ideals of OK of index at most N ; these two components then immediately
imply finiteness of the class number.
Let K be an aglebraic number field with [K : Q] = n, and as in the last lecture we shall let KR = {(zσ ) : zσ = zσ }, where
σ run over embeddings K ,→ Q fixing Q. We also defined the map j : K → KR , j(x) = (σ1 (x), . . . , σn (x)), with the indices
appropriately ordered.
Proposition 7.2.1 Let cσ ∈ R&gt;0 be positive real numbers indexed by the embeddings σ of K into Q fixing Q, such that
cσ = cσ for every σ. If
n
Y
s p
|dK |[OK : a],
ci &gt; π2
i=1
then there exists α ∈ a \ {0} such that, for all σ, one has |σ(α)| &lt; cσ .
p
Recall from last time that vol (j(α)) = |dK |[OK : a].
Proof : Index the embeddings so that σi = τi for i = 1, . . . , r are the real embeddings and σr−1+2t = γt , σr+2t = γt , 1 ≤
t ≤ s are the complex embeddings. Let X ⊆ KR be defined
X = {(zσ ) ∈ Cn : |zσ | &lt; cσ }.
Clearly it is equivalent to show that X contains a nonzero point of j(a), and it is equally obvious that X is both convex
and centrally symmetric. Thus by Minkowski’s lattice point theorem, it suffices to show that vol(X) &gt; 2n vol(j(a)). The
∼
linear isomorphism f : KR −→ Rn from last lecture gives us
volKR (X) = 2s volRn (f (X)).
Furthermore, we know
f (X) = {(x1 , . . . , xn ) : |xi | &lt; ci , 1 ≤ i ≤ r; x2r+1 + x2r+2 &lt; c2r+2 , . . . , x2n−1 + x2n &lt; c2n }
⊆ R &times; &middot; &middot; &middot; &times; R &times; R2 &times; &middot; &middot; &middot; &times; R2 ,
{z
} |
{z
}
|
r copies
s copies
and so
volRn (f (X)) =
r
Y
(2ci )
i=1
s
Y
(πc2r+2j ) = 2r π s
j=1
Y
cσ ,
σ
where e.g. πc2r+2 = volR2 ({(x, y) : x2 + y 2 &lt; c2r+1 = c2r+2 }). Thus
volKR (X) = 2s volRn (f (X)) = 2r+s π s
Y
cσ &gt; 2r+s π s
p
2 s
|dK |[OK
π
: a] = 2n volKR (j(a)),
σ
and the result is now immediate.
Corollary: Under the same hypotheses, there exists α ∈ a \ {0} such that
Y
s p
|N (α)| =
|σ(α)| ≤ π2
|dK |[OK : a].
σ
Proof : Fix ε &gt; 0 and some cσ ∈ R&gt;0 for every σ : K ,→ Q such that cσ = cσ and
Y
s p
cσ = π2
|dK |[OK : a] + ε.
σ
29
By proposition 7.2.1 there is some α ∈ a \ {0} such that
Y
|NK/Q (α)| &lt;
cσ ≤
p
2 s
|dK |[OK
π
: a] + ε,
σ
from which we deduce that
lim inf |NK/Q (α)| ≤
α∈a\{0}
p
2 s
|dK |[OK
π
: a].
Since every NK/Q (α) ∈ Z, we know that the limit infimum is attained by the well-ordering axiom.
Notation: If a / OK is nonzero, we write N (a) = [OK : a].
Proposition 7.2.2 Let α ∈ OK \ {0}; then N (α) = |NK/Q (α)|.
For example, if OK = Z[i], α = 3. We have
3 0
NK/Q (α) = det
= 9,
0 3
N (3) = Z[i] : 3Z[i] = #{a + bi0 ≤ a, b ≤ 2} = 9.
Proof : Let β1 , . . . , βn be an integral basis for OK over Z; then (α) = αβ1 Z + &middot; &middot; &middot; + αβn Z. For i = 1, . . . , n, write
αβi =
n
X
aij βj ,
aij ∈ Z.
j=1
Let A = (aij ) so that | det A| = [OK : a] = N (a); then since by construction we have that A corresponds to the
multiplication-by-alpha map we have also det A = NK/Q (α), from which we deduce the result.
Proposition 7.2.3 The norm function N (&middot;) is totally multiplicative. That is, N (ab) = N (a)N (b) for any a, b / OK .
Proof : It is equivalent to prove that if a = pe11 &middot; &middot; &middot; pekk , then N (a) = N (p1 )e1 &middot; &middot; &middot; N (pk )ek . By definition we have N (a) =
[OK : a] and by the Chinese remainder theorem we know that
OK /a ∼
=
k
M
OK /pei i ,
i=1
which implies that N (a) =
k
Y
N (pei i ) and it remains only to show that N (pe ) = N (p)e for any prime ideal p and
i=1
nonnegative integer e. One has
OK ⊃ p ⊃ p2 ⊃ &middot; &middot; &middot; ⊃ pe−1 ⊃ pe ,
and we claim that [pi : pi+1 ] = [OK : p] = N (p) for every i = 0, . . . , e − 1. Indeed, pi /pi+1 has the structure of
an OK /p-module (indeed, of a vector space, since p is maximal) by defining scalar multiplication OK /p &times; pi /pi+1 via
(x + p, y + pi+1 ) 7→ xy + pi+1 ; it is an exercise to check that this is well-defined.
Thus both OK /p and pi /pi+1 are OK /p-vector spaces, and thus to show [OK : p] = [pi : pi+1 ] it suffices to show that the
two are isomorphic, and since OK /p is clearly one-dimensional over itself it remains only to show that dimOK /p pi /pi+1 = 1
for every i.
Therefore fix x ∈ pi /pi+1 ; we claim that (x) + pi+1 = pi . Indeed, by construction pi |(x) and so we can write
(x) = pi q1 &middot; &middot; &middot; qm 6⊆ pi+1 ,
where the qj are prime and not necessarily distinct (and of course, with no qj = p). Thus (x) + pi+1 = gcd (x), pi+1 = pi ,
as claimed. Thus pi /pi+1 is spanned by x+pi+1 and is thus one dimensional, hence OK /p ∼
= pi /pi+1 and we are done.
30
8
8.1
Week Eight
Lecture Fourteen
[The lecture begins with an exposition of problem 2 from the homework.]
Recall the norm of a nonzero ideal a / OK , defined N (a) = [OK : a]. We saw in proposition 7.2.3 that the norm function
is totally multiplicative.
Proposition 8.1.1 For every N ∈ N, there exist only finitely many ideals a of OK with N (a) ≤ N .
Qr
Proof : Suppose N (a) ≤ N and write a = i=1 pei i , so that
r
Y
N (a) =
N (pi )ei ≤ N.
i=1
If p ∈ Spec OK , then p ∩ Z = (p) for some prime p ∈ Z and so OK /p is some finite extension of Z/pZ ∼
= Fp ; thus in
particular N (p) = [OK : p] = pf for some f ∈ N. We have
N (a) =
r
Y
N (pi )ei =
i=1
r
Y
piei fi ,
i=1
and it suffices to show that for any prime p ∈ Z there are only finitely many p ∈ Spec OK such that p ∩ Z = (p). Indeed,
if p ∈ p ∩ Z then
pOK ⊆ p =⇒ p|pOK = q1 &middot; &middot; &middot; qs ,
the qi prime ideals of OK , not necessarily distinct. It follows that p|qi and thus p = qi for some i, and since there are only
finitely many such factors, there can be only finitely many such p, and we are done.
A corollary of this proposition is that
N (pOK ) =
r
Y
N (pi )ei =
i=1
and since N (pOK ) = NK/Q (p) = pn we obtain the identity
r
Y
pei fi ,
i=1
r
X
ei fi = n.
i=1
Proposition 8.1.2 Eveny ideal class ā contains an integral ideal of norm at most
p
2 s
|dK |.
π
Proof : Fix ā; we know ā contains an integral ideal b, so for some fixed β ∈ b \ {0} one has βb−1 = c / OK . By proposition
7.2.1 that we can find a nonzero γ ∈ c such that
s p
|N (γ)| ≤ π2
|dK |N (c).
Indeed, since (γ) ⊆ c, hence c|(γ) and so there exists some c0 / OK such that (γ) = cc0 ; we claim that b and c0 lie in the
same ideal class. But this is clear, because
c0 = (γ)c−1 = (γ)(βb−1 )−1 = βγ b,
and so c&macr;0 = b̄ as claimed. Now, N (c)N (c0 ) = N (γOK ) = NK/Q (γ), and from our bound on N (γ) we have
s p
N (c0 ) = |NK/Q (γ)|N (c)−1 ≤ π2
|dK |,
and we are done.
Corollary: The ideal class group is finite.
√
Example: If K = Z[ −5], then n = 2 = 2s, dK = −20. Thus every ideal class of ClK contains an integral ideal of norm
less than
q
p
2 s
80
|d
|
=
K
π
π 2 &lt; 3.
31
∼
If N (a) = 1, then a is in the identity class;
√ if 2N (a) = 2 then OK /a = F2 so a is prime and a ∩ Z = 2Z.√ Thus
2
2 ∈ a, [OK : a] = 2, and since 2OK = (2, 1 + −5) as seen in the homework, we know 2O
√K ⊆ a and thus a|(2, 1 + −5) ,
and since the latter is prime as also shown in the homework, we know that a = (2, 1 + −5), and that this exhausts all
cases. That is, ClK ∼
= Z/2Z.
∗ ∼
Theorem 8.1.3 (Dirichlet’s unit theorem) There is an isomorphism of groups OK
= &micro;K &times; Zr+s−1 , where &micro;K is the group
of roots of unity contained in K, r is the number of real embeddings K ,→ Q and 2s the number of complex embeddings.
√
For example,
if K = Q( d) where d ∈ N is squarefree and
√ not congruent to 1 modulo 4. Then r = 2, s = 0, and
√
∗ ∼
OK = Z[ d], hence OK
= (Z/2Z) &times; Z. Thus all units in Z[ d] are of the form &plusmn;εn , where ε is some fundamental unit.
If we factor Pell’s equation over OK , we have
√
√
x2 − dy 2 = (x + y d)(x − y d) = &plusmn;4,
√
and indeed it can be shown that ε = x0 + y0 d, where x0 , y0 is the minimal solution solution over Z of the same
equation.
32
8.2
Lecture Fifteen
Let K/Q be a number field of degree n = r + 2s, with τ1 , . . . , τr the real embeddings and γ1 , γ1 , . . . , γs , γs the complex
embeddings. Denote by &micro;(K) the set (it is, in fact, a group) of roots of unity in K; we know that &micro;(K) is finite. Recall
from last lecture theorem 8.1.3, i.e. Dirichlet’s unit theorem.
Let KR be as before; we may define multiplication of elements of KR componentwise, yielding an R-algebra, and thus we
obtain the map l : KR∗ → Rr+s given by
(zσ ) = (zτ1 , . . . , zγs ) 7→ (log |zτ1 |, . . . , log |zτr |, 2 log |zγ1 |, . . . , 2 log |zγs |).
Composing this with the embedding j : K ,→ KR from before, we obtain the commutative diagramme
j
K∗
NK/Q
Q
σ
Rr+s
N (&middot;)
Q∗
where N ((zσ )) =
l
KR∗
Σ
log | &middot; |
R∗
R
zσ , so that in particular
N ((zτ1 , . . . , zτr , zγ1 , zγ1 , . . . , zγs , zγs )) =
r
Y
zτi
i=1
s
Y
|zγj |2 ,
j=1
and Σ((x1 , . . . , xr+s )) = x1 + x2 + &middot; &middot; &middot; + xr+s . We denote the composition l ◦ j by λ.
∗
if and only if NK/Q (α) = &plusmn;1; therefore define
Clearly α ∈ OK
S = {(zσ ) ∈ KR∗ : N ((zσ )) = &plusmn;1};
∗
we have an obvious inclusion j : OK
,→ S, and we also define H = {~x ∈ Rr+s : Σ(~x) = 0}, a hyperplane in Rr+s . Now,
∗
put Γ := λ(OK ) ⊆ H; we will show that Γ is a complete lattice in H.
Proposition 8.2.1 The sequence
1
&micro;(K)
∗
OK
λ
Γ
0
is exact.
∗
Proof : It is clear that &micro;(K) ,→ OK
is an inclusion, and λ is surjective by construction; thus it remains only to prove that
ker λ = &micro;(K). The inclusion &micro;(K) ⊆ ker λ is trivial: one has
ε ∈ &micro;(K) =⇒ |σ(ε)| = 1 ∀ σ =⇒ λ(ε) = l((log 1, . . . , log 1)) = 0,
and we need only show that ker λ ⊆ &micro;(K).
Let ε ∈ ker λ; then log |zσ | = 0 and so |zσ | = 1 for every σ. For any m ∈ N, consider the polynomial
Y
Pm (X) =
(X − σ(εm )),
σ
which has degree n; since σ(Pm (X)) = Pm (X) for any σ we know that the coefficients of Pm lie in Q, and since σ(εm ) ∈ OK
for every σ we have that Pm (X) ∈ (Q ∩ OK )[X] = Z[X]. We write
Pm (X) = X n + cm,n−1 X n−1 + &middot; &middot; &middot; + cm,1 X + cm,0 , cm,i ∈ Z;
33
we claim that |cm,i | ≤
n
i
; indeed, Vieta’s formulas imply for instance that
X
X
n
σ(εm ) ≤
|σ(ε)|m = n =
|cm,n−1 | = −
,
n−1
σ
σ
and similarly for the other terms. Now, we claim that the set {Pm (X)}m∈N consists of only finitely many polynomials;
but this is clear, since every polynomial has degree n and integer coefficients, and there is an absolute bound on the
size of these coefficients, whence the claim follows easily. By the pigeonhole principle, there exists some subsequence
m1 &lt; m2 &lt; &middot; &middot; &middot; of positive integers such that Pm1 (X) = Pm2 (X) = &middot; &middot; &middot; . One has
Pm1 (X) = Pm2 (X) =⇒ {σ(εm1 )}σ = {σ(εm2 )}σ ,
and since both sets are determined by the value of σ(ε) we must have that εm1 = εm2 . Ordering he roots of Pm (X) as
σ1 (εm ), . . . , σn (εm ), we know there exist mi1 &lt; mi2 such that the roots of Pm1 (X) and Pm2 (X) are the same and occur
in the same order, again by the pigeonhole principle. That is, εmi1 = εmi2 , from which it follows that ε ∈ &micro;(K), and we
are done.
We still have not yet shown that Γ is a complete lattice in H, but assuming this result, Dirichlet’s unit theorem will follow
immediately.
Proof : (of theorem 8.1.3) We will assume the yet-unproven fact that Γ is a complete lattice in H, so that Γ ∼
= Zr+s−1 .
−1
Fix a basis {v1 , v2 , . . . , vr+s−1 } of Γ and for i = 1, . . . , r + s − 1 let εi ∈ λ (vi ); we claim that
∗
(ε1 , . . . , εr+s−1 , &micro;(K)) = OK
,
∗
the left-hand side understood as the group generated by the parenthetised set. Indeed, let α ∈ OK
so that λ(α) ∈ Γ and
we may write
r+s−1
X
λ(α) =
ai vi , ai ∈ Z,
i=1
εa1 1
ar+s−1
&middot; &middot; &middot; εr+s−1
and so if we define α0 =
the sequence from proposition 8.2.1
∗
, we have that
∈ OK
α
that α0 ∈ &micro;(K) = ker Γ,
λ(α0 ) = λ(α) and thus αα0 = 0. We deduce by exactness of
∗
)tor = &micro;(K), from which we deduce
thus that (OK
∗
∗
OK
= (OK
)tor &times; Zr+s−1 = &micro;(K) &times; Zr+s−1 ,
as claimed.
Proposition 8.2.2 Let a &gt; 1 be an integer. There are at most finitely many x ∈ OK such that NK/Q (x) = &plusmn;a, up to
multiplication by a unit of OK .
Proof : We will show that in every coset of OK /aOK , there is at most one solution x to the equation NK/Q (x) = &plusmn;a, up
to multiplication by a unit. Let x, y ∈ OK be such that x − y ∈ aOK and suppose
NK/Q (x) = &plusmn;a and NK/Q (y) = &plusmn;a.
We write x = y + az for some z ∈ OK , so that
&plusmn;NK/Q (y)
x
a
=1+z&middot; =1+z
∈ OK ,
y
y
y
where the last equality holds since y|NK/Q (y) for any y ∈ OK . Thus
∗
argument shows us that x ∈ OK
, hence a = 1, and we are done.
x
y
∗
lies in OK and we deduce that y ∈ OK
; a symmetric
Proposition 8.2.3 As defined above, Γ is a complete lattice in H.
Proof : First, we show that Γ is a lattice, for which it suffices to show that it is discrete, for which it suffices to show that
Γ contains only finitely many elements of the unit ball B in H. Therefore let v ∈ B ∩ Γ so that v = λ(α) = l(j(α)) for
∗
some α ∈ OK
; then
v = (log |σ(α)|)σ ∈ B ⇐⇒ log |σ(α)| ∈ (−1, 1) for every σ ⇐⇒ |σ(α)| ∈ ( 1e , e).
34
Since j(OK ) is discrete in KR , there can only exist finitely many such α; that is, Γ is discrete and is therefore a lattice.
Now, observe that Γ is complete if and only if there is a bounded set T ⊆ H such that H = ∪γ∈Γ T + γ; necessity was
proven at the end of our discussion on lattices by taking T to be the fundamental parallelipiped of Γ; for sufficiency, and
thus for the rest of the proof, we wait until next time.
35
9
Week Nine
9.1
Lecture Sixteen
We begin in medias res in the proof of proposition 8.2.3: observe that if there exists a bounded set T ⊆ H such that
H = ∪γ∈Γ T + γ, then Γ is complete, else W = spanR Γ ⊂ H is a proper subspace, so since Γ = spanZ (Γ) one has
[
H=
γ + T = {γ + t : γ ∈ Γ, t ∈ T } = spanZ (Γ) + T ⊆ spanR (Γ) + T = W + T ⊂ H + T = H,
γ∈Γ
where the last inclusion is proper since the inclusion W ⊂ H is proper; that is, H ⊂ H, a contradiction. Thus to complete
our proof it suffices to show the existence of such a T ; and for this, it suffices to find a bounded set T̃ ⊆ KR∗ such that
[
j(ε) ? T̃ ,
H=
∗
ε∈OK
where j : K ∗ ,→ KR∗ is our injection from before, j(ε) ? T̃ = {j(ε) ? t : t ∈ T }, and ? denotes componentwise multiplication;
clearly ? is associative and commutative. For then similarly taking l : KR∗ → Rr+s as in the last lecture, we may take
T = l(T̃ ) so that
[
Rr+s =
γ + l(T̃ ),
γ∈Γ
since (l ◦
∗
j)(OK
)
= Γ.
Therefore let cσ ∈ R&gt;0 for all embeddings σ : K ,→ Q, such that cσ = cσ for every σ and
Y
s p
cσ &gt; π2
|dK |.
σ
Define X = {(zσ ) ∈ KR : |zσ | &lt; cσ for every σ}. By proposition 7.2.1 we know that there exists some α ∈ OK \ {0} such
that j(α) ∈ X. Thus for every σ we have
Y
|σ(α)| &lt; cσ =⇒ |NK/Q (α)| &lt;
cσ =: C.
σ
Now, it is an immediate corollary of proposition 8.1.1 that, given C ∈ R&gt;0 , there exist finitely many integers α1 , . . . , αm ∈
OK such that 0 &lt; |NK/Q (α)| ≤ C implies α ∼ αi for some i = 1, . . . , m, where ∼ denotes associativity (that is,
∗
α = uαi , u ∈ OK
). For i = 1, . . . , m, put
j(αi )−1 := j(αi−1 ) = (σ1 (αi )−1 , . . . , σn (αi )−1 ) ∈ KR∗ ,
so that
X ? j(α)−1 = {x ? j(αi )−1 : x ∈ X}
and consider
m
[
∗
X ? j(α)−1 . We know there exists α ∈ OK such that |NK/Q (α)| &lt; C, and so there exists some ε ∈ OK
i=1
and some i ∈ 1, . . . , m such that α = εαi , and thus j(ε) ∈ X ? j(αi )−1 . Finally, define
[
m
T̃ = S ∩
X ? j(αi )−1 ;
i=1
[
where S = {y ∈ KR∗ : N (y) = &plusmn;1} as before, N : KR∗ → R∗ from before; we claim that S =
j(ε) ? T̃ . Since one
∗
ε∈OK
∗
inclusion follows by the definition of T̃ , we need only show that for any (yσ ) ∈ S, there exists some ε ∈ OK
such that
(yσ ) ∈ j̃(ε) ? T̃ ⇐⇒ ∃ i ∈ {1, . . . , m} with (yσ ) ∈ (X ? j(αi )−1 ) ? j(ε).
Moreover, this last condition is equivalent to the condition that there exists some (zσ ) ∈ X with
(yσ ) = (zσ ) ? j(αi )−1 ? j(ε) ⇐⇒ yσ = zσ σ(αi )−1 σ(ε) ∀ σ
⇐⇒ σ(αi )σ(ε−1 ) = zσ yσ−1 ∀ σ
⇐⇒ j(αi ε−1 ) = (zσ yσ−1 ).
36
∗
Since ε ∈ OK
we have
|NK/Q (αi ε−1 )| = |NK/Q (αi )||NK/Q (ε)|−1 = |NK/Q (αi )| &lt; C,
Q
and since (yσ ) ∈ S we have N (y) = σ yσ = &plusmn;1. One has
X ? (yσ )−1 = {(zσ yσ−1 ) : (zσ ) ∈ X} = {(wσ ) : |wσ | &lt; cσ |yσ | ∀ σ},
and clearly vol(X ? (yσ )−1 ) = vol(X). Thus there exists some nonzero α ∈ OK such that j(α) ∈ X ? (yσ )−1 and so
Y
|NK/Q (α)| &lt;
cσ .
σ
∗
Hence there exists i ∈ {1, . . . , m} and ε ∈ OK
such that α = αi ε−1 , so we have the reverse inclusion, and we are finally
done.
Now, let d &gt; 0 be squarefree and congruent to one modulo 4. Then there exist x0 , y0 ∈ N such that x20 − dy02 = 1, and
moreover, for any such solution (x, y) there exists n ∈ N such that
√
√
√
x + y d = (x0 + y0 d)n ∈ Z[ 1+2 d ].
√
√
√
There are two embeddings of K = Q( d) ⊆ R into Q, namely id and d 7→ − d, both of them real. Thus r = 2, s = 0, and
√
√ (2x+y)+y d
∗ ∼
by Dirichlet’s unit theorem we have that OK
,
= (Z/2Z) &times; Z. Indeed, writing elements of OK as x + y 1+2 d =
2
we see
√
OK = x+y2 d : x, y ∈ Z, x ≡ y mod 2 .
√ ∗
6⊆ Q, we know that there exist x, y ∈ Z such that NK/Q x+y2 d = &plusmn;1, y 6= 0, and so x2 − dy 2 = 4. By the
Since OK
pigeonhole principle, there must exist some δ ∈ {0, 1} and distinct pairs of positive integers (x1 , y1 ), (x2 , y2 ) ∈ N2 such
that
x21 − dy12 = x22 − dy22 = (−1)δ 4,
x1 ≡ x2 mod 4, y1 ≡ y2 mod 4.
Hence
16 = (x21 − dy12 )(x22 − dy22 ) = (x1 x2 − dy1 y2 )2 − d(x1 y2 − x2 y1 )2 .
Since x1 y2 ≡ x2 y1 mod 4 and x1 x2 ≡ y1 y2 mod 4, one sees that
x1 x2 − dy1 y2 ≡ x1 y2 − x2 y1 ≡ 0 mod 4,
and thus we obtain the integral solution to Pell’s equation
x1 x2 − dy1 y2
4
2
x1 y2 − x2 y1
−d
4
2
= 1.
1 y2
2 y1
That is, writing x̃ = x1 x2 −dy
, ỹ = x1 y2 −x
, we obtain integers x̃, ỹ such that x̃2 − dỹ 2 = 1. Moreover, we see that
4
4
ỹ 6= 0: indeed, since xi , yi 6= 0, we have
ỹ = 0 ⇐⇒ x1 y2 = x2 y1 ⇐⇒
x1
y1
=
⇐⇒ x1 = ax2 , y1 = ay2 some a ∈ Q∗ ⇐⇒ x21 − dy12 = a2 (x22 − dy22 ) = a2 ,
x2
y2
thus a2 = 1 and so a = 1, x1 = x2 , y1 = y2 .
These calculations show that
√
G = {x + y d : x, y ∈ Z, x2 − dy 2 = 1}
∗
is a nontrivial multiplicative subgroup of OK
= (Z/2Z)&times;Z, and in fact, it is not difficult to show that they are equal.
37
9.2
Lecture Seventeen
We begin by reviewing briefly some material from the first few weeks.
Definition: Let L/K be an extension of number fields and let p ∈ Spec OK , q ∈ Spec OL . We say that q lies above p if
pOL ⊆ q.
√
√
For example, if K = Q, L = Q( −5), p = 3OK , q = (3, 2 + −5)OL , then q lies above p since one clearly has pOL =
3OL ⊆ q.
We observe that q lies above p if and only if q ∩ OK = p; indeed, if q ∩ OK = p then one certainly has pOL ⊆ q, and
conversely if q lies above p and q ∩ OK = p0 for some p0 ∈ Spec OK , p0 6= p (note that q ∩ OK must be a prime ideal), then
q|pOL , q|p0 OL and so q contains their gcd, which is OK since they are distinct maximal ideals, a contradiction.
Furtheremore, we observe that q ∩ OK = p if and only if pOL ⊆ q: that pOL is an ideal of OL is clear, and with p = q ∩ OK
one has
pOL = (q ∩ OK )OL ⊆ qOL ;
and conversely, pOL ⊆ q implies pOK ⊆ q and thus q ∩ OK = p. Hence
q ∩ OK = p ⇐⇒ pOL ⊆ q ⇐⇒ pOL ⊆ q ⇐⇒ q|pOL .
By unique factorization of ideals, it follows that there are only finitely many ideals lying above pOK , as long as pOK 6=
OL .
Proposition 9.2.1 If p ∈ Spec OK , then pOL 6= OL .
Proof : Let α ∈ p \ p2 ⊂ OK . We have (α) = αOK ⊆ p, and so (α) = pa for some a ∈ Spec OK , and since α ∈
/ p2 we
know that p - a; thus there exists some γ ∈ a \ p.
Then if pOL = OL , one has
γOL = γ(pOL ) = (γ)pOL =⇒ γOL ⊆ (α)OL ,
and so there exists some β ∈ OL suchthat γ = αβ ⇐⇒ β =
contradiction. It follows that pOL 6= OL .
γ
α
∈ K. Thus β ∈ OK , and so γ = αβ ∈ (α) ⊆ p, so γ ∈ p, a
Theorem 9.2.2 Given p ∈ Spec OK , write
pOL =
r
Y
qei i .
i=1
If we define fi = [OL /qi : OK /p] for i = 1, . . . , r to be the so-called inertia degree of qi over p, then
r
X
ei fi = n.
i=1
We have already proven the result when K = Q; now we prove the general case.
Proof : By the Chinese remainder theorem we know that
OL /pOL ∼
=
r
M
(OL /qei i ).
i=1
Let m = dimOK lp (OL /pOL ), with the module structure induced by the injection OK /p ,→ OL /pOL . Thus [OL : pOL ] =
[OK : p]m for some positive integer m. Since
[OL : qei i ] = [OL : qi ][qi : q2i ] &middot; &middot; &middot; [qei i −1 : qei i ];
from the proof of proposition 7.2.3 we know that [OL : qi ] = [qki : qk+1
] = N (qi ) for any i ≥ 0, hence [OL : qei i ] = [OL : qi ]ei ,
i
ei fi
which by definition equals [OK : p] . Thus
r
X
m=
ei fi ,
i=1
38
and it suffices to show that m = n.
To this end, let α1 , . . . , αm be a basis of OL /pOL over OK /pOK , and let αi ∈ π −1 (αi ), where π is the projection
OL → OL /pOL ; we aim to show that α1 , . . . , αm form a basis of L over K, which implies that m = n = [L : K].
First of all, we have that the αi are K-linearly independent: if not, we would have a nontrivial solution
a1 α1 + &middot; &middot; &middot; + am αm = 0, ai ∈ K.
By clearing denominators if necessary, we may assume that ai ∈ OK ⊆ OL , and so applying π to both sides yields
a1 α1 + &middot; &middot; &middot; + am αm = 0 ∈ OL /pOL .
Since α1 , . . . , αm form a basis, we must have that a1 = &middot; &middot; &middot; = am = 0, from which it follows that ai ∈ pOL for every i.
Therefore let a be the ideal of OK generated by a1 , . . . , am ; note that a 6= {0} since not every ai is zero. We claim that
there exists some β ∈ a−1 \ a−1 p; indeed, one has
a−1 \ a−1 p = ∅ ⇐⇒ a−1 = a−1 p ⇐⇒ p = OK ,
which is not the case. Thus multiplying by β gives
βa1 α1 + &middot; &middot; &middot; + βam αm = 0,
and by construction βαi ∈ OK for every i. If moreover every βai ∈ p, then
β(α1 , . . . , αm ) = βa ⊆ pa = p =⇒ β ∈ pa−1 ,
which is also not the case. Thus for some i, which we may assume by relabelling if necessary equals 1, we have βα1 ∈
/ p,
and so
βa1 α1 + &middot; &middot; &middot; + βam αm = 0,
with βα1 6= 0, contradicting the linear independence of {α1 , . . . , αm }.
Thus {α1 , . . . , αm } is indeed linearly independent over K, and it remains only to show that they span L. For this, it
suffices to show that
d := [OL : OK α1 + &middot; &middot; &middot; + OK αm ]
is finite, as for arbitrary abelian groups H ≤ G one has [G : H] = d =⇒ dg = 0 for every d ∈ G/H, by Lagrange’s
theorem. Therefore let M = OK α1 + &middot; &middot; &middot; + OK αm ; to show that [OL : M ] is finite, it suffices to show that
N := OL /M = p(OL /M ),
for if γ1 , . . . , γs generate N over OK (note that N is finitely generated, since [OL : OK ] is finite), then
γi ∈ pN =⇒ γi =
s
X
aij γj , some aij ∈ p,
j=1
and so with A = (aij ), ~γ = (γ1 , . . . , γs )t , one has
(I − A)~γ = ~0,
where I is the identity matrix; thus by Cramer’s rule
1 − a1,1
−a1,2
−a2,1
1
− a2,2
det(I − A) = ..
..
.
.
−an,1
−an,2
&middot;&middot;&middot;
&middot;&middot;&middot;
..
.
&middot;&middot;&middot;
≡ 1 mod p,
1 − an,n −a1,n
−a2,n
..
.
and so in particular det(I − A) 6= 0, from which we deduce that det(I − A)γi = 0 for every i, hence det(A − I)N = 0 and
so N is finitely-generated over OK , by theorem 1.2.3.
Now we need only show that N = pN , i.e. that
OL /M = p(OL /M ) = pOL /pM.
39
Our final claim is that OL = M + pOL ; indeed, one has
x ∈ OL =⇒ x ∈ OL /pOL =⇒ x = a1 α1 + &middot; &middot; &middot; + am αm
for some ai ∈ OK (so that ai ∈ OL /pOL ). Thus
x − (a1 α1 + &middot; &middot; &middot; + am αm ) ∈ pOL ∩ M = pOL + M ;
finally, we show that pOL = M , for then
OL /M = (M + pOL )/M ∼
= pOL /pOL ∩ M.
The inclusion pM ⊆ pOL ∩ M is clear, so for the converse write
M = OK α1 + &middot; &middot; &middot; + OK αm .
Given x ∈ M ∩ pOL , we ave
x = a1 α1 + &middot; &middot; &middot; + am αm , some ai ∈ OK ,
so
x = a1 α1 + &middot; &middot; &middot; + am αm = 0 in OL /pOL .
By linear independence we deduce that a1 = &middot; &middot; &middot; = am = 0, hence ai ∈ pOL for every i, and so x ∈ pOL , and we are finally
done.
40
10
10.1
Week Ten
Lecture Eighteen
Let Q ⊆ K ⊆ L be a tower of number fields and let n = [L : K], p ∈ Spec OK , writing pOL in its prime factorization
pOL =
r
Y
qei i .
i=1
Last time we defined the inertia degree fi = [OL /qi : OK /p] and proved the identity
r
X
ei fi = n.
i=1
Now, let α ∈ OL such that L = K(α) (such an α exists by the theorem of the primitive element), and let
d = OL : OK [α] .
Definition: The conductor of K(α), denoted F(α), is defined F(α) = {x ∈ OL : xOL ⊆ OK [α]}.
Proposition 10.1.1 The conductor F(α) is the largest ideal of OL contained in OK [α].
Proof : That F(α) lies in OK [α] is clear, as is the fact that it is an ideal of OL ; finally if a is an ideal of OL lying in
OK [α], then aOL = a ⊆ OK [α], and so a ⊆ F(α), and we are done.
That the conductor is nonzero follows from the fact that OL : OK [α] = d implies that dOL ⊆ OK [α], and so (d) ⊆
F(α).
Proposition 10.1.2 Let K, L, n, p, and α be as before, so that L = K(α), and assume that (p, F(α) ∩ OK ) = 1. Let
f (X) ∈ OK [X] be the minimal polynomial of α over K. Let f (X) be its image modulo p and factor
f (X) =
r
Y
ei
pi (X) ,
i=1
where pi (X) is the reduction modulo p of a monic, irreducible polynomial pi (X) ∈ OK [X]. Write fi = deg pi (X) or every
i; then
r
Y
pOL =
qei i , n
i=1
where qi = pOL + pi (α)OL , and fi = [OL /qi : OK /p].
√
√
As an example, we take K = Q, L = Q( −5), α = −5, and p = (2), so that n = 2, F(α) = OL , and f (X) = X 2 + 5.
Then
f (X) ≡ X 2 + 1 mod 2 ≡ (X + 1)2 mod 2.
Thus r = 1, p1 (X) = X + 1, and
√
√
√
(2)Z[ −5] = (2, 1 + −5)2 =⇒ OL /(2, 1 + −5) ∼
= Z/(2) ∼
= F2 .
If we instead take p = (3), then
f (X) ≡ X 2 − 1 mod 3 ≡ (X − 1)(X + 1) mod 3,
so r = 2, p1 (X) = X − 1, p2 (X) = X + 1. We have
√
√
√
√
√
√
√
(3)Z[ −5] = (3, −5 − 1)(3, −5 + 1) =⇒ Z[ −5]/(3, −5 − 1) ∼
= Z[ −5]/(3, 1 + −5) ∼
= F3 .
For p = (5):
√
√
f (X) ≡ X 2 mod 5 =⇒ (5)Z[ −5] = (5, −5)2 .
41
And finally for p = (13) we have that X 2 + 5 ≡ 0 mod 13 has no solution, since 5 is not a quadratic residue modulo 13.
Thus
√
√
(13)Z[ −5] = (13, ( −5)2 + 5) = (13),
√
so p remains prime in Z[ −5] and thus
√
Z[ −5]/(13) ∼
= F132 ,
and thus f1 = 2.
Proof : (of proposition 10.1.2) In the case OL = OK [α], we start by factoring p as indicated. By the Chinese remainder
theorem we know that
OL /pOL = OK [α]/pOK [α] ∼
= OK [X]/(p, f (X)) ∼
= (OK /p)[X]/(f (X)) ∼
=
r
Y
ei
(OK /p)[X]/(pi (X) ).
i=1
ei
We have that each factor (OK /p)[X]/(pi (X) ) contains a unique maximal ideal (pi (X)), and hence their pre-images
pOL + pi (α)OL are exactly the maximal ideals of OL containing pOL . Thus
OL /(pOL + pi (α)OL ) ∼
= (OL /p)[X]/(pi (X))
is a field containing OK /p and having degree fi := deg pi over OK /p = F. Thus there is an injection F ,→ F [X]/(pi (X)),
and
F [X]/(pi (X)) : F = deg(pi (X)) = deg(pi (X)) = fi .
Now, we have
r
Y
(pOL + pi (α)OL )
ei
⊆ pOL +
Y
r
ei
pi (α)
OL ⊆ pOL ;
i=1
i=1
the first inclusion is clear by expanding the product, and the second follows from the fact that
q(X) := f (X) −
r
Y
pi (α) ∈ pOK [X]
i=1
satisfies
0 = f (α) −
r
Y
pi (α)ei = q(α) ∈ pOL .
i=1
Hence
(pOL )|
r
Y
ei
(pOL + pi (α)OL )
=
i=1
r
Y
qei i ,
i=1
where qi = pOL + pi (α)OL . Thus [OL /qi : OK /p] = fi = deg pi , and so the degree n polynomial f (X) has
n = deg f (X) =
r
X
ei deg pi =
i=1
Since (pOL )|
Qr
i=1
r
X
ei fi .
i=1
qei i , we know that
pOL =
r
Y
qεi i , some 0 ≤ εi ≤ ei , i = 1, . . . , r.
i=1
But since
r
X
εi fi = n ≤
i=1
r
X
ei fi = n,
i=1
we must have that εi = ei for every i. This proves the case OL = OK [α].
For the general case, it suffices to show that
OL /pOL ∼
= OK [α]/pOK [α].
42
[Aside: If pi (X), pj (X) are relatively prime in (OK /p)[X], so there exist qi (X), qj (X) such that
pi (X)qi (X) + pj (X)qj (X) = 1,
hence pi (X)qi (X) + pj (X)qj (X) ∈ 1 + pOK [X] and so pi (α)qi (α) + pj (α)qj (α) ∈ 1 + pOL , from which we deduce that
qi 6= qj and that (qi , qj ) = 1]
We have a canonical injection OK [α]/(pOL ∩ OK [α]) ,→ OL /pOL , and we claim that pOL ∩ OK [α] = pOK [α]. The
inclusion
pOK [α] ⊆ pOL ∩ OK [α]
is clear, and for the converse, let x ∈ pOL ∩ OK [α], and write
x=
m
X
ai bi , some ai ∈ p, bi ∈ OL .
i=1
We have that (p, F(α) ∩ OK ) = OK by assumption, and so in particular there exist t ∈ p and u ∈ F(α) ∩ OK such that
t + u = 1, and so tx + ux = x ∈ OK [α]. We know that tx ∈ pOK [α], and since ai ∈ p and
ux =
m
X
ai (uβi ) ∈ pOK [α],
i=1
we deduce that uβi ∈ OK [α] for every i, and the claim is proven. Thus there is an injection π : OK [α]/pOK [α] ,→ OL /pOL ,
and it remains only to show that π is a surjection, for which we wait until the next lecture.
43
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