Algebra Qual Solutions
September 12, 2009
UCLA ALGEBRA QUALIFYING EXAM
Solutions
JED YANG
F. Fields
Errors. We exclude trivial extensions in Qual Problem F6s3 and Qual Problem
F3f3.
F9s1.
F9s2.
F9s3.
F8f1. Let L/F be a cubic (of degree 3) field extension of characteristic zero. Prove
that there is an element a ∈ F and a cubic field extension L0 of the field F0 = Q(a)
such that L is the composite F L0 of F and L0 over F0 .
Proof. Pick b ∈ L\F , which is necessarily of degree 3 over F . Let P be its minimal
polynomial. Recall that the disciminant D of P is a square in F , or else F (b)
blah blah blah, oops. We claim that there is some a ∈ F without cubic root NOT
(necessarily) TRUE
F8f2. A field extension L/F is said to be balanced if every field homomorphism
L −→ L over F is an isomorphism.
(a) Prove that every algebraic (possibly infinite) field extension is balanced.
Proof. Let L/F be an algebraic extension. Let f : L −→ L be a homomorphism fixing F . Recall that field homomorphisms are always injective, it
remains to show that it is surjective. Let a ∈ L. As L/F is algebraic, there
exists a1 , . . . , ad ∈ F such that a satisfy p(x) = xd + a1 xd−1 + . . . + ad . Let
S = {s ∈ L : p(s) = 0}. As f is a homomorphism fixing the coefficients of
the polynomial p(x), if s ∈ S, then f (s) ∈ S as well. Thus we may consider
f : S −→ S as a set map. Since it is still injective, as |S| ≤ d is finite, it
is also surjective. Thus a is in the image of f , but as a is arbitrary, we are
done.
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(b) Give an example of a balanced non-algebraic field extension.
Proof. Consider R/Q. The number of algebraic numbers over Q is countable, so R/Q is non-algebraic. By Qual Problem F5w1, if a homomorphism
fixes Q, then it fixes R. Thus the only field homomorphism of R over Q is
the identity, which is an isomorphism.
¤
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F8f3. Let p be a prime integer and F a field such that the degree of every nontrivial
finite field extension of F is divisible by p. Prove that for any fintie field extension
L/F , there exists a tower of field extensions F = F0 ⊂ F1 ⊂ . . . ⊂ Fn = L such
that [Fi+1 : Fi ] = p for any i = 0, . . . , n − 1.
Proof.
F8s1. Consider the polynomial P (x) = x5 − 4x + 2 in Q[x].
(a) Show that P is irreducible and has 3 real roots and 2 complex ones.
Proof. The polynomial P is irreducible by Eisenstein. By Descartes’ rule of
signs, P has 2 or 0 positive roots and 1 negative root. Now since P (0) = 2,
P (1) = −1, and P (2) = 26, we conclude that there are two positive roots,
in (0, 1) and (1, 2), respectively. We conclude that P has 3 real roots and 2
complex ones.
(b) Show that the Galois group of P is S5 .
Proof. Let G be the Galois group, and as usual, consider it a subgroup of
S5 . Since there are precisely two complex roots, complex conjugation gives
a transposition in G. Since P is irreducible, adjoining one root gives an
extension of degree 5, thus 5 | |G| and there is an element of order 5 by
Cauchy. By Qual Problem G8s1, G = S5 .
¤
F8s2. Let ζn = exp(2πi/n) be a primitive nth root of unity. Let Fn = Q(ζn ), and
dn = [Fn : Q].
(a) Let n = 6. Find an irreducible polynomial of degree d6 in Q[x] whose roots
generate F6 .
Proof. Recall
Qnthat dn = φ(n), where φ is the Euler function. Recall that
xn − 1 = k=1 (x − ζnk ). Let the cyclotomic polynomial be defined as
Q
n/d
Φn (x) = gcd(j,n)=1 (x − ζnj ). Since ζn is a primitive dth root of unity for
Q
d | n, we conclude that d|n Φd (x) = xn − 1. Notice that Fn is the splitting
field of Φn (x). Let ζ be a root of Φn (x). Since ζ generate Fn , the minimal
polynomial of ζ has degree dn , hence is Φn (x). Thus Φ6 (x) is an irreducible
polynomial of degree d6 that we are looking for.
For p a prime, xp − 1 = Φ1 (x)Φp (x), where Φ1 (x) = x − 1. This gives
Φp (x) = xp−1 + . . . + x + 1. Using this, we get that (x − 1)(x + 1)(x2 + x +
1)Φ6 (x) = x6 − 1. Long division gives Φ6 (x) = x2 − x + 1.
¤
(b) Let n = 12. Find an irreducible polynomial of degree d12 in Q[x] whose
roots generate F12 .
Proof. To calculate Φ12 (x), we first calculate Φ4 (x). Indeed, Φ4 (x) = (x4 −
1)/((x − 1)(x + 1)) = x2 + 1, hence Φ12 (x) = (x12 − 1)/((x − 1)(x + 1)(x2 +
x + 1)(x2 + 1)(x2 − x + 1)) = x4 − x2 + 1.
¤
F8s3.
F7f1. Let F be a field. Show that the unit group F \{0} of F is finitely generated
if and only if F is finite.
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Yang 3
Proof. If F is finite, then its unit group is cyclic by Qual Problem F6f1, hence
finitely generated. Conversely, suppose the unit group F × is finitely generated. By
Qual Problem G5f1, any subgroup is also finitely generated. If ch F = 0, then
F contains Q and thus Q× < F × is finitely generated, a contradiction, as there
are infinitely many primes. Otherwise, suppose ch F = p > 0, then F contains
k = Z/p (see Qual Problem F6f1). Suppose F × is generated by x1 , . . . , xn . If xi
is not algebraic over k, then k(x) ∼
= k(xi ) ⊂ F . Thus the unit group of k(x) is
finitely generated, a contradiction, as there are infinitely many monic irreducible
polynomials in k[x]. Thus we conclude that the xi are algebraic over k, thus F is
a finite algebraic extension of k, hence finite.
¤
F7f2. Let f (x) be the polynomial x4 − 2x2 − 2 over Q and K be a splitting field
of f (x). Determine the Galois group Gal(K/Q), and find the number of Galois
extensions of Q inside K.
p
√
√
Proof. Let
α
=
1+
√ 3, and let K = Q(α, 2i). In K, f factors as (x − α)(x +
√
α)(x − 2i/α)(x + 2i/α); furthermore, the roots generated K, hence K is the
splitting field. Now α is a √
root of f , which is irreducible over Q by Eisenstein,
hence [Q(α) : Q] = 4. Since 2i has minimal polynomial x2 + 2 (in Q), which stays
irreducible over Q(α) ⊂ R, we get that [K : Q] = 8. Moreover, as K is a splitting
field, it is normal; as there are no repeated roots, it is separable, hence Galois. The
Galois group G = Gal(K/Q) has order 8. Since Q(α)/Q is not
√normal,√G cannot be
abelian. Thus G may be D8 or Q8 . An automorphism send 2i to ± 2i and α to
one of the four roots of f . By degree √
considerations,
evidently all eight choices are
√
indeed automorphisms. Notice that 2i 7→ − 2i and α 7→ −α are two different
elements of order 2, which does not happen in Q8 . Thus we conclude that G ∼
= D8 .
By the fundamental theorem of Galois theory, Galois extensions of Q inside K
correspond bijectively with the normal subgroups of D8 . The proper normal subgroups of D8 are the 3 subgroups of index 2, and the centre (TODO: Proof?).
Adding the trivial group and D8 , we get 6 normal subgroups, hence 6 Galois extensions of Q inside K, including Q/Q and K/Q.
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F7f3. Let f (x) be an irreducible polynomial over the field F and let K/F be a
finite extension.
(a) Define what it means for the extension K/F to be normal.
Proof. The extension K/F is normal if, for every irreducible polynomial f
over F , as long as K contains one root of f then it contains all roots of
f.
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(b) Show that if K is normal over F , then, in K[x], f (x) factors into a product
of irreducible polynomials of the same degree.
Proof. Let f1 and f2 be irreducible factors of f in K. It suffices to show
that deg f1 = deg f2 . Let αi be a root of fi in the algebraic closure, i = 1, 2.
Then F (αi ) is isomorphic to F [x]/f (x), i = 1, 2. Let σ : F (α1 ) −→ F (α2 )
be an F -isomorphism sending α1 to α2 . Extend σ to the algebraic closure.
If K/F is normal, σ fixes K. Thus σ(f1 ) is irreducible in σ(K) = K with
σ(α1 ) = α2 as a root. Thus deg f1 = deg σ(f1 ) = [K(α2 ) : K] = deg f2 . ¤
(c) Show by example that this result does not hold for K not normal.
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Yang 4
Proof. Consider
f (x) = x3 − 2, which is irreducible by Eisenstein over Q.
√
3
Now K( 2) is real, hence does not contain the two imaginary roots of f .
Thus f factors into a linear and a quadratic factor.
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F7s1. Let F = Q(ζ) where ζ = e2πi/5 and let E/F be a cyclic
√ Galois extension of
degree 5. Prove that there exists α ∈ F such that E = F ( 5 α).
Proof. Let σ ∈ Gal(E/F ) be a generator, and consider it as a linear transformation
of the vector space E over F . Then σ 5 = id, so the minimal polynomial of the
transformation σ satisfies x5 −1, thus ζ is an eigenvalue. Let β ∈ E be an associated
eigenvector, that is σ(β) = ζβ.
Let α = β 5 . Now σ(α) = σ(β 5 ) = √(σ(β))5 = (ζβ)5 = α, thus α is fixed by
hσi = Gal(E/F ), hence α ∈ F . Now 5 α = ζ i β for some i, which generates an
intermediate field of E/F . But√it is not fixed by σ hence is not of degree 1. So
[E : F ] = 5 being prime forces 5 α to be of degree 5, thus generating E.
¤
√ √
F7s2. Let K = Q( 3, 7 5).
(a) Prove that K has only one subfield F ⊂ K such that [F : Q] = 2.
Proof. a
(b) Find all subfields of K.
Proof. b
(c) Find an element u ∈ K such that K = Q(u).
Proof. c
(d) Describe all elements u ∈ K such that K = Q(u).
Proof. d
F7s3. Let F = Z/3. First explain why F[x]/(x2 − 2) is isomorphic to F[x]/(x2 −
2x − 1). Then find an explicit isomorphism:
ϕ : F[x]/(x2 − 2) −→ F[x]/(x2 − 2x − 1).
∼
Proof. Let f be an irreducible polynomial of degree n and α its root. Then F[x]/f =
F(α), which is the unique field of degree n over F. Notice that x2 −2 and x2 −2x−1
are both irreducible as they have no roots in F. Therefore the two fields in question
are isomorphic. Let α be a root of x2 − 2, then β = α + 1 is a root of x2 − 2x − 1 =
(x − 1)2 − 2. Thus ϕ : α 7→ α + 1 is an explicit isomorphism between F(α) and
F(β).
¤
F6f1. Let F be a finite field of positive characteristic p. Show that the unit group
F \{0} of F is a cyclic group and that F is a Galois extension of Z/p.
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Yang 5
Proof. Let G = F \{0} be the unit group. We immediately get that G = Z/a1 ⊕
Z/a2 ⊕ . . . ⊕ Z/an , with a1 | a2 | . . . | an . Now for g ∈ G, we notice that an g = 0,
which, when written multiplicatively, is g an = 1. Thus each element is a root of
xan − 1, which has (at most) an roots; hence |G| ≤ an . But on the other hand,
|G| = a1 a2 · . . . · an , thus forcing n = 1, as desired.
Now it is apparent that Z/p is a subfield. It remains to show that a finite
extension of a finite field is Galois. Recall that a finite field F is normal as it is
the splitting field of x|F | − x. Also, recall that the Frobenius map x 7→ xp is an
endomorphism in characteristic p. Moreover, it is injective (obvious) and surjective
(by finiteness and injectivity), hence finite fields are perfect, and any (algebraic)
extension of a finite field is separable.
¤
F6f2. Let f (x) be the polynomial x6 + 3 over Q. Determine the Galois group of
f (x), i.e., the Galois group of K/Q where K is a splitting field of f (x).
Proof. See Qual Problem F4s1.
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F6f3. Let f (x) be an irreducible polynomial over F and K/F a normal extension.
Show that f (x) factors into irreducible polynomials over K all of the same degree.
Proof. Repeat of Qual Problem F7f3-b.
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F6s1.
(a) Show that the Galois group of the splitting field of X 4 − 2 over Q has order
8.
√
Proof. Let α = 4 2. Since X 4 − 2 is irreducible (Einsenstein) and separable,
with roots ±α, ±iα, we get a Galois extension. Let K = Q(α, i). Then the
roots are in K, and also generate K. Thus K is the splitting field of X 4 − 2
over Q. The Galois group has order the same as the degree [K : Q]. It is
obvious that [K : Q] = [K : Q(α)][Q(α) : Q] = 2 · 4 since i is degree 2 over
Q(α) ⊂ R and adjoining a root of a fourth degree irreducible gives a fourth
degree extension.
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(b) Is this Galois group isomorphic to the dihedral group, the quarternion group
or one of the three abelian groups of order 8?
Proof. By the discussion above, it is obvious that [Q(α) : Q] is not a normal
extension. Thus by the Fundemental Theorem for Finite Galois Extensions,
the corresponding Gal(K/Q(α)) is not normal in G = Gal(K/Q). This
shows that G is nonabelian. The only nonabelian groups of order 8 are the
dihedral group D8 and the quarternion group Q8 . Since G permute the 4
roots of X 4 − 2, we get G ֒→ S4 , which has order 3 · 8. Thus G is a 2-Sylow
subgroup of S4 , which is D8 . Indeed, notice that Q8 is not a subgroup of
S4 . Indeed, it has 6 elements of order 4 (the same as S4 ), hence it must
contain both (1234) and (1324), whose product is (142), an element of order
3, a contradiction. As such, the Galois group is isomorphic to the dihedral
group.
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F6s2. Let F be a finite field.
(a) Show that more than half the elements of F are squares.
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Proof. Notice 0 = a2 − b2 = (a + b)(a − b) yields a + b = 0 if a 6= b. As such,
each square can be the square of at most 2 elements. But 0 is the square of
only one element, so more than half the elements are squares.
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(b) Show that every element of F is the sum of two squares.
Proof. Let x ∈ F be arbitrary and let S = {a2 : a ∈ F } be the squares.
Consider S ∩ (x − S). Since |S| = |x − S| > 12 |F |, the intersection is
nonempty. That is, there exists a, b ∈ F such that a2 = x − b2 .
¤
F6s3. Let K be a finite extension of the field F with no proper intermediate fields.
(a) If K/F is normal, show [K : F ] must be prime.
Proof. Suppose [K : F ] 6= 1 (otherwise obviously false). Let G = AutF (K)
be the group of automorphism of K fixing F , and let M = K G be the fixed
field of G. Then K/M is Galois with Galois group G. Moreover, we have
K/M/F . Since there are no proper intermediate fields, either M = F or
M = K. If M = F , then K/F is Galois. If [K : F ] is not prime, then there
exists a proper subgroup H < G by Cauchy, whose fixed field K H is a proper
intermediate field of K/F , a contradiction. Otherwise, M = K, then any
automorphism of K fixing F is trivial. As [K : F ] 6= 1 = |AutF (K)|, K/F
is not Galois. But K/F is normal, hence it is not separable. Let α ∈ K\F
be a non-separable element and mα (x) be its minimal polynomial. Since
K/F is normal, mα (x) = (x − α)(x − α2 ) · . . . · (x − αn ) splits in F (α) ⊂ K.
But if α 6= αi , then a nontrivial F -automorphism would exist by switching
the two. Thus α = αi for all i, that is, mα (x) = (x − α)n . Recall that
inseperable polynomials exist only in positive charasteristic p > 0 and are
of the form f (xp ). Thus we conclude that mα (x) = (x − α)pt for some
t ≥ 1. Then the minimal polynomial for αt is mαt (x) = (x − αt )p , hence
[F (αt ) : F ] = p. The nontrivial extension F (αt )/F thus forces K = F (αt ),
yielding [K : F ] = p as well.
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(b) Give an example to show that [K : F ] need not be prime if K/F is not
normal, explaining why your example works.
Proof. There is an irreducible polynomial of degree n with Galois group Sn
over Q. [Indeed, see Dummit–Foote §14.8: Let fp ∈ Fp [x] be polynomials
of degree n, p = 2, 3, 5, such that f2 is irreducible, f3 is the product of an
irreducible quadratic with irreducibles of odd degree, and f5 is the product
of x with an irreducible of degree n − 1. And then f ∈ Z[x] be a polynomial
such that f (x) ≡ fp (x) (mod p) for p = 2, 3, 5. Then f is irreducible, and
the Galois group contains a transposition and an (n − 1)-cycle hence is Sn .]
Take such a polynomial f of degree 4, and let α be a root. Then by
the fundamental theorem of Galois theory, Q(α) corresponds to a subgroup
H < S4 of index 4. If Q(α)/Q admit an intermediate field, then it must
correspond to A4 , the only subgroup of index 2 (see Qual Problem G3w3).
Then [A4 : H] = 2, a contradiction. Indeed, a subgroup of index 2 is normal
(see Qual Problem G4s2), but A4 is simple.
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Remark. To avoid using the simplicity of An , n ≥ 5, we could investigate the two
possibilities for groups of order 6 by hand. Indeed, S3 and Z/6 contain an odd cycle
and an element of order 6, respectively.
September 12, 2009
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F5f1. Let Fq be the finite field with q = pn elements and let N : Fq −→ Fp be the
norm map, defined by
Y
Nx =
σ(x)
where σ runs over the Galois group G = Gal(Fq /Fp ). Prove that N is surjective.
Proof. This is a special case of Qual Problem F5w3.
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F5f2. Let ϕ(x) = x4 +a3 x3 +a2 x2 +a1 x+a0 be an irreducible polynomial of degree
4 in Q[x] and let K be the field generated by the complex roots α1 , α2 , α3 , α4 of ϕ.
Let F be the field generated by:
β1 = (α1 + α2 )(α3 + α4 ),
β2 = (α1 + α3 )(α2 + α4 ),
β3 = (α1 + α4 )(α2 + α3 ).
Prove that K/F is an abelian extension, that is, the Galois group H = Gal(K/F )
is abelian.
Proof. Recall that in characteristic 0, irreducible polynomials are separable. Hence
the roots αi are distinct. If β1 = β2 , then (α1 − α4 )(α2 − α3 ) = 0, a contradiction.
We conclude by symmetry that the βi are all distinct. Let S4 act on the roots
αi by permuting them in the natural way. We may consider H as a subgroup of
S4 , consisting of the elements that fix each of the βi . By running through the
cycle types, one easily checks that H consists precisely of the identity and the
(2, 2)-cycles. So H has order 4 and is abelian.
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F5f3. Let K be the splitting field of f (x) = x11 − 7 over Q. Describe the Galois
group G = Gal(K/Q) by giving generators and relations. Determine the number
of quadratic subfields of K (a quadratic subfield is a subfield E ⊂ K such that
[E : Q] = 2).
√
Proof. Let ζ be a primitive 10th root of unity and let α = 11 7. The roots of f are
αζ i for i = 0, . . . , 10. It is obvious that these root generate K = Q(α, ζ), and f
split in that field. Therefore K is indeed the splitting field of f over Q.
Now [Q(α) : Q] = 11 and [Q(ζ) : Q] = 10 are relatively prime, hence |G| =
[Q(α, ζ) : Q] = 110. The Galois group is determined by the action on the generators α and ζ. Now the possibilities are α 7→ ζ a α for a = Z/11 and ζ 7→ ζ b for
b ∈ (Z/11)× . Since this gives 110 possibilities, the order of G, all these possibilities
are evidently valid elements. Let σ be given by α 7→ ζα and ζ 7→ ζ, and let τ be
given by α 7→ α and ζ 7→ ζ 2 . Then σ and τ generate the Galois group. Indeed,
notice that σ 11 = τ 10 = 1. Also τ στ −1 fixes ζ and sends α 7→ αζ, hence τ σ = τ σ 2 .
We thus have two­ generators, know their orders, and
® know how they commute. In
conclusion: G = σ, τ : σ 11 = τ 10 = τ στ −1 σ −2 = 1 . TODO: Actually, we probably need to prove that this actually yield 110 elements. Group presentations is an
unsolvable problem in formal logic.
A quadratic subfield E of K corresponds to the (normal) subgroups of G with
index 2. Indeed, E is the fixed field of Gal(K/E) < G, and [G : Gal(K/E)] = [K :
Q]/[K : E] = [E : Q] = 2. TODO: The rest is left as exercise to the reader. (See
Dummit–Foote §14.2.)
F5w1. Show that the identity map is the only field automorphism of the real
numbers. Show this is not true of the complex numbers.
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Yang 8
Proof. Let f : R −→ R be a field automorphism, implying f (0) = 0 and f (1) = 1.
Thus f (n) = n for n ∈ Z. Then b · f ( ab ) = f (b · ab ) = f (a) = a for a, b ∈ Z, b 6= 0.
Thus f ( ab ) = ab . Thus f fixes Q.
A homomorphism fixing Q fixes R. [Indeed, first notice that f is order-preserving.
Indeed, if a > b, then a − b = c2 for some c 6= 0. Thus f (a) − f (b) = f (a − b) =
f (c2 ) = (f (c))2 > 0 as f (c) 6= 0. Suppose, towards a contradiction, that f (x) 6= x.
WLOG, let x < f (x). As Q is dense in R, there exists y ∈ Q such that x < y =
f (y) < f (x). Then x < y but f (y) < f (x), a contradiction.]
Recall that complex conjugation is a non-identity field automorphism of the
complex numbers. It is trivial to check.
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F5w2. Let F be a field of positive characteristic p and f the polynomial xp −x−a ∈
F [x]. Let K/F be a splitting field of f . Show that K/F is Galois and determine
explicitly the Galois group of K/F .
Proof. (Artin–Shreier). Let α be a root of f in the algebraic closure of F . Notice
that α + i is a root of xp − x − a for all i ∈ Z/p. So f splits completely in F (α),
which is therefore normal. If α ∈ F then [K : F ] = 1 and we are done. Otherwise,
Qp−1
f (x) = i= (x − (α + i)) isQirreducible. Indeed, if f (x) factors in F [x], then there
exists I ⊂ Z/p such that i∈I (x − (α + i)) ∈ F [x]. Let d = |I| and consider
P
coefficient of xd−1 , which is −dα − i∈I i. This proves that dα ∈ F , which implies
(since α ∈
/ F ) that p | d ≤ p, hence I = Z/p, and thus P is irreducible. All this
implies [F (α) : F ] = deg f = p. Since we have enumerated p different roots of f ,
it is separable. Thus K/F is Galois with G = Gal(K/F ) of order p. Explicitly, we
get Z/p −→ Gal(K/F ) given by i 7→ ϕi , where ϕi : α 7→ α + i.
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F5w3. Let K/F be a finite extension of finite fields. Prove that the norm map
NK/F : K −→ F is surjective.
Proof. Let |F | = q, then |K| = q m for some m. Recall that the multiplicative group
K × is cyclic (see Qual Problem F6f1); let g be its generator. The subgroup F × is
m
the unique subgroup of order q − 1, hence it is generated by h = g (q −1)/(q−1) .
a
×
Recall that NK/F (0) = 0. It sufficesQto prove that any h ∈ F is in the image
of N = NK/F . By definition, N (x) = σ(x), where σ runs over the Galois group
G = Gal(K/F ). Recall that the Gal(K/F ) is generated by the q-Frobenius auto2
m−1
morphism x 7→ xq (see Dummit–Foote §14.3). Thus N (x) = x1+q+q +...+q
=
m
x(q −1)/(q−1) . Thus N (g a ) = ha , as desired.
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F4f1. Let a be an integer and let p be a prime. Show that if a is not a pth power,
then xp − a is irreducible over Q.
Qn
√
Proof. Let b = p a be the real root, then xp − a factors as i=1 (x − ζ i b) in Q.
Suppose, towards a contradiction, that a is not a pth power, that is, b ∈
/ Q, but
p
x
−
a
is
reducible
over
Q.
Then
for
some
nontrivial
I
⊂
{1,
2,
.
.
.
,
n}
we have
Q
i
(x
−
ζ
b)
∈
Q[x],
where
k
=
|I|
satisfies
1
≤
k
<
p.
In
particular,
the
constant
i∈I
P
P
term (−1)k ζ I bk ∈ Q. Since b is real, we conclude that ζ I is real as well. As it
has magnitude 1, it is necessarily ±1. Therefore bk ∈ Q. But if 1 ≤ k < p, then
(k, p) = 1, so by the Euclidean algorithm, b ∈ Q, a contradiction.
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September 12, 2009
Yang 9
F4f2. Show that if K and L are finite separable extensions of F with K Galois
over F , such that K ∩ L = F , then [KL : F ] = [L : F ][K : F ]. Show that if neither
K nor L are Galois over F , then this fact need not be true.
Proof. (See Dummit–Foote, §14.4, Prop. 19.) Let K and L be finite separable
extensions of F with K Galois over F , such that K ∩ L = F . Notice that if K is the
splitting field of a polynomial f over F , then KL is the splitting field of f over L, and
as such, KL/L is indeed Galois. It suffice to prove that Gal(KL/L) ∼
= Gal(K/F ).
As K/F is Galois (in particular, normal), an embedding of K that fixes F is an
automorphism of K. Thus the restriction map ϕ : Gal(KL/L) −→ Gal(K/F )
is well-defined. It is a homomorphism (easy), with a trivial kernel. Indeed, if
σ ∈ ker ϕ, then σ fixes L (σ ∈ Gal(KL/L)) and K (σ ∈ ker ϕ), and thus fixes KL.
Now let H = ϕ(Gal(KL/L)) < Gal(K/F ), and let KH be its fixed field. As H
fixes L, KH contains F = K ∩ L. On the other hand, KH L is fixed by Gal(KL/L),
thus by Fundamental Theorem, KH L ⊂ L (obviously, with equality), and hence
KH ⊂ L, concluding that KH ⊂ K ∩L = F . Finally, KH = F gives H = Gal(K/F )
by Fundamental√Theorem, as desired.
√
Let K = Q( 3 2) and L = Q(ω 3 2), where ω is a primitive cubic root of unity.
It is obvious that [K : F ] = [L : F ] = 3, while KL is the splitting field of x3 − 2,
which has degree (dividing) 6.
¤
F4f3. By using several quadratic extensions of the rational function field in two
variables F = F2 (x, y) where F2 is the field with 2 elements, give an example of
a field extension of finite degree of F that possesses infinitely many intermediate
fields.
√ √
Proof. By Qual Problem F0f3-d, the biquadratic extension F ( x, y)/F has infinitely many intermediate fields.
¤
F4s1. Let K be a finite extension of Q obtained by adjoining to Q a root of
f (x) = x6 + 3.
(a) Show that K contains a primitive 6th root of unity.
Proof. Let α be a root of x6 + 3 and K = Q(α). Notice that ζ = (α3 + 1)/2
is a primitive 6th root of unity. Indeed, by direction computation, ζ 6 = 1,
so ζ is a 6th root. Also, ζ 6= 1, ζ 2 = (α3 − 1)/2, ζ 3 = −1, showing that it is
indeed primitive.
¤
(b) Show that K is a Galois extension of Q.
Proof. Notice that x6 + 3 splits in Q(α), since it has roots ±α, (±α ± α4 )/2.
Also, notice that all six roots are distinct, so K is the splitting field of f
and is separable, hence K/Q is Galois.
¤
(c) Determine the number of fields F of degree 3 over Q with F ⊂ K.
Proof. Notice that the extension is of degree 6, hence the Galois group
G = Gal(K/Q) is of order 6. Since there is only one generator, the automorphisms fixing Q are determined by the action on α, which necessarily
is one of the six roots of the polynomial for which it is the splitting field
for. This explicitly gives us the 6 elements of G. Indeed, e : α 7→ α is the
identity, σ : α 7→ −α has order 2, τ : α 7→ (α + α4 )/2 also has order 2 by
direct computation. This means that G ∼
= S3 (and not Z/6, the other group
of order 6).
September 12, 2009
Yang 10
Since K/Q is of degree 6, if F/Q is of degree 3, then F is a subfield of K
such that [K : F ] = 2. These correspond to subgroups H < G of order 2.
Recall that in S3 , there are 3 elements of order 2, so there are 3 subgroups
of G of order 2, and hence the number of fields is also 3.
¤
F4s2. Suppose that f (x) is a polynomial in Q[x] of degree d > 1 with d roots
x1 , . . . , xd in C. If x2 = ax1 for a ∈ Q different from −1, prove that f (x) is
reducible.
Proof. Let f (x) = a0 + a1 x + a2 x2 + . . . + ad xd , with ad 6= 0. If a = 0 then 0 is a
root and f (x) is reducible. Otherwise, suppose a 6= 0 and, towards a contradiction,
that f is irreducible. There must be some ai 6= 0 for i < d, lest f (x) = ad xd be
reducible. Consider f (ax) = a0 + a1 ax + a2 a2 x + . . . + ad ad xd . Since f (ax) and
ad f (x) have the same leading coefficient and both have x1 as a root, they must
be the same. Comparing coefficient for xi , we get ai ai = ai ad , thus a is a root of
unity. But a ∈ Q is different from −1, thus a = 1. But then x1 = x2 is a repeated
root, and all irreducible polynomials of Q (being characteristic 0) are separable, a
contradiction.
¤
Remark. This is false if a = −1, as seen in (x + i)(x − i).
F4s3. Let K be a field and L a finite extension of K. Consider the set A of all
elements x ∈ L with the property that K[x] is a Galois extension of K with an
abelian Galois group Gal(K[x]/K). Show that A is a subfield of L containing K.
Proof.
F3f1. Determine the Galois group of the polynomial x4 + 3x2 + 1 over Q.
q
q
√
√
−3± 5
. Let α = −3+2 5 ,
Proof. Using the quadratic formula, the roots are ±
2
then by direct computation we see that the other 3 roots are −α, 1/α, and −1/α.
Therefore the splitting field of the polynomial is simply Q(α) of degree 4, since
the polynomial is irreducible, by direct computation. Each element of the Galois
group necessarily sends α to one of the roots of its irreducible polynomial. This also
completely determines the action of that element. So evidently, α 7→ −α, α 7→ 1/α,
and α 7→ −1/α all give Q-automorphisms. It is also easily seen that they are all
of order 2. Therefore the Galois group is Z/2 × Z/2, not Z/4, the other group of
order 4.
¤
F3f2. Let f (x) be a polynomial of degree n > 0 over a field F .
(a) Prove that there is a field homomorphism α : F (x) −→ F (x) such that
α(x) = f (x).
Proof. Simply define α such that α(x) = f (x), and extend by linearity. It
is simple to check that this is a well-defined field homomorphism.
¤
(b) Let L be the image of α. Prove that the field extension F (x)/L is finite and
find its degree.
September 12, 2009
Yang 11
Proof. Now L contains polynomials in f (x), that is, L = {a0 + a1 f (x) +
a2 (f (x))2 + . . . + ad (f (x))d }. Notice that L is obviously an intermediate
field of F (x)/F . It remains to find the minimal polynomial of x to get the
degree of F (x)/L. First, notice that x satisfies f (T ) − f (x) as a polynomial
in L[T ], so the minimal polynomial has degree at most n. Suppose, towards
a contradiction, that the minimal polynomial has degree d < n. Then it is
of the form pd T d +pd−1 T d−1 +. . .+p1 T +p0 , where the pi ∈ L. Since pi is a
polynomial in f (x), the degree with respect to x is a multiple of n. As such,
the leading term of each monomial pi xi has degree i modulo n. Since d < n,
all d + 1 monomials pi xi have different leading term degrees, and hence
has no chance of cancellation, a contradiction. We thus establish that the
minimal polynomial of x over L is f (T )−f (x), and hence [F (x) : L] = n. ¤
(c) Find the minimal polynomial of x over L.
Proof. See part (b).
¤
F3f3. Let p be a prime integer. Suppose that the degree of every finite extension
of a field F is divisible by p. Prove that the degree of every finite extension of F is
a power of p.
Proof. Suppose the degree of every non-trivial (otherwise obviously false) finite
extension of F is divisible by p. Let E/F be a finite extension. Let L be the
separable closure of F in E, then E/L is purely inseparable, hence its degree is a
power of the characteristic q. But if q 6= p, then F is perfect, hence [E : L] = 1.
Indeed, otherwise, some element a ∈ F has no q th root in F , then a root of xq − a
generate an extension of degree q, not divisible by p. It remains to show that [L : F ]
is a power of p. Since it is separable, there is a field E containing L such that E/F
is Galois. Since [L : F ] divides [E : F ], it remains to show that [E : F ] is a power of
p. Let P be a p-Sylow subgroup of Gal(E/F ), and consider the corresponding fixed
field E P . Recall that [E P : F ] = [Gal(E/F ) : P ], which is not a multiple of p by
definition, so E P /F must not be a non-trivial finite extension. Thus we conclude
Gal(E/F ) = P is a p-group, hence [E : F ] is a power of p.
¤
F3w1. Let Fq be the finite field of q elements.
(a) List all subfields of Fp6 for a prime p.
Proof. Let Fq ⊂ Fp6 . We know that q is a prime power, and that p6 is a
power of q (which is the degree of the extension). As such, it is obvious that
q = p, p2 , p3 , and p6 .
¤
(b) Find a formula for the number of monic irreducible polynomials of degree
6 in Fp [x].
n
Proof. Recall that xp −x is the product of the monic irreducible polynomials of degree d, where d runs through (positive) divisors of n. Let f (d) denote
the degree of the product of the monic polynomials of degree d. Then it is
evident that f (1) = p, f (2) = p2 − f (1) = p2 − p, f (3) = p3 − f (1) = p3 − p,
and finally f (6) = p6 − f (3) − f (2) − f (1) = p6 − p3 − p2 + p. Therefore the number of monic irreducible polynomials of degree 6 is 61 f (6) =
(p6 − p3 − p2 + p)/6.
¤
September 12, 2009
Yang 12
Remark. By the MöbiusP
inversion formula, the number of monic irreducible polynomials of degree n is n1 d|n µ(d)pn/d , where µ is the Möbius function that is 0 if
not square-free and (−1)r if it has r distinct prime factors. See, say, Dummit–Foote
§14.3 for details.
F3w2. Let K/F be a quadratic extension of fields and M/F be a Galois extension
over F containing K such that Gal(M/K) is a cyclic group of odd prime order p.
(a) Determine the possible groups Gal(M/F ) up to isomorphisms.
Proof. Let G = Gal(M/F ), with |G| = 2p, p an odd prime. Let H = hgi be
a 2-Sylow subgroup of G. Let P be a p-Sylow subgroup of G, it is normal
as it has index 2 (see Qual Problem G4s2). Thus we get G = P ⋊ϕ H, with
ϕ : H −→ Aut(P ). Now ϕ(g) has order dividing 2. As Aut(P ) ∼
= Z/(p − 1)
is cyclic of even order, there is precisely one element of order 2. Thus
there are two possible Gal(M/F ), namely, the abelian group Z/2p and the
dihedral group D2p .
¤
(b) Find the number of intermediate fields L between F and M with [L : F ] = p.
Proof. By the fundamental theorem of Galois theory, these intermediate
fields are in bijective correspondence with the subgroups H < Gal(M/F )
of index p, which have order 2, and hence in bijective correspondence with
the elements of order 2. Since Z/2p is cyclic of even order, it has precisely
one element of order 2. As for D2p , since p is odd, the elements of order 2
are the p reflections.
¤
F3w3. Find the degree of the splitting field E of x6 − 3 over the following fields.
√
(a) Q[ −3].
√
Proof. The degree is 6. Let α = 6 3 and ζ = e2πi/6 a primitive 6th root of
unity. Notice that f = x6 − 3 has roots αζ j , j = 0, . . . , 5, which generate
E = Q(α, √
ζ), and splits in E. Thus E is the splitting field of f over Q. As
ζ − ζ 5 =√ −3 (say, by using Euclidean geometry on the complex plane),
E is also the splitting
M = Q[ −3] is an intermediate field of E/Q, and thus √
field of f over M . It is apparent that [M : Q] = 2 as −3 satisfies x2 + 3,
which is irreducible over Q by Eisenstein. As x6 − 3 is irreducible over
Q by Eisenstein, [Q(α) : Q] = 6. As Q(α) is real, ζ is not in it; but ζ
satisfies x2 − x + 1 (by Euclidean geometry), hence [E : Q(α)] = 2. Thus
[E : M ] = [E : Q(α)][Q(α) : Q]/[M : Q] = 6, as desired.
¤
(b) F7 .
Proof. The degree is 6. Recall that any field extension of F7 is of the form
n
F7n , which is the splitting field of x7 − x. Since x6 − 3 is separable (its
n
derivative is nonzero), if it splits in F7n , we have x6 − 3 | x7 − x. If
P
n
n
k
(x6 − 3)f = x7 − x, then f must have the form i=0 3i x7 −6i−6 . Then
n
n
(x6 − 3)f = x7 − 3k+1 x7 −6k−6 , thus we must have 3k+1 ≡ 1 (mod 7) and
7n − 6k − 6 = 1. Solving the first equation we get k ≡ 5 (mod 6). Now
6k = 7n − 7 = 6(7n−1 + 7n−2 + . . . + 1), hence k ≡ n − 1 (mod 6). Thus n
is a multiple of 6. That is, x6 − 3 splits in F7n if and only if n is a multiple
of 6. Thus the splitting field is F76 .
¤
(c) F5 .
September 12, 2009
Yang 13
n
Proof. The degree is 2. Similarly, if (x6 − 3)f = x5 − x, then f has the
Pk
n
form i=0 3i x5 −6i−6 , yielding 3k+1 ≡ 1 (mod 5) and 5n − 6k − 6 = 1. The
first gives k ≡ 3 (mod 4). Since n = 2 is the smallest value of n that works,
the splitting field is F52 .
¤
F2f1.
p
√
(a) Determine the minimal polynomial of u = 3 + 2 2 over Q.
q
p
√
√
√
Proof. Notice that u = 3 + 2 2 = (1 + 2)2 = 1 + 2. It evidently
satisfies (x − 1)2 − 2, which is clearly irreducible. Thus it is the minimal
polynomial.
¤
(b) Determine the minimal polynomial of u−1 over Q.
√
Proof. Notice that u−1 = 2 − 1, which evidently satisfies (x + 1)2 − 2,
again irreducible.
¤
F2f2.
(a) Let F be the field generated by the roots of the polynomial x6 + 3 over Q.
Determine the Galois group of F/Q.
Proof. See Qual Problem F4s1. The Galois group is S3 .
(b) Describe all subfields of F .
¤
Proof. Let α be a root of x6 + 3 and ζ a primitive 6th root of unity. By
the fundamental theorem of Galois theory, the subfields correspond to the
subgroups of S3 . Notice that S3 consists of the identity, three transpositions
of order 2, and two 3-cycles. Besides the trivial cases, there are obviously
three subgroups of order 2 and 1 subgroup of order 3. They evidently
correspond to the subfields Q(ζ i α2 ) for i = 0, 2, 4 and the subfield Q(α3 ),
respectively. Details are left as an exercise to the reader.
¤
F2f3. Let p be a prime integer such that p ≡ 2 or 3 (mod 5). Prove that the
polynomial 1 + x + x2 + x3 + x4 is irreducible over Z/p.
Proof. Let f = 1 + x + x2 + x3 + x4 . Recall that finite fields are unique. Thus if
f were reducible, then it would have a root in Fp2 , the degree 2 extension of Z/p.
2
Any element in there satisfy xp − x. Let α be a root of f , notice that it satisfies
2
(x − 1)f = x5 − 1. Since p2 ≡ 4 (mod 5), αp − α = α4 − α, yielding α3 = 1 as
α 6= 0. But combining with α5 = 1 gives α = 1, which is not a root of f .
¤
F2s1. Let F7 be the field with 7 elements and let L be the splitting field of the
polynomial x171 − 1 over F7 . Determine the degree of L over F7 .
Proof. The degree is 3. Indeed, a finite extension of F7 is necessarily F7n for some
n
n, in which all elements satisfy x7 − x. Let f = x171 − 1, which has nonzero
n
derivative hence is separable. As such, if it splits in F7n , we have f | x7 −1 − 1.
n
But that happens if and only if 171 | 7 − 1. So we seek the smallest n such that
it is true.
¤
F2s2. Show that there exists a Galois extension of Q of degree p for each prime p.
September 12, 2009
Yang 14
Proof. This is a baby Inverse Galois Theory problem. Let q be a prime, to be
chosen later, and let ζ be a primitive q th root of unity. We know that Gal(Q(ζ)/Q)
is cyclic of order q − 1. If p | q − 1, then there is a subgroup of index p (see Qual
Problem G0f3). Then by the fundamental theorem of Galois theory, this subgroup
corresponds to an intermediate field extension of degree p over Q. It remains to
pick prime q such that q ≡ 1 (mod p). This is possible by the famous Dirchlet’s
theorem in number theory, which says there are infinitely many primes congruent
to a modulo b, given (a, b) = 1.
¤
√
√
F2s3. Let α = i + 2 where i = −1.
(a) Compute the minimal polynomial of α over Q.
Proof. Evidently α satisfies (x2 − 2)2 + 1 = x4 − 4x2 + 5, which is irreducible
by Eisenstein (shift x 7→ x + 1), hence is the minimal polynomial of α. ¤
(b) Let F be the splitting field and compute the degree of F over Q.
√
√
Proof. Notice that the roots are ± 2 ± i, which generate Q(α, 5). Notice
√
that the minimal polynomial of α over Q remains irreducible over Q( 5),
which
√ can √be checked by direct computation. As such [F : Q] = [F :
¤
Q( 5)][Q( 5) : Q] = 8.
(c) Show that F contains 3 quadratic extensions of Q.
√
√
Proof. Notice that F contains i, 5, and i 5, each of order 2, and evidently
generate distinct fields.
¤
(d) Use this information to determine the Galois group.
Proof. Since the subextension [Q(α) : Q] is not normal, the Galois group G
is not abelian. Thus G is either the dihedral group D8 or the quarternion
group
√ Q8 . By the fundamental theorem of Galois theory, the extensions
[Q( 2 ± i) : Q] correspond to subgroups of G with index 4, namely of order
2. But Q8 only has one element of order 2, thus we conclude G ∼
= D8 . ¤
F2w1. The discriminant of the special cubic polynomial f (x) = x3 +ax+b is given
by −4a3 − 27b2 . Determine the Galois group of the splitting field of x3 − x + 1 over
the following fields:
(a) F3 , the field with 3 elements.
Proof. If a cubic is reducible, it must have a root. Since x3 − x + 1 has no
root in F3 , it is irreducible. Adjoining one root gives an extension of degree
3. Since finite fields are perfect, that is the splitting field. The Galois group
is of order 3, hence necessarily is Z/3.
¤
(b) F5 , the field with 5 elements.
Proof. Notice that in F5 the polynomial factors as (x − 3)(x2 + 3x − 2). So
the extension is necessarily of degree 2, yielding Z/2 as the Galois group. ¤
(c) Q, the rational numbers.
Proof. Let G be the Galois group, which we identify with a subgroup of
S3 as usual. Since the polynomial has no rational root, it is irreducible,
thus 3 | |G|. The disciminant is −23, hence not all roots (in C) are real.
Thus, complex conjugation is an automorphism of order 2, so 2 | |G|. Thus
¤
|G| = 6, hence G ∼
= S3 .
September 12, 2009
Yang 15
F2w2.√A field extension K/Q is called biquadratic if it has degree 4 and if K =
√
Q( a, b) for some a, b ∈ Q.
(a) Show that a biquadratic extension is normal with Galois group Gal(K/Q) ∼
=
Z/2 × Z/2 and list all sub-extensions.
2
2
Proof. Notice that K is the splitting
√ field of (x − a)(x − b), hence K/Q
√
is normal. The roots are ± a, ± b. But a 6= b, lest [K√: Q] = 2.
√ Thus
the
extension
is
separable
as
well,
hence
Galois.
Since
a
→
7
−
a and
√
√
b 7→ − b are evidently two different elements of G = Gal(K/Q) of degree
2, we conclude that G ∼
= Z/2 × Z/2, not Z/4, the other group of order
4. By the fundamental theorem of Galois theory, the proper subextensions
√
corresponds
to √
the nontrivial subgroups of G, which are evidently Q( a),
√
Q( b), and Q( ab).
¤
(b) Prove that if K/Q is a normal extension of degree 4 with Gal(K/Q) ∼
=
Z/2 × Z/2 then K/Q is biquadratic.
Proof. Let K/Q be a normal extension of degree 4 with Gal(K/Q) ∼
=
Z/2 × Z/2. Then K is the splitting field of a separable polynomial f over
Q. We may, of course, replace f by the product of its irreducible factors.
The degree of each irreducible factor must divide [K : Q] = 4. If there were
an irreducible factor of degree 4, then Gal(K/Q) ∼
= Z/4 if it has order 4, a
contradiction. Thus f is a product of irreducible quadratics. Furthermore,
each quadratic with discriminant d can be replaced by x2 − d, while preserving
factor x2 − a from f , we have
√
√ the splitting field. Take an irreducible
[Q( a) : Q] = 2. Consider f over Q( a), it cannot split completely, lest
Gal(K/Q) be√of order 2. Thus some factor, say x2 −√b, remains irreducible,
√
√
√
¤
then [Q( a, b) : Q( a)] = 2, yielding K = Q( a, b).
F2w3. Let K be a finite extension of the field F with no proper intermediate
subfields.
(a) If K/F is normal, show that the degree [K : F ] is a prime.
(b) Give an example to show that [K : F ] need not be prime if K/F is not
normal.
Proof. Repeat of Qual Problem F6s3.
¤
F1f1. Let f (x) = x3 − 2x − 2.
(a) Show that f (x) is irreducible over Q.
Proof. As f is degree 3, if it is reducible, it must have a root. But the
rational roots of a monic polynomial must divide the constant term. It is
easily seen that ±1, ±2 are not roots, hence f is irreducible.
¤
(b) Let α be a complex root of f (x). Express α−1 as a polynomial in α with
coefficients in Q.
Proof. As α is a root of f , we have α3 − 2α − 2 = 0. Multiplying by α−1 ,
¤
we get α2 − 2 − 2α−1 = 0. Thus α−1 = 21 α2 − 1.
Remark. Seriously, what kind of qual question is this?
F1f2. Let f (x) = x3 + nx + 2, where n is an integer. Determine the (infinitely
many) values of n for which f is irreducible over Q.
September 12, 2009
Yang 16
Proof. If f were reducible, it would then have a root. The rational roots of a monic
polynomial must divide the constant term. Thus n must be such that f has a root
in {±1, ±2}. Solving, n must be in {−3, 1, −5}. Thus f is irreducible over Q if and
only if n ∈ Z\{−3, 1, −5}.
¤
F1f3. Let G be the Galois group of xp − 2 over Q where p is a prime. Show that
G is isomorphic to the group of matrices of the form
µ
¶
a b
0 1
where a ∈ (Z/p)× and b ∈ Z/p.
√
Proof. Let ζ be a primitive pth root of unity, and α = p 2 be the real root. Notice
that the roots of xp −2 are precisely ζ b α for b ∈ Z/p. Thus the roots generate Q(α, ζ)
and xp − 2 split in it. Further notice that [Q(α) : Q] = p and [Q(ζ) : Q] = p − 1
are coprime, so |G| = p(p − 1) by order considerations. Let ϕ(a, b) be defined by
ζ 7→ ζ a and α 7→ ζ b α. Since each element of G must map the generators ζ, α
to their conjugates, we have at most p(p − 1) possible automorphisms. As this
is also the order of |G|, we conclude that ϕ(a, b) for a ∈ (Z/p)× and β ∈ Z/p
are precisely the distinct elements of G. Now notice that by direct computation,
ϕ(a, b) ◦ ϕ(c, d) = ϕ(ac, ad + b), exactly how the first row
¶ matrices multiply.
µ of the
a b
is indeed a group
Thus it is a trivial matter to check that ϕ(a, b) 7→
0 1
isomorphism.
¤
F1s1. Let F be a prime field, i.e., the rationals or a field with p elements. Prove
that an algebraic closure of F has infinite degree over F .
√
Proof. Let F = Q. Notice [Q( n 23) : Q] = n for any n. The algebraic closure
contains all these field extensions, hence must be infinite.
Let F = Fp . If the algebraic closure E has finite degree over the finite field
Fp , then
Q E is also finite. But a finite field cannot be algebraically closed. Indeed,
¤
1 + a∈E (x − a) is well-defined yet has no root.
F1s2. Let f ∈ Q[x] be a polynomial of degree three. Let K = Q(α) be a splitting
field of f . Determine all the possibile Galois groups of K/F , prove these are all
such, and give explicit examples of K, i.e., determine a α or f .
Proof. Recall that in characteristic 0, all finite extensions are simple, so the requirement that K = Q(α) is simple is automatically satisfied. The Galois groups
are (isomorphic to) subgroups of S3 , which are 1, Z/2, Z/3, and S3 . If f splits in
Q, say, f = x3 , then the Galois group is trivial. If precisely one root is in Q, say
f = x(x2 + 23), then the splitting field is the same as the irreducible quadratic,
hence Galois group is Z/2. Otherwise, no roots are in Q, so adjoining any root
give an extension of degree 3. If there are two complex roots, say f = x3 − 23,
then complex conjugation is an element of order 2 in the Galois group, hence is S3 .
Finally, we want to exhibit Z/3 as a Galois group. Let ζ be a primitive 7th root
of unity. Then [Q(ζ) : Q] = 6, and complex conjugation generates a (normal) subgroup of order 2. Thus its fixed field K is Galois over Q of degree 3. Now complex
i
7−i
conjugation
Q3 fixes ζi + ζ7−i for i3= 1,22, 3. We thus get that K is the splitting field
of f = i=1 (x − ζ − ζ ) = x + x − 2x − 1.
¤
September 12, 2009
Yang 17
F1s3. Prove that the polynomial x4 + 1 is not irreducible over any field of positive
characteristic.
Proof. It suffices to prove that x4 + 1 is not irreducible over any field Fp . If p = 2,
then x4 + 1 = (x + 1)4 is reducible. Otherwise, p is odd, so p2 ≡ 1 (mod 8). Then
2
2
x4 + 1 | x8 − 1 | xp −1 − 1 | xp − x. If α were a root of x4 + 1, then it satisfies
2
xp − x, whose solutions form the field Fp2 . Thus we conclude Fp (α) is a subfield
of Fp2 . Thus [Fp (α) : Fp ] ≤ 2 is not 4, hence x4 + 1 is not irreducible.
¤
√
√
F0f1. Find the minimal polynomial of α = 2 + 3 3 over the field of rational
numbers Q.
Proof. Let a and b be degrees p and q over a field F , respectively. Then [F (a, b) :
F ] ≤ pq. If p and q are coprime, then obviously since p and q both divide [F (a, b) :
F ], we have equality. If p and q are distinct primes, then we get F (a, b) = F (a + b).
Indeed, one inclusion is obvious; it remains to show a + b has degree pq. Since
F (a, a + b) = F (a, b), we get pq = [F (a, b) : F ] = [F (a, a + b) : F ] ≤ [F (a) :
F ] · [F (a + b) : F ]. Thus [F (a + b) : F ] ≥ q. And similarly, [F (a + b) : F ] ≥ p. Since
a + b ∈ F (a, b), it has degree dividing pq, thus it is pq, as desired.
This shows that α has degree 6 over Q, now it remains
√ 3 to find a monic polynomial
of degree 6 that α satisfies. Notice
that
(α
−
2) − 3 = 0. Expanding and
√
3
2
rearranging, we get α + 6α − 3 = 2(2 + 3α ). Squaring and rearranging, we get
α6 − 6α4 − 6α3 + 12α2 − 36α + 1 = 0. Therefore x6 − 6x4 − 6x3 + 12x2 − 36x + 1
is the minimal polynomial.
¤
Remark. Actually, as long as p and q are coprime, and F has characteristic 0,
then F (a, b) = F (a + b). See I. M. Isaacs, Degrees of Sums in a Separable Field
Extension, http://www.jstor.org/stable/2036661.
F0f2. Let ζ be a primitive 16th root of unity over the field F . Determine the
dimension [F (ζ) : F ] when F is the following:
(a) The field of 9 elements.
Proof. The degree is 2. Notice ζ satisfies x16 − 1. Since 16 | 92 − 1, by Qual
Problem F0s2, the splitting field of x16 − 1 is F92 . So [F (ζ) : F ] | 2. But
since ζ is primitive, it cannot satisfy x8 − 1, so it is not in F .
¤
(b) The field of 7 elements.
Proof. The degree is 2. Again, the splitting field of x16 − 1 is F72 of degree
2; and ζ is not in F .
¤
(c) The field of 17 elements.
Proof. The degree is 1. The elements of F are precisely the roots of x17 − x,
which includes ζ.
¤
(d) What can you say in the case the characteristic p = 2? The degree is 1.
In characteristic 2, by the Freshmen’s dream, we have x16 − 1 = (x − 1)16 ,
hence 1 is the only 16th root of unity.
¤
F0f3. Let F be an infinite field of characteristic p > 0. Recall that a finite dimensional extension L/F is said to be simple if L = F (u) for some element u ∈ L.
(a) Suppose L/F has a finite number of intermediate fields. Show that L/F is
simple.
September 12, 2009
Yang 18
Proof. Clearly L is finitely generated over F , thus with induction, it suffices
to show that F (α, β) is simple over F . Consider intermediate fields F (α+cβ)
for c ∈ F . Since F is infinite, and each of these is an intermediate field of
L/F , we conclude that for some c 6= d, K = F (α + cβ) = F (α + dβ). This
means (c − d)β ∈ K, and as c − d ∈ F × , we get β ∈ K and finally α in K.
So F (α, β) = F (α + cβ) is simple over F , as desired.
¤
(b) Let K be an intermediate field F ⊂ K ⊂ L, and suppose that L = F (u)
with xr + a1 xr−1 + . . . + ar the monic irreducible polynomial of u over K,
ai ∈ K. Show that K = F (a1 , a2 , . . . , ar ).
Proof. Since L = K(u), we have [L : K] = r. Now F ′ = F (a1 , . . . , ar ) is a
subfield of K. But since L = F ′ (u), and u satisfies a degree r polynomial,
it follows that [L : F ′ (u)] ≤ r. This implies K = F ′ , as desired.
¤
(c) Conclude that L/F is simple if and only if there are a finite number of
intermediate fields.
Proof. By part (a), we have the backward implication. Conversely, let L =
F (u), and f be the minimal polynomial of u over F . Take intermediate
subfield K, then the minimal polynomial g of u in K is a factor of f . By
part (b), g determines completely the subfield K. Since there are finitely
many factors of f , the number of subfields is finite.
√ √¤
(d) Let E = F (x, y) where x and y are indeterminates, and set M = E( p x, p y).
Show that M/E has an infinite number of intermediate fields.
√
√
Proof. Let α = p x and β = p y and follow the proof of part (a). If
intermediate fields E(α + cβ) = E(α + dβ) for c 6= d in E, then M =
E(α + cβ). But in characteristic p, we have Freshmen’s dream, (α + cβ)p =
x + cp y ∈ E. So α + cβ has degree (at most) p over E, a contradiction to
[M : E] = p2 . Thus the fields E(α + cβ) for c ∈ E are all different. Since
|E| is infinite, there are infinitely many intermediate fields in M/E.
¤
F0s1. Let α = 1 +
√
√
3
2 + 3 4.
(a) Find the degree of α over Q.
Proof. By direct computation, we see that α satisfies x3 − 3x2 − 3x − 1.
Since tihs is degree 3, it must have a root if reducible, which is not the case
by rational root theorem. Thus α is of degree 3 over Q.
¤
(b) Find a normal closure of Q(α)/Q.
Proof. The normal closure must contain the roots of the minimal polynomial
x3 − 3x2 − 3x − 1. Thus we calculate its splitting field. Factoring in C, we
2
get (x − α)(x2 − (3 − α)x +
√1/α). The disciminant is d = (3 − α) − 4/α < 0.
The roots generate Q(α, d), and the polynomial split in this, thus this is
the splitting field.
¤
F0s2. Let q be a power of a prime integer, n ∈ N. Let k be the least positive
integer such that q k ≡ 1 (mod n). Prove that the finite field Fqk is a splitting field
of the polynomial xn − 1 over Fq .
September 12, 2009
Yang 19
Proof. Recall that the any finite field extension of Fq is of the form Fqm , whose
m
nonzero elements are the roots of xq −1 −1. If n is a multiple of p, the characteristic
of Fq , then q k is a power of p, thus p | n | q k − 1, a contradiction. Thus p ∤ n, so
the derivative nxn−1 6≡ 0, hence xn − 1 is separable over Fq . So if xn − 1 splits in
m
Fqm , we get xn − 1 | xq −1 − 1. But this happens if and only if n | q m − 1. Thus
¤
the smallest field that xn − 1 split over is Fqk , as desired.
F0s3.