1 Spring 2002 – Galois Theory

Problem 1.1. Let F7 be the field with 7 elements and let L be the splitting field of thepolynomial X171 − 1 = 0 over F7. Determine the degree of L over F7, explaining carefullythe principles underlying your computation.

Solution: Note that 73 = 49 · 7 = 343, so

[x ∈ (F73)×] =⇒ [x342 − 1 = 0],

since (F73)× is a multiplicative group of order 342. Also, F73 contains all the roots ofx342 − 1 = 0 since the number of roots of a polynomial cannot exceed its degree (by theDivision Algorithm). Next note that 171 · 2 = 342, so

[x171 − 1 = 0] ⇒ [x171 = 1] ⇒ [x342 − 1 = 0].

This implies that all the roots of X171− 1 are contained in F73 and so L ⊂ F73 since L canbe obtained from F7 by adjoining all the roots of X171 − 1. We therefore have

F73——L——F7︸ ︷︷ ︸3

and so L = F73 or L = F7 since 3 = [F73 : F7] = [F73 : L][L : F7] is prime. Next ifα ∈ (F73)×, then α2 is a root of X171 − 1. Also, (F73)× is cyclic and hence isomorphic toZ73 = {0, 1, 2, . . . , 342}, so the map α 7→ α2 on F73 has an image of size bigger than 7:

2α = 2β ⇔ 2(α− β) = 0 in Z73 ⇔ α− β = 171.

We therefore conclude that X171−1 has more than 7 distinct roots and hence L = F73 .

• Splitting Field: A splitting field of a polynomial f ∈ K[x] (K a field) is an extensionL of K such that f decomposes into linear factors in L[x] and L is generated over Kby the roots of f .

• Irreducible Polynomial: An irreducible polynomial is a polynomial of positive degreethat does not factor into two polynomials of positive degree.

• Separable Polynomial: A polynomial f ∈ K[x] is called separable if it does not havemultiple roots in any extension of K.

• Irreducible ⇒ Separable: If f is an irreducible polynomials over the following fieldsthen f is separable:

(i) Field of zero characteristic

(ii) Field of characteristic p - degf

(iii) Finite field

1

• Field Extensions: If L/K is a field extension, then L can be regarded as a vectorspace over K. The dimension of L over K is also called the degree of the extensionand denoted [L : K].

• Multiplicativity of Degree: If L is a finite extension over K and M is a finite extensionover L, then M is a finite extension over K and

[M : K] = [M : L][L : K].

• Finite Fields: For any prime p and any natural number n, there exists a field with pn

elements which is unique up to isomorphism. These are the only finite fields.

• Cyclicity of F×: If F is a finite field, then F× is cyclic (Use the Fundamental Theoremof Finitely Generated Abelian Groups I and consider the biggest invariant factor nk

and the polynomial xnk − 1).

• Euler Function: The Euler function φ assigns to each positive integer n the numberφ(n) of integers k such that 1 ≤ k ≤ n and (k, n) = 1.

– A cyclic group of order n has φ(n) generators.– (k, n) = 1 if and only if k is a unit in the ring Zn.

Problem 1.2. Show that there exists a Galois extension of Q of degree p for each primep. State precisely all results which are needed to justify your answer.

Solution: The solution that was here before was incorrect. xp− a is irreducible over Qby Eisenstein’s Criterion (and hence separable since Q has characteristic 0) for any a whichis square–free. Now let E be the splitting field of such an xp − a.

• Roots of Unity: The (complex) nth roots unity form a cyclic group Cn. The generatorsof Cn are the nth primitive roots of unity. These are roots of the form µk = e(2πk/n)i

for (n, k) = 1.

• Algebraic/Transcendental Elements: Let E/F be fields. An element u ∈ E is algebraicover F if there exists f ∈ F [x] such that f(u) = 0; otherwise u is transcendental. Ifu is transcendental over F , then

F [u] ∼= F [x], where F [u] = {f(u) | f ∈ F [x]}

• Irreducible Polynomial Finite Extension: If h ∈ F [x] is irreducible of degree n:

(i) x+ (h) is a root of h in E = F [x]/(h) = F [x+ (h)]

(ii) E is a field. (This is NOT true if h is not irreducible!)

(iii) E/F is finite and [E : F ] = n.

2

• Minimal Polynomial: If u ∈ E is algebraic over F , then

I ≡ {f ∈ F [x] | f(u) = 0}

is an ideal in the PID F [x]. The generator of I is called the minimal polynomial of uand denoted mu.

(i) mu is irreducible.

(ii) The degree of mu is called the degree of u over F .

• Algebraic ⇐⇒ Finite: An element u ∈ E is algebraic over F if and only if F [u] is afinite–dimensional vector space over F . In this case F [u] is a field and

[F [u] : F ] = deg(mu).

More precisely, we haveF [u] ∼= F [x]/(mu)

andF [u] ∼= F (u).

As a consequence, every finite field extension is algebraic (E/F is algebraic if everyelement of E is algebraic over F : E/F finite implies that for all u ∈ E, E[u] is finiteand hence u is algebraic).

• Cyclotomic Extension (prime case): Let p be a prime number. Consider the polyno-mial over Q

xp − 1 = (x− 1)(xp−1 + xp−2 + · · ·+ x+ 1) ≡ (x− 1)Φp,

the roots of which are the pth roots of unity. If ξ is a primitive root of unity, thenQ(ξ) would be a splitting field of Φp (since it will contain all pth roots of unity). Sincep is prime, any root of Φp is a primitive root of unity. Φp is irreducible:

Over Zp[x], xp−1 = (x−1)p, so [Φp]p = (x−1)p−1. If Φp = gh, then [Φp]p = [g]p[h]p =(x− 1)k(x− 1)l, with k, l > 0, and k + l = p− 1. But then [g(1)]p = [g]p(1) = 0 andsimilarly [h(1)]p = 0 so that Φp(1) = g(1)h(1) is divisible by p2, which is impossiblesinec Φp(1) = p.

From this we conclude that [Q(ξ) : Q] = p− 1.

• Automorphism Groups: Let E/F be fields. The automorphisms of E over F (i.e.automorphisms of E which fixes F ) form a group under composition denoted AutFE.

– |AutFE| ≤ [E : F ]

3

– EG = F if and only if |G| = [E : F ], where G ⊂ AutFE is a subgroup and

EG = {u ∈ E | σ(u) = u, ∀σ ∈ G}.

• Galois Extension: A finite extension E of a field F is a Galois extension if

|AutFE| = [E : F ].

AutEF is called the Galois group of the extension E and is denoted Gal(E/F ).

• Separable Galois: Let f ∈ F [x] be such that all its irreducible factors are separable.Then its splitting field is a Galois extension of F .

• Galois Correspondence: Let E/F be Galois and let G = Gal(E/F ). We have

Φ : {subfields of E containing F} −→ {subgroups of G} : K 7→ GK ,

where GK = {g ∈ G | g |K= id} ⊂ G, is a bijection, with inverse

Ψ : {subgroups of G} −→ {subfields of E containing F} : H 7→ EH ,

where EH = {α ∈ E | σ(α) = α, ∀h ∈ H} ⊂ F . Also

|GK | = [E : K],

[E : EH ] = |H|.

Finally, a Galois extension of F contained in E corresponds to a normal subgroup ofG and vice versa.

Problem 1.3. Let α =√i+ 2 where i =

√−1.

(a) Compute the minimal polynomial of α over Q.(b) Let F be the splitting field and compute the degree of F over Q;(c) Show that F contains 3 quadratic extensions of Q;(d) Use this information to determine the Galois group.

Solution: (a) α2(α2 − 4) = (i+ 2)(i− 2) = −5, so α is a root of

f ≡ x4 − 4x2 + 5.

We claim f is irreducible and hence f = mQ(α): Making a change of variable y = x2, weknow that (2 + i) is a root of the polynomial y2 − 4y + 5, so it must be the case that itscomplex conjugate (2− i) is also a root, so

f = (x2 − (2 + i))(x2 + (2− i)) = (x− (√

2 + i))(x+ (√

2 + i)(x− (√

2− i))(x+ (√

2− i).

4

If f = gh, then the degree of g and h must add up to 4, but if say g is linear, then it is clearfrom the above that g /∈ Q[x], so it must be the case that both g and h are quadratic. Butit is equally easy to see that any product of two linear factors also does not lie in Q[x].

(b) The splitting field of mQ(α) can be found by adjoining all the roots, hence is equalto

K ≡ Q(√

2 + i,√

2− i).

We have that [K : Q] = [K : Q(√

2 + i)][Q(√

2 + i) : Q]. By (a), the latter is equalto the degree of mQ(α) and hence is 4. Next observe that K = [Q(

√2 + i)](

√2− i),

so we look for the minimal polynomial of√

2− i over Q(√

2 + i). To this end note that(√

2 + i)(√

2− i) =√

5 so√

2− i is a root of

((√

2 + i)x)2 − 5 = ((√

2 + i)x−√

5)((√

2 + i)x+√

5),

which is irreducible. So [K : Q(√

2 + i)] = 2 and [K : Q] = 8.(c) We again note that (

√2 + i)(

√2− i) =

√5 and 1

2 [(√

2 + i)2 − (√

2− i)2] = i, sothat

√5, i

√5 and i are all in K. These are all quadratic extensions of Q since they satisfy

polynomials x2−5 = (x−√

5)(x+√

5), x2+5 = (x−i√

5)(x+i√

5) and x2+1 = (x−i)(x+i),respectively, all of which are irreducible of degree 2 (they are irreducible since they clearlyhave no root in Q).

(d) mQ(α) is separable so K/Q is Galois. Let G = Gal(K/Q). Then |G| = [K : Q] = 8.By part (c), there are 3 quadratic extensions. In addition, all 3 extensions are in fact alsoGalois over Q (since they are all splitting fields of the corresponding minimal polynomials,all of which are separable). By the Galois Correspondence, these correspond to 3 distinctnormal subgroups of G of index 2 (hence order 4). There are 3 abelian groups of order 8:Z/8Z, Z/2Z × Z/2Z × Z/2Z and Z/2Z × Z/4Z, none of which has 3 subgroups of order4. This leaves us with D4 and Q8. The quarternion group has only one subgroup of order2, namely 〈−1〉, but again by the Galois Correspondence, G has at least 2 subgroups oforder 2, corresponding to Q(

√2 + i) and Q(

√2− i). So we are finally left to conclude that

G = D4.

• Irreducibility – Degree 2 or 3: A polynomial of degree 2 or 3 is reducible if and onlyif it has a root in the field in question (hence irreducible if it has no root).

• Polynomials Over Q Galois: The splitting field of any polynomial over Q is Galois.

2 Winter 2002 – Fields

Problem 2.1. The discriminant of the special cubic polynomial f(x) = x3 +ax+b is givenby −4a3 − 27b2. Determine the Galois group of the splitting field of x3 − x+ 1 over

5

(a) F3, the field with 3 elements.(b) F5, the field with 5 elements.(c) Q, the rational numbers.

Solution: (a) Since x3 − x+ 1 is a cubic, it is reducible if and only if it has a root. Aquick check shows that it has no root in F3 so it is irreducible over F3, hence separable andgenerates a Galois extension of degree divisible by 3 over F3 (the extension generated by asingle root has degree 3). Therefore there are only two possibilities for the Galois group:A3 or S3. The Galois group is A3 if and only if the discriminant is the square of an elementin F3 (the Galois group is contained in An if and only if each σ ∈ G fixes

√D if and only

if√D ∈ F since F is the fixed field of the Galois group). In this case the discriminant is

calculated to be−4 · (−1)3 − 27 = −23 = 1 = 12 in F3.

So the Galois group is indeed A3. (Or just note that over a finite field all finite extensionsare cyclic, so the Galois group cannot be S3 since it is not cyclic.)

(b) In this case a check shows that 3 is a root of x3 − x + 1 over F5 and there are noother roots. So the polynomial splits into a linear term and an irreducible quadratic term.This generates a quadratic extension and the Galois group is Z/2Z (the order of the Galoisgroup is equal to the degree of the extension).

(c) By the Rational Root Test, the only possible roots are ±1 neither of which are roots,so x3 − x+ 1 is irreducible over Q. The discriminant is −23 which is not a square in Q, sothe Galois group is S3.

• Polynomials Over Finite Fields Galois: The splitting field of any polynomial overa finite field is Galois (the splitting field of a polynomial f is the splitting field of theproduct of irreducible factors of f , which is separable).

• Characterization of Galois (Polynomial): An extension E/F is Galois if and only ifE is the splitting field of some separable polynomial over F . If this is true, everyirreducible polynomial in F [x] which has a root in E is separable and has all its rootin E.

• Characterization of Galois (Fixed Field): E/F is Galois if and only if the fixed fieldof AutEF is exactly F (in general it is bigger).

• Finite Extensions of Finite Fields are Cyclic: If F is a finite field, E/F is finite,then E/F is Galois with cyclic Galois group (E× is cyclic so let θ be a generator,then E = F (θ) and [E : F ] = deg(mF (θ)). By uniqueness of finite fields and theirreducibility of mF (θ), mF (θ) must split in E, so that E/F is Galois. Assuming for

6

simplicity that F = Fp, some prime p, we see that the Galois group is then generatedby the Frobenius automorphism).

• Permutation of Roots: Let E/F be a field extension and α ∈ E algebraic over F . Ifσ ∈AutEF , then σ(α) is a root of mF (α) (this is obvious since σ fixes F ).

• Gal(E/F ) ↪→ Sn: Let E/F be Galois. Then E is the splitting field of some separablef ∈ F [x] with n roots. So any σ ∈ Gal(E/F ) defines a permutation of the n roots.

• Discriminant: The discriminant of a polynomial with roots x1, . . . , xn is given by

D =∏i<j

(xi − xj)2.

• Gal(E/F ) ↪→ An?: The Galois group G of f ∈ F [x] (the Galois group of a polynomialover F is the Galois group of the splitting field of f over F ) is a subgroup of An ifand only if the discriminant D ∈ F is the square of an element of F .

• Rational Root Test: Suppose P (x) = anxn+· · ·+a1x+a0 is a polynomial with integer

coefficients, and x = p/q is a rational root of P (x), then

p | a0 and q | an.

Problem 2.2. 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/2Z×Z/2Z and list all sub–extensions.

(b) Prove that if K/Q is a normal extension of degree 4 with Gal(K/Q) ∼= Z/2Z×Z/2Zthen K/Q is biquadratic.

Solution: (a) First note that a 6= b since otherwise [K : Q] = 2, which implies thatf ≡ (x2 − a)(x2 − b) is separable. K is clearly the splitting field of f so K/Q is normal.Next note that Q(

√a) ⊂ Q(

√a,√b) corresponds to a subgroup H ⊂ Gal (K/Q) of order 2

by the Galois Correspondence. Similarly, by considering Q(√b), there is another subgroup

H ′ of order 2. H 6= H ′ by the Galois Correspondence, so the Galois group cannot be Z/4Z(which has only one subgroup of order 2). The only other possibility is Z/2Z× Z/2Z.

To describe the subfields, let σ, τ ∈ Gal(K/Q), with

σ :√a 7→ −

√a,√b 7→

√b

andτ :

√a 7→

√a,√b 7→ −

√b.

7

Then we see that στ = τσ, so

Gal(K/Q) = {σ, τ | σ2 = τ2 = e, τσ = στ}.

The subfields are thenK〈σ〉,K〈τ〉,K〈στ〉,

where KH = {k ∈ K | σ(k) = k,∀σ ∈ H}. These are equal to

Q(√b),Q(

√a),Q(

√ab).

(b) By the Galois Correspondence, K contains two distinct quadratic extensions of Q.These extensions are Galois since Gal(K/Q) is abelian so they must be splitting fields ofirreducible quadratics, say Q(

√a) and Q(

√b) (here a 6= b are the discriminants of the

respective quadratics). [Q(√a)](

√b)/Q(

√a) = Q(

√a,√b)/Q(

√a) has degree 1 or 2, but

cannot be 1 since a 6= b, so Q(√a,√b)/Q has degree 4, which implies K = Q(

√a,√b).

• Normal Extension: If E/F is algebraic, then E is a normal extension of F if E is thesplitting field over F of a collection of polynomials f ∈ F [x].

Problem 2.3. Let K be a finite extension of the field F with no proper intermediatesubfields.

(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, and

justify your answer.

Solution: (a) Let E be the fixed field of AutFK. Since there are no proper inter-mediate fields, E = F or E = K. If E = F , then K/F is Galois and by the GaloisCorrespondence, each subgroup of the Galois group corresponds to a subfield and we aredone by Cauchy’s Theorem: If the degree is n not a prime, then ∃p < n, p prime, suchthat p | n, Cauchy’s theorem says there is an element g of order p, but then the subfieldcorresponding to 〈g〉 would be a proper intermediate field, a contradiction.

If E = K, then AutFK is trivial. If F were a perfect field, then it is the case that anyirreducible polynomial is separable. K/F is normal so it is the splitting field of a collectionof polynomials over F . If f is one of the polynomials and α is a root of f which does notlie in F , then F (α) = K, which implies that K is the splitting field of mF (α), which isirreducible, hence separable, so that K/F is Galois and we are done as before. We maytherefore assume that F is not perfect and so in particular char(F ) = p, where p is a prime.

With α as before, we have that [K : F ] = deg(mF (α)) and that α is the only root of fwhich does not lie in F (if β were another one, F (α) = F (β) = K and there would be an

8

automorphism of K sending α to β and fixing everything in F , but since AutFK is trivial,we conclude that α = β). In the splitting field K, we have

mF (α) = (x− α)(x− α1) . . . (x− αn−1).

Since mF (α) is irreducible, none of the αi’s is in F , so it must be the case that αi = α, ∀i.Therefore

mF (α) = (x− α)n.

This is a polynomial over F so in particular αn ∈ F . If αk ∈ F for any k < n, thenmF (α) | (xk − αk) would be of lower degree, a contradiction. Since

(x− α)n =n∑

k=0

(n

k

)xkαn−k,

it must be the case that p |(nk

), 1 ≤ k < n. In particular p | n so n = γp, some γ. Now we

haveF——F (αp)——F (α) = K.

Since there are no intermediate fields, F (αp) = F (α) = K, so that γ = 1 and [K : F ] = p.(b) Let F be a field and let α be an element of order 4 over F such that the splitting

field L of mF (α) has S4 as its Galois group. Let E = F (α), then [E : F ] = 4 which is notprime and [L : F ] = 4! = 24. We claim there is no intermediate field between E and F .

By Galois Correspondence, E corresponds to some subgroup H ⊂ Gal(L/F ) ∼= S4

such that [S4 : H] = 4. Showing that there is no intermediate field is then equivalent toshowing that H ⊂ S4 is maximal. To this end, observe that A4 has no subgroup of index2: |A4| = 24/2 = 12 so a subgroup of index 2 is a group of order 6, and hence isomorphicto either Z/6Z or S3. It cannot be S3 since S3 contains odd permutations. It also cannotbe Z/6Z since S4 does not contain any element of order 6: First notice that S4 does notcontain any 6–cycles. Next if σ is an element of order 6, then writing σ = τ1 . . . τn, wherethe τi are disjoint cycles, we see that the order of σ is equal to the least common multipleof the τi’s. Cycles in S4 can have lengths 1, 2, 3 or 4, so the only possibility is a cycle oflength 2 and a cycle of length 3, but these cannot be disjoint (there are only 4 letters topermute), a contradiction.

Next observe that A4 is the only subgroup of index 2 in S4 (more generally, An is theonly subgroup of index 2 in Sn: Notice that any subgroup A of index 2 is normal andhence Sn/A ∼= Z/2Z. Therefore σ2A = A,∀σ ∈ Sn ⇒ σ2 ∈ A,∀σ ∈ Sn. This in particularimplies that all three cycles are in A, since if τ is a three cycle, then τ2 = τ−1. But An isgenerated by three cycles, so A = An.) This shows that any subgroup of S4 of index 4 mustbe maximal: Suppose [Sn : A] = 4 and A is not maximal, then ∃A ⊂ B a subgroup suchthat [B : A] > 1. So 4 = [Sn : A] = [Sn : B][B : A] > [Sn : B] ⇒ [Sn : B] = 2 ⇒ B = An

9

which is a contradiction, because then [An : A] = 2. We conclude that H is maximal sinceit has index 4.

• Perfection: A field of characteristic p is called perfect if every element of K is a pth

power in K, i.e. K = Kp. Any field of characteristic 0 is also called perfect. Any finitefield is perfect, as is Q. Every irreducible polynomial over a perfect field is separable.

3 Fall 2002 – Fields

Problem 3.1. a) Determine the minimal polynomial of u =√

3 + 2√

2 over Q.b) Determine the minimal polynomial of u−1 over Q.

Solution: a) u2 = 3 + 2√

2, so (u2 − 3)2 = 8 and therefore u is a root of

f ≡ x4 − 6x2 + 1.

Making the change of variable y = x2 and using the quadratic equation, we see that fcannot factor as a product of two quadratics. On the other hand, the rational root testshows that it is impossible for f to factor as a linear term times a cubic. We conclude thatf is irreducible so f = mQ(u).

b) (u−1)2 = 13+2

√2. So 3(u−1)2 +2

√2(u−1)2 = 1. Subtracting and squaring to get rid of

√2, we see that u−1 is also a root of x4 − 6x2 + 1, which we have already determined to be

irreducible, so mQ(u−1) = x4 − 6x2 + 1. (Or just note x4( 1x4 − 6

x2 + 1) = x4 − 6x2 + 1.)

Problem 3.2. a) Let F be the field generated by the roots of the polynomial X6 + 3 overQ. Determine the Galois group F/Q.

b) Describe all subfields of F .

Solution: a) Let ξ = e2πi/6. Then the roots of X6 + 3 are ξi 6√−3, 1 ≤ i ≤ 6. So

F = Q(ξ, 6√−3). Now notice that

ξ = eπi/3 =12

+ i

√3

2∈ Q( 6

√−3),

since ( 6√−3)3 =

√−3 = i

√3. So we conclude F = Q( 6

√−3). By Eisenstein’s Criterion,

X6 + 3 is irreducible and so it is the minimal polynomial of 6√−3 and therefore [F : Q] =

[Q( 6√−3 : Q] = 6. This extension is Galois (since X6 + 3 has 6 distinct roots, hence

separable) so let G denote the Galois group. Then we have that |G| = 6, which implies thatG is either S3 or Z/6Z (these are the only groups of order 6). To decide which one it is,

10

notice that ( 6√−3)2 = (−3)1/3 = − 3

√3, so 3

√3 ∈ F . Consider Q( 3

√3)/Q. This extension is

not Galois, since 3√

3 is a root ofX3−3 but Q( 3√

3) does not contain any other roots ofX3−3,which is also irreducible by Eisenstein. By the Fundamental Theorem of Galois Theory,there is a correspondence between subfields which are Galois over Q and normal subgroupsof G. We therefore conclude that G cannot be abelian since it contains a subgroup whichis not normal (the subgroup fixing the subfield Q( 3

√3), so the only possibility is G = S3.

(If we had wished to be more explicit, we could have used the fact that an element ofthe Galois group maps a root to a root to make computations. First it is clear that G ⊂ S6.Since 6

√−3 generates Q( 6

√3) over Q, any σ ∈ G is completely determined by σ( 6

√−3). Label

the roots 1 through 6 such that ξi 6√

3 is labeled i. Let G = {σ1, σ2, . . . , σ6} such that σi

maps 6√−3 to the ith root. Consider σ1 : 6

√−3 7→ ξ 6

√3. We will know everything about σ1

once we figure out where it maps ξ = 12 +

√−32 . To this end notice we calculate:

σ1(√−3) = σ1((

6√−3)3) = ξ3

√−3 = −

√−3 ⇒ σ

(12

+√−32

)=

12−√−32

= ξ5.

From this we see that σ21(ξ

6√−3) = σ1( 6

√−3) = σ(ξ5 · ξ 6

√−3) = (ξ5)5 6

√−3 = ξ 6

√−3 (since

ξ6 = 1). So we conclude that 〈σ6〉 is a subgroup of order 2. On the other hand, it is clearthat σ3 : 6

√−3 7→ ξ3 6

√−3 = − 6

√−3 also has order 2. Therefore G contains two subgroups

of order 2 and hence cannot be cyclic (a cyclic group of order n contains a unique subgroupof order d for each d | n), so G = S3.)

b) By the Fundamental Theorem of Galois Theory, subfields of F corresponds to sub-groups of G = S3 in an order reversing way. S3 has 3 subgroups of order 2, generated bythe three transpositions:

{e, (12)}, {e, (13)}, {e, (23)}.

These correspond to the three subfields of F which has degree 6/2 = 3 over Q:

Q(ξ2 3√−3),Q(ξ4 3

√3),Q( 3

√3).

S3 has one subgroup of order 3 (3 is prime so this subgroup is cyclic. The only elements inS3 of order 3 are the cyclic permutations of order 3), namely

A3 = {e, (123), (123)2},

which correspond to the subfield of F which has degree 2 over Q:

Q(√−3).

There are no other non–trivial subfields.

Suppose now instead we want to find the Galois group of X6−3: Let ξ denote a primitive6th root of unity and α = eπi/6 6

√3. Then F = Q(ξ, α). We have [F : Q] = [F : Q(α)][Q(α) :

11

Q]. X6 − 3 is irreducible over Q by Eisenstein’s Criterion, so it is the minimal polynomialof α and therefore [Q(α) : Q] = 6. The minimal polynomial of ξ over Q is x2 − x + 1 =(x− eiπ/3)(x− e5πi/3). Since the degree of the polynomial is 2 and neither of the root is inQ(α), x2 − x + 1 remains irreducible over Q(α) and we conclude that [F : Q(α)] = 2 and[F : Q] = 12.

The subextension Q(ξ)/Q is Galois (it is the splitting field of x2 − x+ 1 over Q), so bythe Galois Correspondence this corresponds to a normal subgroup H of Gal(F/Q) of order6 fixing Q(ξ), which can be either Z/6Z or S3 (notice that If E/K/F is such that E/F isGalois, then E/K is also Galois). The roots of X6 − 3 are

α, ξα, ξ2α, ξ3α, ξ4α, ξ5α.

Any σ ∈ AutQ(ξ)F = Gal(F/Q(ξ) must map a root to a root and hence is determined byσ(α) (remember that σ fixes ξ, so if σ(α) = ξmα, then σ(ξkα) = ξkσ(α) = ξm+kα, wherem and k are defined modulo 6). We see that γ : AutQ(ξ)F −→ Z/6Z by σ 7→ n whereσ(α) = ξnα is an isomorphism (If σ and τ are such that σ(α) = ξmα and τ(α) = ξnα,then στ(α) = ξm+nα and στ(ξkα) = ξkστ(α) = ξm+n(ξkα), so γ(στ) = γ(σ) + γ(τ)) andtherefore H ∼= Z/6Z. By considering Q(α)/Q, we also see that there is a subgroup H2 oforder 2. Therefore Gal(F/Q) ∼= H oH2, which gives us two possibilities: Z/6Z× Z/2Z orD6 (symmetries of a hexagon). Since Q(α) is not normal, Gal(F/Q) cannot be abelian andwe are forced to conclude that Gal(F/Q) = D6 (symmetries of a regular hexagon).

• Eisenstein’s Criterion: Let p be a prime in Z and let

f(x) = xn + an−1xn−1 + · · ·+ a1x+ a0 ∈ Z[x], n ≥ 1.

Supposep | ai, 0 ≤ i < n but p2 - a0.

Then f(x) is irreducible in both Z[x] and Q[x]. Sometimes a change of variable isuseful if this does not apply directly, e.g. if one can show that the polynomial withy = x+ 1 is irreducible, then the polynomial must also be irreducible in the originalvariable(just replace x by y in the factorization if it were reducible).

• Trivial Observation: If E/K/F is such that E/F is Galois, then E/K is also Galois.

• Dihedral Group: The dihedral group Dn is the group of symmetries of a regular n–gon. A regular n–gon has 2n symmetries: n rotations and n reflections. A quickobservation is then that there are n subgroups of order n if n is odd, correspondingto the n reflections and n+ 1 subgroups of order 2 if n is even, corresponding to then reflections plus rotation by π radians about the origin. The presentation for Dn is

Dn = {r, s | rn = s2 = 1, rs = sr−1},

12

where r is rotation counterclockwise about the origin through 2π/n radians and s isone of the reflections.

• Semidirect Product:

– Let G be a group. Then G decomposes as a semidirect product of subgroups Nand H, write G = N oH, if(i) N CG; (ii) N ∩H = {e}; (iii) NH = G.(i) ensures that NH = HN is a group and (ii) and (iii) says every element canbe written uniquely as nh with n ∈ N , h ∈ H.

– On the other hand, if there are groups H and N such that there is a homomor-phism

α : H −→ Aut(N),

then N ×H is a group with multiplication given by

(n1, h1)(n2, h2) = (n1α(h1)n2, h1h2).

Denote this group by N oα H. We have

N C (N oα H),

where N = {(n, e) | n ∈ N}. Define H = {(e, h) | h ∈ H}, then N∩H = {(e, e)}.We see that

N oα H = NH

as a semidirect product.

– If G splits as a semidirect product of N and H, define

Γ : H −→ Aut(H) : h 7→ conjugation by h,

thenN oΓ H −→ G : (n, h) 7→ nh

is a group isomorphism.

– Now suppose G is a finite group, N C G, N ∩H = {e} and |N ||H| = |G|, thenit is the case that NH = G and so G splits as a semidirect product of N and H.Explicitly, we have that

G = NH ∼= N oΓ H,

where Γ is conjugation by elements of H.

13

∗ We can in fact determine all possible isomorphism classes of NH. NH lookslike N oΓ H, where

(n1, h1)(n2, h2) = (n1Γ(h1)(n2), h1h2) = (n1(h1n2h−11 ), h1h2).

Each Γ(h) is an automorphism of N . Note that by associativity and the factthat Γ is a homomorphism

(n1, h1)(n2, h2)(n3, h3) = (n1Γ(h1)(n2)Γ(h1h2)(n3), h1h2h3)= (n1Γ(h1)(n2Γ(h2)(n3)), h1h2h3)= (n1Γ(h1)(n2)Γ(h1)(Γ(h2)(n3)), h1h2h3),

soΓ(h1h2) = Γ(h1) ◦ Γ(h2)

and thereforeα : H −→ Aut(N) : h 7→ Γ(h)

is a homomorphism. This shows that each isomorphism class of NH arosefrom some homomorphism α : H −→ Aut(N). Conversely, given α : H −→Aut(N), we can form N oα H and we have as before

N oα H ∼= NH ∼= NH,

where N = {(n, 0) | n ∈ N} and H = {(0, h) | h ∈ H}.∗ Note that Γ is given by

Γ(h)(n) = hnh−1 = α(h)(n).

∗ It is not the case that different α’s necessarily lead to distinct isomorphismclasses.

Problem 3.3. Let p be a prime integer such that p ≡ 2 or 3 (mod 5). Prove that thepolynomial

1 +X +X2 +X3 +X4

is irreducible over Z/pZ.

Solution: 1 + X + X2 + X3 + X4 = X5−1X−1 ≡ Φ5(X), the 5th cyclotomic polynomial

and has as roots all the primitive 5th roots of unity. If ξ is a primitive 5th root of unity,we would like to find the least n such that Fpn contains ξ. It will be enough to show thatn = 4 (if Φ5(X) = f(X)g(X) were reducible, then say η is a root of f which has degreek, then η is a primitive root of unity and Fp(η) would be isomorphic to Fpk with k < 4, acontradiction).

14

First suppose n is such that ξ ∈ Fpn . (Fpn)× is cyclic so let θ be a generator. Thenθpn−1 = 1 and θ raised to a lower power is not equal to 1. If ξ ∈ Fpn , then (∃k ∈ N)(θk = ξ)so

(θk)5 = 1 in Fpn and (θk)l 6= 1 in Fpn for l < 5.

This implies that5k = α(pn − 1), some α ∈ N.

If α | 5 then α = 5 since 5 is prime so we conclude k = pn − 1 and ξ = 1, a contradiction.If α | k, then we can cancel α on both sides of the equation to get that

5k′ = (pn − 1) =⇒ pn ≡ 1 (mod 5).

Conversely, suppose pn ≡ 1 (mod 5). Then pn = 5k + 1, some k ∈ N. Again let θ be agenerator of (Fpn)×. Then (θk)5 = θpn−1 = 1 and so θk is a primitive 5th root of unity.The roots of unity form a cyclic group, so that if Fpn contains one primitive root of unityit must contain all of them, which implies that ξ ∈ Fpn .

Now let n be the order of p in F5, i.e. n is minimal such that pn ≡ 1 (mod 5). Then fromthe above we know that n is also the minimal n such that ξ ∈ Fpn . For both p ≡ 2 (mod 5)and p ≡ 3 (mod 5) we have that n = 4 so Φ5 is irreducible over Fp (here we use the factthat Z/nZ is a ring so that a1 = b1, a2 = b2, then a1 + a2 = b1 + b2 and a1a2 = b1b2).

• Factorization of Φq over Fp: Let p 6= q be primes. If n is the least integer such thatpn ≡ 1 mod q, then the minimal polynomial of ξq has degree n over Fp. Moreover, Φq

is the product of q−1n distinct irreducible polynomials of degree n in Fp.

• Z/nZ: Z/nZ = {0, 1, . . . , n− 1} form a commutative ring with 1 under addition andmultiplication modulo n (these operations are of course both well–defined).

4 Winter 2003 – Fields

Problem 4.1. Let Fq be the finite field of q elements. Answer the following questions:(a) List all subfields of Fp6 for a prime p. Justify your answer.(b) Find a formula for the number of monic irreducible polynomials of degree 6 in Fp[X].

Justify your answer.

Solution: (a) Fp6 is the splitting field of xp6 − x over Zp (xp6 − x is separable sincethe derivative is -1 and has no root and therefore has p6 roots, so by order counting, Fp6 iscontained in the splitting field. On the other hand, each element in F×

p6 satisfies xp6−1 = 1

and so each element in Fp6 is a root of xp6 − x and we conclude the splitting field is equal

15

to Fp6). The Galois group Gal(Fp6/F)p) is generated by the Frobenius automorphism andis cyclic of order 6, hence isomorphic to Z/6Z. The two non–trivial subgroups of Z/6Z,namely Z/2Z and Z/3Z correspond to intermediate fields between Fp6 and Fp of degree(over Fp) 3 and 2, respectively. These are the unique splitting fields of xp3 − x and xp2 − xover Fp, namely Fp3 and Fp2 , respectively.

(b) Let f be an irreducible polynomial over Fp of degree d | 6. If α is a root of f , thenFp(α)/F has degree d and hence is equal to Fpd by uniqueness of splitting fields. Fpd/Fp isGalois, so Fpd(⊂ Fp6) in fact contains all roots of f . Since Fp6 is equal to the set of rootsof xp6 − x, it must be the case that f | xp6 − x. Now let g be the product of all (monic)minimal polynomials of elements in Fp6 with no repeats, then it follows that g | xp6 − x.Conversely, it is clearly that xp6−x | g and so we conclude g = xp6−x. Since any irreduciblepolynomial of degree d is the minimal polynomial of (each of) its roots, we see that in factxp6 − x is the product of all irreducible polynomials over Fp of degree d | 6.

Now let ψ(d) = number of irreducible polynomials of degree d over Fp. Then (since thedegree of xp6 − x is p6) we have

p6 =∑d|6

dψ(d).

By the Mobius Inversion Formula, we get

6ψ(6) =∑d|6

µ(d)p6/d,

where µ(1) = 1, µ(n) = 0 if n contains a square factor, and µ(n) = (−1)r if n has r distinctprime factors. Calculating, we get

6ψ(6) = µ(1)p6 + µ(2)p3 + µ(3)p2 + µ(6)p

= p6 − p3 − p2 + p.

So ψ(6) = 16(p6 − p3 − p2 + p).

Instead of using the Mobius Inversion Formula we could have proceeded as follows. Firstnote that if f is an irreducible polynomial over Fp, then f is separable. Say the degree of fis 6, then f has 6 distinct roots and it is the case that none of these roots can lie in Fp2 orFp3 (if α is a root of f , then f = mFp(α) has degree 6. If say α ∈ Fp2 , then the degree of αover Fp would be 2, a contradiction). Now observe that Fp2 and Fp3 intersect in Fp since 2and 3 are relatively prime, so inclusion–exclusion gives that there are

p6 − p3 − p2 + p

elements of degree 6 over Fp in Fp6 . It suffices to count the number of minimal polynomialsof these elements of order 6 (as before if f is irreducible of degree 6, then it is the minimal

16

polynomial of (each of) its roots). Each such minimal polynomials has 6 distinct roots,so there must be 1

6(p6 − p3 − p2 + p) monic irreducible polynomials over Fp of degree 6.(Explicitly, pick an element α1 of degree six, write down its minimal polynomial f1. Nowpick some α2 which is not equal to any of the roots of f1 and write down its minimalpolynomial f2. Continue this way, we will eventually exhaust all elements of degree 6and will have found all irreducible polynomials of degree 6. Notice this implies that 6 |p6 − p3 − p2 + p.)

• xpn − x ⇔ Fpn : The finite field Fpn is precisely the set of roots of the polynomialxpn − x over Fp.

• xpn − x: The polynomial xpn − x over Fp is in fact the product of all irreduciblepolynomials of degree d | n over Fp.

• Mobius Inversion Formula: Let f(n) be a function defined for all nonnegative integersn and let

F (n) =∑d|n

f(d).

Then we havef(n) =

∑d|n

µ(d)F (n

d),

where µ(1) = 1, µ(n) = 0 if n has a square factor, and µ(n) = (−1)r if n has r distinctprime factors.

• Splitting Fields of Irreducible Polynomials over Fp: Let f be irreducible of degree nover Fp. Then the splitting field of f is exactly Fpn .

Problem 4.2. Let K/F be a quadratic extension of fields and M/F be a Galois extensionover F containing K such that Gal(M/K) is a cyclic group of odd prime order p. Answerthe following two questions:

(a) Determine the possible groups Gal(M/F ) up to isomorphisms, and justify youranswer.

(b) Find the number of intermediate fields L between F and M with [L : F ] = p. Justifyyour answer.

Solution: (a) Let G = Gal(M/F ). Then

|G| = [M : F ] = [M : K][K : F ] = 2p.

Let H = Gal(M/K). First we claim H CG: H has order p so [G : H] = 2. Now let G acton the cosets of H by left multiplication. This induces

λ : G −→ S2 : g 7→ λg : aH 7→ haH.

17

We have ker(λ) =⋂

g∈G gHg−1 and by the First Isomorphism Theorem and Lagrange’s

Theorem, |G/ker(λ)| | |S2| = 2, which forces |ker(λ)| = p and therefore H = ker(λ) C G.Next by Cauchy’s Theorem, there is a subgroup H2 ⊂ G of order 2. Hence

G = HH2∼= Z/pZ o Z/2Z

as a semidirect product. This gives us two possibilities: Z/pZ × Z/2Z or Dp (symmetriesof a p–gon).

(b) By the Galois Correspondence, intermediate field L such that [L : F ] = p correspondto subgroups of G of order 2. If G ∼= Z/pZ × Z/2Z, then there is exactly one subgroup oforder 2 since p is odd. If G ∼= Dp, then there are p subgroups of order 2, correspondingto the p reflections (there is a reflection through each line joining a vertex and the sideopposite to it).

Problem 4.3. Find the degree of the splitting field E of X6 − 3 over the following fields:(a) Q(

√−3) (Q: the field of rational numbers);

(b) F7, the field with 7 elements;(c) F5, the field with 5 elements,and justify your answer.

Solution: (a) X6 − 3 is irreducible over Q by Eisenstein’s Criterion applied with theprime 3. The splitting field is given by

E = Q(ξ, 31/6),

where ξ = e2πi6 is a primitive 6th root of unity. We claim that [Q(ξ, 31/6) : Q] = 12:

First we have that [Q(31//6) : Q] = 6. Next it is clear that the cyclotomic polynomialX2 − X + 1 = Φ6 (which has ξ and ξ5 as roots) remains irreducible over Q(31/6), so[Q(31/6, ξ) : Q(31/6] = 2. Finally, X2 + 3 is irreducible over Q so is the minimal polynomialof√−3. Hence [Q(

√−3) : Q] = 2 and

12 = [E : Q] = [E : Q(√−3]2,

so [E : Q(√−3)] = 6.

(b) First note that 36 ≡ 1 (mod 7). So

x6 − 3 | (x6)6 − 36 = x36 − 1.

F7n is the (unique) splitting field of x7n − x, so if the roots of x6 − 3 are contained in F7n ,then it must be the case that x6−3 | x7n−x = x(x7n−1−1). Since (xm−1) | (xn−1) if andonly if m | n, we now wish to find the least integer m such that 7m ≡ 1 (mod 36). To this

18

end notice first that 7 ∈ (Z/36Z)× ∼= (Z/22Z)×× (Z/32Z)× ∼= Z/2Z×Z/6Z, by the ChineseRemainder Theorem. So 7 has either order 2 or order 6 in (Z/36Z)×. It clearly does nothave order 2 so it must be the case that 76 ≡ 1 (mod 36). So x36 − 1 | x76−1 − 1 | x76 − xand hence all roots of x6 − 3 are contained in F76 which implies that E ⊂ F76 and so[E : F] = 1, 2, 3, or 6 (since it must divide [F76 : F7] = 6).

E 6= F7 since a6 = 1 6= 3 for all a ∈ F7.E 6= F72 since a72−1 = a48 = (a6)8 = 1 for all a ∈ F72 , but if a is a root of x6 − 3, then

a6 = 3 so (a6)8 = 38 = 32(36) ≡ 2 (mod 7) (we have used the fact that 36 ≡ 1 (mod 7) andis not equal to 1.

E 6= F73 since a73−1 = a342 − 1 = (a6)57 − 1 = 1 for all a ∈ F73 , but if a is a root ofx6 − 3, then a6 = 3 so (a6)57 = 357 = (33) = 27 = 6 and is not equal to 1.

We are therefore left to conclude that E = F76 and [E : F7] = 6.(c) By the same reasoning as in (b), we calculate:

x6 − 3 | (x6)4 − 34 = x24 − 1 | x25 − x = x52 − x,

so all the roots of x6 − 3 are contained in F52 . Hence E ⊂ F52 . x52 − x has distinct rootssince it has derivative −1 so x6 − 3 must also have distinct roots since it divides x52 − x.Therefore x6 − 3 must have 6 roots and so E 6= F5 (which only has 5 elements) and weconclude E = F52 and [E : F5] = 2.

• Important Fact About Finite Fields: Every a ∈ Fpn (p a prime integer) satisfiesapn

= a and every 0 6= a ∈ Fpn satisfies apn−1 = 1.

• Cyclotomic Polynomials: The nth cyclotomic polynomial Φn(x) is defined to be thepolynomial whose roots are the primitive nth roots of unity:

Φn(x) =∏

a<n; (a,n)=1

(x− ξan),

where ξn = e2πin . Clearly deg(Φn) = φ(n), where φ is the Euler φ–function. Notice

thatxn − 1 =

∏d|n

Φd(x).

This allows us to compute Φn(x) recursively. It can also be shown inductively thatΦn(x) ∈ Z[x].

Note that comparing degrees in the last equation gives

n =∑d|n

φ(d).

19

Applying the Mobius Inversion Formula to n = F (n) =∑

d|n φ(d), we get

φ(n) =∑d|n

µ(d)(nd

).

• Cyclotomic Extension (general case): If we can now show that Φn(x) is irreducible,then we will be able to conclude that

[Q(ξn) : Q] = φ(n).

To this end supposeΦn(x) = f(x)g(x),

with f, g ∈ Z[x], f irreducible. Let ξ be a root of f and p - n a prime. Then ξp is aroot of Φn(x). If g(ξp) = 0, then ξ is a root of g(xp). Reducing mod p and using thefact that f is the minimal polynomial of ξ, we get that

(g(x))p = f(x)h(x),

some h(x) ∈ Z[x]. This implies f and g have a common root, but this implies thenΦn(x) and therefore xn − 1 has a multiple root over Fp. This is a contradiction sincexn− 1 is separable (it has derivative nxn−1 which is relatively prime to xn− 1). So ξp

is a root of f for each p - n. Now given any a such that (a, n) = 1, write a = p1 . . . pk

with pi primes, then we see that ξa = (((ξp1)p2)...)pk is a root of f and therefore allprimitive nth roots of unity are roots of f and we conclude Φn(x) = f(x) is irreducible.The map

(Z/nZ)× −→ Gal(Q(ξn)/Q) : a 7→ σa : σa(ξn) = ξan

is an isomorphism. SoGal(Q(ξn)/Q) ∼= (Z/nZ)×.

• (a− b) | (an − bn):

(a− 1)(an−1 + an−1 + · · ·+ a+ 1) = an − 1

since all the intermediate terms telescope. So

an − bn = bn((a

b

)n− 1

)= bn

((ab

)− 1

) ((ab

)n−1+

(ab

)n−2+ · · ·+

(ab

)+ 1

)= (a− b)(an−1 + an−2b+ · · ·+ abn−1 + bn−1).

20

• (xm−1) | (xn−1) ⇔ m | n: By the previous item, (xm−1) | ((xm)k−1) = xkm−1 forany k ∈ N. Conversely, suppose (xm− 1) | (xn− 1). The roots of xm− 1 are preciselythe mth roots of unity. Let ξ be a primitive mth root of unity. Since (xm−1) | (xn−1),ξ must also be a nth root of unity. By the Division Algorithm, write n = mk + r,where r < n. Then we have that

1 = ξn = ξmk+r = ξmkξr = ξr,

from which we conclude that ξr = 1, but since ξ is a primitive root of unity, it mustbe the case that r = 0 and therefore m | n.

21