5.2 Galois Theory

In this section, L/KL / K always denotes a finite field extension.
Know the BASIC DEFINITIONS AND FACTS: The Galois group of L/KL / K is G(LK):={σAut(L)σ(a)=aG(L \mid K):=\{\sigma \in \operatorname{Aut}(L) \mid \sigma(a)=a for all aK}a \in K\}; recall that we always have |G(LK)|[L:K]|G(L \mid K)| \leq[L: K]; for any subgroup HH of Aut(L)\operatorname{Aut}(L), the fixed field of HH is Fix(H):={αLσ(α)=α\operatorname{Fix}(H):=\{\alpha \in L \mid \sigma(\alpha)=\alpha for all σH}.L/K\sigma \in H\} . L / K is Galois’ iff it is normal and separable iff LL is the splitting field over KK of a separable polynomial fK[x]f \in K[x] iff |G(LK)|=[L:K]|G(L \mid K)|=[L: K] iff Fix(G(LK))=K\operatorname{Fix}(G(L \mid K))=K.

Know Artin’s Theorem: [L:Fix(H)]=|H|[L: \operatorname{Fix}(H)]=|H| for any finite subgroup HH of Aut(L)\operatorname{Aut}(L). Consequence: If L/KL / K is any finite field extension, then |G(LK)||G(L \mid K)| divides [L:K][L: K].

Know the Fundamental Theorem of Galois Theory: If L/KL / K is Galois, the maps MG(LM)M \mapsto G(L \mid M) and HFix(H)H \mapsto \operatorname{Fix}(H) are inverse (and inclusion reversing) bijections between {MM\{M \mid M is an intermediate field, L/M/K}\mathrm{L} / \mathrm{M} / \mathrm{K}\} and {HH\{H \mid H is a subgroup of G(LK)}G(L \mid K)\}. Explicitly:
Fix(G(LM))=M\operatorname{Fix}(G(L \mid M))=M and G(LFix(H))=H;G(L \mid \operatorname{Fix}(H))=H ;
keep in mind that an intermediate field MM corresponds to a subgroup HH of index [M:K][M: K] in G(LK)G(L \mid K), and that |H|=|G(LK)|÷[M:K]=[L:M]|H|=|G(L \mid K)| \div[M: K]=[L: M].

Know the correspondence between normal subextensions and normal subgroups: M/KM / K is normal iff σ(M)=M\sigma(M)=M for all σG(LK)\sigma \in G(L \mid K) iff G(LM)G(L \mid M) is a normal subgroup of G(LK)G(L \mid K); if this is the case, then any element of G(MK)G(M \mid K) can be extended to an element of G(LK)G(L \mid K), and G(MK)G(LK)/G(LM)G(M \mid K) \cong G(L \mid K) / G(L \mid M).

Consequence: If L/KL / K is Galois’ and αL\alpha \in L, then G(LK)G(L \mid K) permutes the roots of the minimal polynomial μαK\mu_{\alpha \mid K} transitively (choose MM to be the splitting field over KK of μαK\mu_{\alpha \mid K} ).

Know that extensions of finite fields are cyclic: An extension L/KL / K of finite fields is always Galois’; if K=𝔽qK=\mathbb{F}_{q} and [L:K]=n[L: K]=n, then G(LK)G(L \mid K) is the cyclic group of order nn generated by the Frobenius automorphism σ\sigma with σ(α)=αq\sigma(\alpha)=\alpha^{q} for all αL\alpha \in L.

Know the basic facts about roots of unity: xn1x^{n}-1 is separable over KK iff char( KK ) does not divide nn (hence always if char(K)=0\operatorname{ch} \operatorname{ar}(K)=0 ); denote by KnK_{n} its splitting field over KK. Then the roots of xn1x^{n}-1 form a cyclic subgroup UnU_{n} of order nn of Kn*K_{n}^{*}; any generator ζn\zeta_{n} of UnU_{n} is called a primitive nth n^{\text {th }} root of unity, and we have Kn=K(ζn).Kn/KK_{n}=K\left(\zeta_{n}\right) . K_{n} / K is Galois’ with G(KnK)G\left(K_{n} \mid K\right) isomorphic to a subgroup of n*\mathbb{Z}_{n}^{*}. In particular, G(KnK)G\left(K_{n} \mid K\right) is ALWAYS ABELIAN but NOT necessarily cyclic. If K=K=\mathbb{Q}, the irreducibility of the cyclotomic polynomials implies G((ζn))n*G\left(\mathbb{Q}\left(\zeta_{n}\right) \mid \mathbb{Q}\right) \cong \mathbb{Z}_{n}^{*}. Hence for n=pmn=p^{m} with an odd prime pp and m,G((ζn))m \in \mathbb{N}, G\left(\mathbb{Q}\left(\zeta_{n}\right) \mid \mathbb{Q}\right) is cyclic ( (p1)pm1\cong \mathbb{Z}_{(p-1) p^{m-1}} ); note that this is not true for p=2p=2. (ζn)\mathbb{Q}\left(\zeta_{n}\right) is usually called a cyclotomic field.

  1. (May 78 #11) Describe all intermediate fields of E/FE / F if E/FE / F is Galois with group Gal(E/F)=S3\operatorname{Gal}(E / F)= S_{3}.

  2. (May 80#680 \# 6 ) If F=(β)F=\mathbb{Q}(\beta) for a primitive nn-th root of unity β\beta, and bn=aFb^{n}=a \in F where aFna \notin F^{n} is not an nn-th power in FF, show G(F(b)/F)G(F(b) / F) is abelian.

  3. (Mar 83 #7) (a) SHOW E=𝔽p6E=\mathbb{F}_{p^{6}} is Galois over F=𝔽pF=\mathbb{F}_{p}.
    (b) Express the trace TrE/F(x)\operatorname{Tr}_{E / F}(x) as a polynomial in xx, and show there is an xx with TrE/F(x)\operatorname{Tr}_{E / F}(x) \neq 0 .
    (c) Show B(x,y):=TrE/F(xy)B(x, y):=\operatorname{Tr}_{E / F}(x y) is a nondegenerate bilinear form on E×EE \times E to FF.

  4. (Sep 83 #8) If E/E / \mathbb{Q} is Galois and B(x,y):=TrE/(xy)B(x, y):=\operatorname{Tr}_{E / \mathbb{Q}}(x y) (you may assume this is a nondegenerate bilinear form on E×EE \times E to \mathbb{Q} ), find the adjoints σ*\sigma^{*} of the elements σ\sigma of the Galois group G(E/)G(E / \mathbb{Q}) with respect to the bilinear form BB.

  5. (1985 #4) Let EnE_{n} be the splitting field of xn1x^{n}-1 over \mathbb{Q}.
    (a) What is [En:]\left[E_{n}: \mathbb{Q}\right] ?
    (b) What is |Gn|\left|G_{n}\right| for Gn=Gal(En/)G_{n}=\operatorname{Gal}\left(E_{n} / \mathbb{Q}\right) ? PROVE GnG_{n} is abelian.
    (c) SHOW G16G_{16} is not cyclic.

  6. (Fall 87#887 \# 8 ) If E/E / \mathbb{Q} is a splitting field of an irreducible polynomial ff of degree 8 , and aEa \in E is a root of ff so that ff splits over (a)\mathbb{Q}(a) into 2 linear and 3 quadratic factors, find the possible orders of Gal(E/)\operatorname{Gal}(E / \mathbb{Q}) and show that this group is always solvable.

  7. (May 89#689 \# 6 ) If E/FE / F is Galois with Gal(E/F)G a l(E / F) simple, for any element aEa \in E which is not in FF show that EE is a splitting field for the minimum polynomial of aa over FF.

  8. (May 90#390 \# 3 ) If E/E / \mathbb{Q} is a finite Galois extension inside \mathbb{C} with Gal(E/)\operatorname{Gal}(E / \mathbb{Q}) simple of order >2>2, show the imaginary unit ii CANNOT belong to EE.

  9. (Jan 92#392 \# 3 ) If ω\omega is a primitive cube root of 1 , determine whether (ω23)\mathbb{Q}(\omega \sqrt[3]{2}) is a Galois extension of \mathbb{Q}. Give reasons.

  10. (Aug 94#794 \# 7 ) Let ζn\zeta_{n} be a primitive nn-th root of unity in \mathbb{C} for n>2n>2. Show that the fixed field of (ζn)\mathbb{Q}\left(\zeta_{n}\right) under complex conjugation is (ζn+ζn)=(ζn)\mathbb{Q}\left(\zeta_{n}+\bar{\zeta}_{n}\right)=\mathbb{Q}\left(\zeta_{n}\right) \cap \mathbb{R}. (Hint: write ζn\bar{\zeta}_{n} as a power of ζn\zeta_{n} and find a polynomial of low degree satisfied by ζn\zeta_{n} over (ζn+ζn)\mathbb{Q}\left(\zeta_{n}+\bar{\zeta}_{n}\right).)

  11. (Jan 95 #8) Give an example of a polynomial f(x)[x]f(x) \in \mathbb{Q}[x] having all these properties: (1) degree 4; (2) no rational roots; (3) no repeated factors in [x]\mathbb{Q}[x]; (4) its Galois group over \mathbb{Q} is cyclic of order 2.

  12. (Aug 95#695 \# 6 ) Let E/E / \mathbb{Q} be a splitting field of x39x+12x^{3}-9 x+12. Show that there is a single normal extension F/F / \mathbb{Q} with EFE \supsetneq F \supsetneq \mathbb{Q}. Find [F:][F: \mathbb{Q}].

  13. (Aug 96#696 \# 6 ) Let E/FE / F be a finite Galois extension with Galois group GG. The Normal Basis Theorem states that there is an element uu in EE whose images under the elements of GG form an FF-basis of EE. Prove that for any subgroup HH of GG, the subfield corresponding to HH in the Galois correspondence is F(uH)F\left(u_{H}\right) for uH=hHh(u)u_{H}=\sum_{h \in H} h(u).

  14. (Aug 97#1ac)97 \# 1 \mathrm{ac}) Let pp be a prime number and FF a field containing pp distinct pp-th roots of unity. Let E/FE / F be a Galois extension for which [E:F]=p[E: F]=p.
    (a) Prove that the Galois group Gal(E/F)\operatorname{Gal}(E / F) is cyclic of order pp.
    (b) Prove that there is an element bEFb \in E \backslash F with bpFb^{p} \in F.

  15. (Aug 98 #7) Suppose that KK is a finite Galois extension of the rational field \mathbb{Q} which contains 3\sqrt{3} and has cyclic Galois group Gal(K/)\operatorname{Gal}(K / \mathbb{Q}). Show that L=(3)L=\mathbb{Q}(\sqrt{3}) is the only quadratic extension of \mathbb{Q} contained in KK.

  16. Let EE be a separable extension of the field FF, with [E:F]=n[E: F]=n. Use Galois theory to find an upper bound B(n)B(n) for the number of intermediate fields K,FKEK, F \subseteq K \subseteq E, that depends only on nn. You don’t need to make the bound B(n)B(n) very tight!

  17. (Aug 02 #9) Construct a Galois extension of \mathbb{Q} of degree 3.

  18. (Jan 04#504 \# 5 ) Let ζ\zeta be a primitive 9 -th root of unity. Let K=(ζ)K=\mathbb{Q}(\zeta) and F=(ζ+ζ1)F=\mathbb{Q}\left(\zeta+\zeta^{-1}\right).
    (a) Show that [K:F]=2[K: F]=2.
    (b) Show that the extension FF \mid \mathbb{Q} is normal.

  19. (Aug 04 #4) Let L/KL / K be a finite Galois extension. Suppose there exists an element αL\alpha \in L and another root α\alpha^{\prime} of the minimal polynomial μαK\mu_{\alpha \mid K} of α\alpha over KK such that the difference αα\alpha^{\prime}-\alpha is an element of K{0}K \backslash\{0\}.
    (a) Prove that the characteristic pp of KK is different from 0 and that pp divides [L:K][L: K].
    (b) Give an example of an extension L/KL / K and elements α,α\alpha, \alpha^{\prime} as described above.

  20. (Jan 06 #10) Find all subfields of the field Q(ζ3,23)Q\left(\zeta_{3}, \sqrt[3]{2}\right) for ζ3\zeta_{3} \in \mathbb{C} a primitive cube root of unity.

  21. (Aug 06 #8) Let MKM \mid K be a Galois extension of degree 270. Show that there is an intermediate extension M|L|KM|L| K with [L:K]=30[L: K]=30.

  22. (Aug 09 #6) Let K/FK / F be a finite extension of fields, and let α,βK\alpha, \beta \in K be such that K=F(α,β)K= F(\alpha, \beta). Let n=[F(α):F]n=[F(\alpha): F] and m=[F(β):F]m=[F(\beta): F], and assume that nn and mm are relatively prime.
    (a) Prove that [K:F]=nm[K: F]=n m.
    (b) Assume that K/FK / F is Galois. Let μα,F(x)\mu_{\alpha, F}(x) and μβ,F(x)\mu_{\beta, F}(x) be the minimal polynomials of α\alpha and β\beta over FF, respectively. Let αK\alpha^{\prime} \in K be a root of μα,F(x)\mu_{\alpha, F}(x), and let β\beta^{\prime} be a root of μβ,F(x)\mu_{\beta, F}(x). Prove that there exists a unique σGal(K/F)\sigma \in \operatorname{Gal}(K / F) such that σ(α)=α\sigma(\alpha)=\alpha^{\prime} and σ(β)=β\sigma(\beta)=\beta^{\prime}.
    (c) Again assume that K/FK / F is Galois. Let SS be the set of all elements cFc \in F such that F(α+cβ)KF(\alpha+c \beta) \neq K. Prove that |S|nm|S| \leq n m.

  23. (Aug 10#610 \# 6 ) Let K=(33,55)K=\mathbb{Q}(\sqrt[3]{3}, \sqrt[5]{5}), the field obtained from \mathbb{Q} by adjoining 33\sqrt[3]{3} and 55\sqrt[5]{5}.
    (a) Prove that [K:]=15[K: \mathbb{Q}]=15.
    (b) Let LL \subseteq \mathbb{C} be the Galois closure of KK over \mathbb{Q}, that is, LL is the minimal Galois extension of \mathbb{Q} which contains KK. Describe LL explicitly in the form (α1,,αt)\mathbb{Q}\left(\alpha_{1}, \ldots, \alpha_{t}\right), determine [L:][L: \mathbb{Q}] and describe the elements of the Galois group Gal(L/)\operatorname{Gal}(L / \mathbb{Q}) by their actions on α1,,αt\alpha_{1}, \ldots, \alpha_{t}.
    (c) Prove that K=(33+55)K=\mathbb{Q}(\sqrt[3]{3}+\sqrt[5]{5}).

  24. (Aug 11#4b11 \# 4 \mathrm{~b} ) Let K/FK / F be a finite extension of finite fields. Show the norm map NK/FN_{K / F} : KFK \rightarrow F is surjective.

  25. (Aug 12 #8) In this problem you may use the following fact without proof: for any finite group GG there exists a Galois extension M/LM / L with Gal(M/L)G\operatorname{Gal}(M / L) \cong G.
    (a) Prove that there exists a field extension K/FK / F such that [K:F]=4[K: F]=4 and there are no intermediate fields between FF and KK other than FF and KK. Hint: First reduce the question to a purely group-theoretic problem. Partial credit will be given for such reduction.
    (b) Is it possible to construct an extension satisfying (a) if FF is finite? Justify your answer.
    (c) Is it possible to construct an extension satisfying (a) if KK is contained in a cyclotomic field (ζn)\mathbb{Q}\left(\zeta_{n}\right) for some nn (where ζn\zeta_{n} is a primitive nth n^{\text {th }} root of unity)? Justify your answer.

  26. (Jan 13 #5) Let ω=e2πi/3\omega=e^{2 \pi i / 3} and consider the field K=(23,ω)K=\mathbb{Q}(\sqrt[3]{2}, \omega).
    (a) Prove that [K:]=6[K: \mathbb{Q}]=6.
    (b) Prove that K/K / \mathbb{Q} is a Galois extension.
    (c) Let M/LM / L be any finite Galois extension. Prove that an element γM\gamma \in M is primitive for M/LM / L (that is, L(γ)=ML(\gamma)=M ) if and only if σ(γ)γ\sigma(\gamma) \neq \gamma for any σGal(M/L){1}\sigma \in \operatorname{Gal}(M / L) \backslash\{1\}.
    (d) Now prove that γ=23+ω\gamma=\sqrt[3]{2}+\omega is a primitive element for K/K / \mathbb{Q}.
    (e) Let x6+a5x5++a0x^{6}+a_{5} x^{5}+\ldots+a_{0} be the minimal polynomial of γ\gamma over \mathbb{Q}. Prove that a5=3a_{5}=3 without actually computing the minimal polynomial.

  27. (Jan 16#616 \# 6 ) Let K=¯K=\overline{\mathbb{Q}} be the algebraic closure of the rationals in \mathbb{C}, i.e., the set of elements in the complex numbers which are algebraic over the rationals. By Zorn’s lemma, there exists a maximal subfield of KK, say EE, which does not contain the square root of 2 . Prove that every finite normal extension of EE has cyclic Galois group. (Hint: reduce this to a question about groups.)

  28. (Jan 16#716 \# 7 ) Let ϵ\epsilon be a primitive 16 th root of unity in the complex numbers. Set s=2ϵs=\sqrt{2} \cdot \epsilon. Let E=[ϵ]E=\mathbb{Q}[\epsilon], where \mathbb{Q} is the field of rational numbers, and set f(X)=X8+16[X]f(X)=X^{8}+16 \in \mathbb{Q}[X]. Show that ss is a root of f(X)f(X). Prove that 2[ϵ]\sqrt{2} \in \mathbb{Q}[\epsilon], and hence that f(X)f(X) splits completely over EE. If G=Gal(E/)G=\operatorname{Gal}(E / \mathbb{Q}), prove that no nonidentity element of GG fixes ss. Prove that f(X)f(X) is irreducible over \mathbb{Q}.

  29. (Aug 17#617 \# 6 ) Let M|K,M|L,K|F,L|FM|K, M| L, K|F, L| F be finite field extensions. Assume that for α,βM\alpha, \beta \in M, K=F(α),L=F(β)K=F(\alpha), L=F(\beta) and M=F(α,β)M=F(\alpha, \beta). Set a=[K:F]a=[K: F] and b=[L:F]b=[L: F].
    (a) If KFK \mid F and LFL \mid F are Galois, show that also MFM \mid F is Galois.
    (b) If KFK \mid F and LFL \mid F are Galois, prove that [M:F][M: F] divides aba b.
    (c) Give an example of M,K,L,FM, K, L, F as above (but without the Galois assumption) such that [M:F][M: F] does not divide aba b.

  30. (Jan 18 #8) Let M|K|FM|K| F be a tower of finite field extensions such that MFM \mid F and KFK \mid F are both Galois. Assume that the Galois group G(MK)G(M \mid K) is cyclic. Prove that LFL \mid F is Galois for every intermediate field LL with M|L|KM|L| K.

  31. (Aug 19#719 \# 7 ) Let pp and qq be distinct primes, let F=(p3,q5)F=\mathbb{Q}(\sqrt[3]{p}, \sqrt[5]{q}) and let KK be the Galois closure of FF over \mathbb{Q}.
    (a) Prove that [F:]=15[F: \mathbb{Q}]=15.
    (b) Prove that [K:]=120[K: \mathbb{Q}]=120.
    (c) Prove that Gal(K/)\operatorname{Gal}(K / \mathbb{Q}) has a normal subgroup of order 15.
    (d) Prove that Gal(K/)\operatorname{Gal}(K / \mathbb{Q}) has no normal subgroup of order 8 .

  32. (Jan 20#820 \# 8 ) Let K/FK / F be a field extension. Suppose that K=F(a,b)K=F(a, b) for some a,bKa, b \in K such that a2Fa^{2} \in F and b2Fb^{2} \in F.
    (a) Prove that [K:F]4[K: F] \leq 4.
    (b) Give a specific example (with full proof) where [K:F]=4[K: F]=4.
    (c) Assume that char(F)2\operatorname{char}(F) \neq 2. Prove that the extension K/FK / F is Galois.
    (d) Now assume that FF is finite. Prove that [K:F]2[K: F] \leq 2.

  33. (Jan 21 #1) Let FF be a field. Let f(x)F[x]f(x) \in F[x] be an irreducible separable polynomial of degree nn, let GG be the Galois group of ff over FF, and assume that GG is abelian.
    (a) Prove that if gGg \in G is any non-trivial element, then gg does not fix any of the roots of ff.
    (b) Now assume that FF \subseteq \mathbb{R} and nn is odd. Prove that all roots of ff must be real.
    (c) Now let F=F=\mathbb{Q}. Prove that there are infinitely many nn for which there exists ff as above with a non-real root and cyclic GG.

  34. (Jan 22 #8) Let FF be a field, and let f(x)F[x]f(x) \in F[x] be a separable irreducible polynomial of degree nn.
    (a) Let αβ\alpha \neq \beta be distinct roots of ff (in some fixed field extension KK of FF ). Prove that [F(α,β):F]n(n1)[F(\alpha, \beta): F] \leq n(n-1).
    (b) Let α\alpha and β\beta be as in (i). Prove that [F(α+β):F](n2)=n(n1)2[F(\alpha+\beta): F] \leq\binom{ n}{2}=\frac{n(n-1)}{2}. Hint: Use the action of a suitable Galois group.
    (c) Assume that F=F=\mathbb{Q} and nn is prime. Give an explicit example of ff and α\alpha and β\beta satisfying the above conditions such that the equality in (a) holds (you are NOT allowed to choose your prime nn ).