Category Archives: Galois Theory

Albert radicals of norm and Hilbert’s Satz 90

\mathbf{Problem(Albert):} Let E be a cyclic extension of dimension n over F and let \eta be a generator of \text{Gal }E/F. Let r|n, n=rm and suppose c is a non-zero element of F such that c^r=N_{E/F}(u) for some u\in E. Show that there exists a v in the (unique) subfield K of E/F of dimensionality m such that c=N_{K/F}(v).

\mathbf{Proof:} G=\text{Gal }E/F is a cyclic group, then it  has a unique subgroup H=\{\eta^m,\eta^{2m},\cdots,\eta^{rm}=1\} of index r. By the Galois corresponding theorem, there exists a unique subfield K=\text{Inv }H such that \text{Gal K/F}\cong G/H. \text{Gal }K/F=\{\eta^1|_K,\eta^2|_K\cdots,\eta^m|_K\}. K has dimensionality m over F.

Consider w=c^{-1}u\eta(u)\eta^2(u)\cdots \eta^{m-1}(u), then \displaystyle\eta(w)=\frac{\eta^m(u)}{u}w. We also have

\displaystyle N_{E/K}(w)=\eta^m(w)\eta^{2m}(w)\cdots \eta^{rm}(w)=c^{-r}\eta(u)\eta^2(u)\cdots\eta^n(u)=1,

by Hilbert’s Satz 90, \exists \, l\in E such that \displaystyle w=\frac{\eta^m(l)}{l}.

Let \displaystyle v=\frac{ul}{\eta(l)}, then v\in K, because

\displaystyle \eta^m(v)=\frac{\eta^m(u)\eta^m(l)}{\eta^{m+1}(l)}=\frac{\eta^m(u)\eta(l)}{\eta^{m+1}(l)}\cdot\frac{\eta^m(l)}{\eta(l)}=\frac{\eta^m(u)}{\eta(w)}\cdot\frac{\eta^m(l)}{\eta(l)}=\frac{\eta^m(u)}{\eta(w)}\cdot\frac{wl}{\eta(l)}=u\frac{l}{\eta(l)}=v

Surprisingly we have

\displaystyle N_{K/F}(v)=v\eta(v)\cdots \eta^{m-1}(v)=u\eta(u)\cdots\eta^{m-1}(u)\frac{l}{\eta^m(l)}=cw\frac{l}{\eta^m(l)}=c.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Jacabson p300. This problem puzzled me for three weeks. Finally it turns out to be very easy.

When the norm map is surjective

\mathbf{Problem:} Suppose F is finite field with q=p^m elements, E an Galois extension of F such that [E:F]=n. Prove that N_{E/F}:E^\ast\to F^\ast is surjective.

\mathbf{Proof:} From theorem 4.26 we know that \text{Gal }E/F is cyclic over F generated by \eta:a\to a^q.

\displaystyle N_{E/F}(a)=aa^qa^{q^2}\cdots a^{q^{n-1}}=a^{\frac{q^n-1}{q-1}}.

N_{E/F}:E^\ast\to F^\ast is a homomorphism. It is well known that  any finite subgroup of the multiplicative group of field is cyclic, which means E^\ast and F\ast is cyclic.

Since |E\ast|=q^n-1 and |F^\ast|=q-1, and N_{E/F}(a)=a^m, where m={\frac{q^n-1}{q-1}}, then |\text{Ker } N|=(q^n-1,m)=m and |\text{Im }N|=\frac{q^n-1}{(q^n-1,m)}=q-1. Thus N is surjective.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Jacobson p300.

Criterion for embedding in cyclic field

\mathbf{Problem:} Assume F has p distinct pth roots of 1. p a prime, and E/F is cyclic of dimensional p^f. Let z be a primitive pth root of 1. Show that if E/F can be imbedded in a cyclic field K/F of dimension p^{f+1}, then z=N_{E/F}(u) for some u\in E.

\mathbf{Proof:} Suppose \sigma is the generating isomorphism of cyclic Galois group K/F. Then \displaystyle E=\text{Inv }\sigma^{p^f}. N_{K/E}(z)=z^p=1, so by Hilbert satz 90, there exists a\in K such that \displaystyle z=\sigma^{p^f}(a)a^{-1} .
\displaystyle \sigma^{p^f}(a^{-1}\sigma(a))=\sigma^{p^f}(a^{-1})\sigma^{p^f+1}(a)=(z a)^{-1}\sigma(z a)=a^{-1}\sigma(a).
So a^{-1}\sigma(a)=a_0\in E, then
N_{E/F}(a_0)=a^{-1}\sigma(a)\sigma(a^{-1})\sigma^2(a)\cdots \sigma^{p^f-1}(a)\sigma^{p^f}(a)=a^{-1}\sigma^{p^f}(a)=z.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Jacobson p300.

Galois group of C(t,u) over C(t^n,u^n) and application to cosnx

\mathbf{Problem:} Let E be an extension of \mathbb{C} such that E=\mathbb{C}(t,u) where t is transcedental over \mathbb{C} and u satisfies the equation u^2+t^2=1 over \mathbb{C}(t). Find the Galois group of \mathbb{C}(t,u) over \mathbb{C}(t^n,u^n) for any n\in \mathbb{N} Show that

\displaystyle u_n=\frac{1}{2} [(t+iu)^n+(t-iu)^n], i=\sqrt{-1}

is contained in \mathbb{C}(t^n,u^n). Use this to prove that the function cosnx is expressible rationally with complex coefficients in terms of \cos^nx and \sin^n x. Does this hold for sin nx.
\mathbf{Proof:} \forall\, \eta\in \text{Gal }\mathbb{C}(t,u)/\mathbb{C}(t^n,u^n), we must have

\displaystyle\eta(t^n)=\eta^n(t)=t^n    (1)
\displaystyle\eta(u^n)=\eta^n(u)=u^n    (2)
\displaystyle\eta^2(t)+\eta^2(u)=1      (3)

This means (\eta(t)/t)^n=1. So there exist w_1, a n-th root of 1, such that \eta(t)=w_1t. Similarly \exists\, w_2 such that \eta(u)=w_2u. (3) becomes

w^2_1t^2+w^2_2u^2=1. Combing with t^2+u^2=1, we get w^2_1=w^2_2=1. Since w^n_1=w^n_2=1, we know that when n is even w_1=w_2=1 and when n is odd w_1=\pm 1 and w_2=\pm 1.

 In conclusion, the Galois group is trivial when n is odd and Klein group when n is even.

No matter n is even and odd, \displaystyle u_n=\frac{1}{2} [(t+iu)^n+(t-iu)^n] is invariant, so u_n is in \mathbb{C}(t^n,u^n).

If n is odd, \displaystyle v_n=\frac{1}{2i}[(t+iu)^n-(t-iu)^n] is in \mathbb{C}(t^n,u^n).  If n is even, v_n is not in the ground field, because \eta(t)=t, \eta(u)=-u can not fix v_n.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} At first I am surprised by the result I got. I can not believe that the galois is trivial when n is odd.  So I conculted to Zhuohui. Thank him for his kind help. I drew the following picture after discussion with him.

case n=4


case n=3

Inverse Galois Problem

\mathbf{Problem:} Use the fact that any finite group G is isomorphic to a subgroup of S_n to prove that given any finite group G there exists fields F and E/F such that

\displaystyle \text{Gal }E/F\cong G

\mathbf{Proof:} Any field K, x_1,x_2,\cdots,x_n are indeterminate. Let E=K(x_1,x_2,\cdots,x_n). D is the field of symmetric rational functions in latex x_1,x_2,\cdots,x_n Or if we let p_1,p_2,\cdots,p_n are basic symmetric polynomials, namely

\displaystyle p_1=\sum x_i, \displaystyle p_2=\sum _{i>j} x_ix_j, \cdots \displaystyle p_n=x_1x_2\cdots x_n,

then D=K(p_1,p_2,\cdots,p_n).

As we all know, \text{Gal }E/D=S_n. Since G\leq S_n, by the Galois corresponding theorem, there exists an intermediate field F such that \text{Gal }E/F\cong G.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} This is the easiest inverse Galois problem since one does not specify the ground field. If you restrict F=\mathbb{Q}, things are much trickier.

Polynomial of prime degree and solvable transitive subgroup

L is the group of transformations of \mathbb{Z}_p of the form x\to ax+b, a\neq 0.
H is all the translations x\to x+b.

\mathbf{Lemma 1:} |L|=p(p-1) and |H|=p, H\lhd L.

\mathbf{Lemma 2:} G is a group of transformation in \mathbb{Z}_p containing the group H of translations as normal subgroup. Show that G is a subgroup of L, this means L is the normalizer of H.
\mathbf{Proof:} Suppose \tau:x\to x+1 and let \eta\in G, \eta \tau\eta^{-1} has the form x\to x+k. Hence \eta(x+1)=\eta\tau(x)=\eta(x)+k. If \eta(0)=b, then \eta(x)=kx+b.
\text{Q.E.D}\hfill \square

\mathbf{Lemma 3:} Any solvable transitive subgroup of S_p, p a prime, is equivalent to a subgroup of L containing the group of translations.

\mathbf{Proof:} Suppose G is this group. Then G has a composition series
\displaystyle G\rhd H_1\rhd H_2\rhd \cdots \rhd H_{n-1}\rhd H_n=1
whose factors are all cyclic groups of prime order.
By lemma 4, we know that H_{n-1} is also a solvable transitive subgroup of S_p. Since H_{n-1} is cyclic of prime order, then it must be generated by some p-cycle, otherwise it can not be transitive.
Apply some conjugation to G in S_p, we can assume H_{n-1} is generated by (12\cdots p). So H_{n-1} is a translation group. We can identify the element of G as the transformation of \mathbb{Z}_p. By lemma 2, we know that G is equivalent to some subgroup of L.
\text{Q.E.D}\hfill \square
\mathbf{Lemma 4:} Let H be a normal nontrivial subgroup of a transitive subgroup G of S_n of transformations of \{1,2,\cdots,n\}. Show that all H-orbits have the same cardinality. Hence show that if n=p is a prime, then H is transitive.

\mathbf{Theorem(Galois):} Let f(x)\in F[x] be irreducible of prime degree over F of characteristic 0, E a splitting field over F of f(x). Show that f(x) is solvable by radicals over F if and only if E=F(r_i,r_j) for any two roots r_i,r_j of f(x).

\mathbf{Proof:} Since F has characteristic 0, f(x) is solvable by radicals if and only if G=\text{Gal }E/F is a solvable subgroup of symmetric group S_p. f(x) is irreducible if and only if G is transitive.
(1) if f is solvable by radicals, then G is a solvable transitive subgroup of S_p.
By lemma 3, we know G is equivalent to a subgroup of L containing H. WLOG, assume H\leq G\leq L. The action of G on roots looks like r_i\to r_{ai+b}, a\neq 0.
Consider \text{Gal }E/F(r_i,r_j), r_i\neq r_j. \forall\,\eta\in \text{Gal }E/F(r_i,r_j), \exists\, a,b unique such that i=ai+b and j=aj+b. Then (a-1)(i-j)=0. Since i\neq j, we get a=1 and b=0, which means \eta=id. So E=F(r_i,r_j).
(2) If E=F(r_i,r_j), then p| |\text{Gal }E/F(r_i,r_j)|. This Galois group contains a normal Sylow p-subgorup which is isomorphic to H. By lemma 2, we know G is equivalent to a subgroup of L which is solvable. So G is solvable, f(x) is solvable by radicals.
\text{Q.E.D}\hfill \square

\mathbf{Remark:} Jacobson p262.

Algebraically closed and Primitive element theorem

\mathbf{Problem:} Let F be a perfect field and F\subset E an algebraic field extension, such that every non-constant polynomial f(x)\in F[x] has a root in E. Show that E is algebraically closed.

\mathbf{Proof:} Suppose f(x)\in F[x] is a non-constant polynomial. Let K be a splitting field of f.
Since F is perfect, K is a separable extension of F. Primitive element theorem implies that any finite separable extension is simple extension. That is \exists\,\alpha\in K such that K\in F(\alpha). Suppose \alpha has a minimal polynomial g(x) over F. By assumption, g(x) has a root \gamma\in E. Since there exists an isomorphism \eta:F(\alpha)/ F\to F(\gamma)/ F\subset E by \eta(\alpha)=\gamma, then E contains a splitting field of f(x) over F.
\forall\, f\in E[x], adjoin a root \xi of f(x), we get E(\xi)/E. Since E(\xi)/E and E/F are algebraic. E(\xi)/F is also algebraic. Hence, \xi is a root of some non-constant polynomial h(x)\in F[x]. By the previous proof, h(x) splits over E. We conclude that \xi\in E and E(\xi)=E.
\text{Q.E.D}\hfill \square


The discriminant of a cubic equation and its relation with roots

\mathbf{Problem:} Let F=\mathbb{R} and let f(x)=x^3-a_1x^2+a_2x-a_3 with discriminant d. Show that f(x) has multiple roots, three distinct real roots, or one real root and two non-real roots according as d=0, d>0, d<0.

\mathbf{Proof:} f has multiple roots if and only if d=0.

Supoose d\neq 0. Since cubic equation always has at least one  root in \mathbb{R}, assume f(x)=(x-r)g(x), r\in \mathbb{R}. Then Galois group G_f of f is equal to that of g(x) over \mathbb{R}, which is the symmetric group S_2 or the alternating group A_2=1.

d>0 if and only if G_f is contained in symmetric group A_3. Since G_f has only two cases. d>0 if and only if G_f is A_2=1 which means f has three distinct roots in \mathbb{R}

Correspondingly, we get d<0 if and only if   one real root and two non-real roots.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Jacobson p260.

Normalized root tower and primitive roots of unity

\mathbf{Problem 1:} Let p be a prime unequal to the characteristic of the field F. Show that, if a\in F, then x^p-a is either irreducible in F[x] or it has a root in F.

\mathbf{Proof:} Let w be a p-th primitive root of unity, K=F(w). Since F has characteristic 0, K contains p distinct pth root of unity.

Let U=\{\text{pth roots of unity contained in } K\}. U is a multiplicative subgroup of K.

Suppose E is a splitting field of x^p-a over K and r is one root of x^p-a=0 in E. All the roots of x^p-a are  zr, z\in U. So E=K(r). \forall\, \xi\in \text{Gal }E/K, \xi(r)=zr, z\in U. We can verify that this induces a monomorphism

\text{Gal }E/K\mapsto U

\xi\to z

Since U is cyclic of order p. G which is a subgroup of U can only be the trivial group \{1\} or the whole group U.

If G=\{1\}, E=K=F(w). And x^p-a splits over F(w). Then it must have a root in F.

If G=U, then [K(r):K]=[E:K]=|G|=p, x^p-a must be irreducible in K[x], hence also in F[x].

\text{Q.E.D}\hfill \square

\mathbf{Problem 2:} Let E/F be the cyclotomic field of the pth roots of unity over the field F of characteristic 0. Show that E can be imbedded in a field K which has a root tower over F such that the integers n_i are primes and [F_{i+1}:F_i]=n_i. Call such a root tower normalized.

\mathbf{Proof:} Since  E is the cyclotomic field over F of characteristic 0, G=\text{Gal }E/F is abelian by lemma 1 in page 252.

Then G is a finite solvable group. So G has a composition series

\displaystyle G=G_1\rhd G_2\rhd \cdots\rhd G_{r+1}=\{1\} 

whose composition factor G_i/G_{i+1} is cyclic of prime order p_i. By Galois corresponding theorem, we have an increasing chain of subfields

F=F_1\subset F_2\subset\cdots\subset F_{r+1}=E

such that G_i=\text{Gal }E/F_i and \displaystyle G_i/G_{i+1}=\text{Gal }F_{i+1}/F_i. So \displaystyle [F_{i+1}:F_i]=p_i is prime. This is a normalized root tower.

\text{Q.E.D}\hfill \square

\mathbf{Problem3:} Obtain normalized root tower over \mathbb{Q} of the cyclotomic fields of 5th and 7th roots of unity.

\mathbf{Proof:} Suppose w is a primitive 5th roots of unity. [\mathbb{Q}(w):\mathbb{Q}]=4 and  G=\text{Gal }\mathbb{Q}(w)/\mathbb{Q} is multiplicative group of \mathbb{Z}_5, because \forall\,\zeta\in G maps \{w,w^2,w^3,w^4,w^5=1\} to itself. The automorphism \zeta:w\to w^3 is a generator of G=\{\zeta,\zeta^2,\zeta^3,\zeta^4=1\}.

G=\langle \zeta\rangle\rhd \langle \zeta^2\rangle\rhd=\{1\}

the corresponding root tower is

\mathbb{Q}\subset \mathbb{Q}(w+w^4)\subset \mathbb Q(w)

It is easy to verify this is a normalized root tower.

For 7th roots of unity, suppose w is a primitive 7th root of unity,

\mathbb{Q}\subset \mathbb{Q}(w+w^2+w^4)\subset \mathbb Q(w)

is a normalized tower where [\mathbb{Q}(w+w^2+w^4):\mathbb{Q}]=2.

\text{Q.E.D}\hfill \square

\mathbf{Problem4:} Prove that, if f(x)=0 has a solvable Galois group over a field F of characteristic 0. Then its splitting field can be imbedded in an extension field which has a normalized tower over F.

Suppose splitting field is E and [E:F]=n, the proof relies on adjoin nth roots of unity to E.

\mathbf{Remark:} Jacobson p256.

One example of inseparable polynomial

\mathbf{Theorem:} F is a field. If F[x] contains an inseparable polynomial, then F can not have characteristic 0 or F is some finite field with characteristic p.

In general, we do not need to worry about the separability of the extension field, because we often deal with finite fields or the one of characteristic 0.

An example of inseparable polynomial
Let K be a field of characteristic p, and F=K(x) is the field of rational polynomials, x is the indeterminate. Consider F[y], y is the indeterminate, x is an irreducible element in this UFD. So f(y)=y^p-x is irreducible in E by the Einsenstein criterion. There exists a splitting field of f(y) over F. Suppose \sigma is root of f(y) then \sigma^p=x. By freshman’s dream f(y)=(y-\sigma)^p. So f has multiple roots. It is not separable.

\mathbf{Remark:} Isaacs, p281