Tag Archives: galois group

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.

Symmetrical polynomial under even permutation

\mathbf{Problem:} Let F be of characteristic \neq 2 and also let H be the subgroup of G=\text{Gal }E/F(p_1,p_2,\cdots,p_n) corresponding to the alternating group, that is the set of \zeta(\pi), \pi\in A_n. Show that \text{Inv}H=F(p_1,p_2,\cdots,p_n,\Delta) where \displaystyle\Delta=\prod\limits_{i<j}(x_i-x_j).

\mathbf{Proof:} It is a basic fact that \zeta(\pi)(\Delta)=\Delta when \pi is an even permutation and \zeta(\pi)(\Delta)=-\Delta when \pi is an odd permutation.

So all \pi\in H preserve \Delta and p_1,p_2,\cdots,p_n, which means F(p_1,p_2,\cdots,p_n,\Delta)\subset \text{Inv}H.

On the other hand no odd permutation can preserve \Delta, then \text{Gal }E/F(p_1,p_2,\cdots,p_n,\Delta)\leq H.

Combining these two, we know \text{Inv}H=F(p_1,p_2,\cdots,p_n,\Delta).

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

\mathbf{Remark:} Jacobson p244.

Automorphism of Z_p(t) and its invariant field

\mathbf{Problem:} Let E=\mathbb{Z}_p(t) where t is transcedental over \mathbb{Z}_p. Let G be the group of automorphism generated by the automorphism of E such that t\to t+1. Determine F=\text{Inv}G and [E:F].

\mathbf{Proof:} Suppose \sigma is the automorphism of E that t\to t+1.  Group G is generated by \sigma. It is easy to verify that G is the cyclic group \{1,\sigma,\cdots,\sigma^{p-1}\}. So If F=\text{Inv}G, then [E:F]=|G|=p.

Let’s determine F. Note that \mathbb{Z}_p(u) where u=t^p-t is invariant under \sigma, because a^p=a, for \forall \,a\in \mathbb{Z}_p.  So \mathbb{Z}_p(u)\subset F.

On the other hand, \mathbb{Z}_p(u) is a field of characteristic p and E is a splitting field of x^p-x-u over \mathbb{Z}_p(u). x^p-x-u has no root in \mathbb{Z}_p(u), thus it is irreducible on \mathbb{Z}_p(u)[x] by a previous conclusion. So [E:\mathbb{Z}_p(u)]=p=[E:F], which means F=\mathbb{Z}_p(u) where u=t^p-t.

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

\mathbf{Problem:} What if G is replaced by the group of automorphism such that t\to at+b, a,b\in\mathbb{Z}_p, a\neq 0

If a=1, b=0, then F=E and [E:F]=1

If a=1, b\neq 0, then the conclusion is the same with previous problem.

If a\neq 1,0, |G|=p-1=[E:F], F=\mathbb{Z}_p(u) where \displaystyle u=\prod\limits_{i=1}^{p-1}\sigma^{i}(t).

\mathbf{Remark:} Jacobson, Algebra I p243.

Galois group of order 6

\mathbf{Problem:} Let E/\mathbb{C}(t) where t is transcedental over \mathbb{C} and let w\in \mathbb{C} satisfy \displaystyle w^3=1, w\neq 1. Let \sigma be the automorphism of E/\mathbb{C} such that \tau(t)=t^{-1}. Show that

\displaystyle \sigma^3=1=\tau^2,\quad \tau\sigma=\sigma^{-1}\tau.

Show that the group of automorphism generated by \sigma and \tau has order 6 and the subfield F=\text{Inv}G=\mathbb{C}(u) where u=t^3+t^{-3}.

\mathbf{Proof:} It is easy to verify the structure of group G. And G is the dihedral group D_6

Let H_1=\langle\sigma\rangle\leq G, H_2=\langle\tau\rangle\leq G. Then \displaystyle \text{Inv}H_1=\left\{\frac{f(t)}{g(t)}|g(t)\neq 0, \frac{f(t)}{g(t)}=\frac{f(\omega t)}{g(\omega t)}\right\}=\left\{\frac{f(t^3)}{g(t^3)}|g(t) \neq 0\right\}.

Consider \displaystyle \text{Inv}H_2=\left\{\frac{f(t)}{g(t)}|g(t)\neq 0, \frac{f(t)}{g(t)}=\frac{f(t^{-1})}{g(t^{-1})}\right\}, any \displaystyle \frac{f(t)}{g(t)}\in \text{Inv}H_2\exists\, h, l\in \mathbb{C}(t) such that

 \displaystyle 2\frac{f(t)}{g(t)}=\frac{f(t)}{g(t)}+\frac{f(t^{-1})}{g(t^{-1})}=\frac{f(t)g(t^{-1})+g(t)f(t^{-1})}{g(t)g(t^{-1})}=\frac{h(t+t^{-1})}{l(t+t^{-1})}

So \displaystyle \text{Inv}H_2=\left\{\frac{f(t+t^{-1})}{g(t+t^{-1})}\lvert g\neq 0\right\}.

So \text{Inv}G=\text{Inv}H_1\cap\text{Inv}H_2=\mathbb{C}(t^3+t^{-3})

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

\mathbf{Remark:}Jacobson p243

Splitting field of f(r)=r^3+r^2-2r-1 over Q

\mathbf{Problem:} Let E=\mathbb{Q}(r) where r^3+r^2-2r-1=0. Verify that r'=r^2-2 is also a root of x^3+x^2-2x-1=0. Determined \text{Gal }E/\mathbb{Q}. Show that E is normal over \mathbb{Q}.
\mathbf{Proof:} r^3+r^2-2r-1=0 means (r+1)(r^2-2)=-1. So we only need to prove \displaystyle \frac{-1}{r+1} is another root of f(x)=x^3+x^2-2x-1=0.
Let y=r+1, f(r)=0 means y^3-2y^2-y+1=0.
Let \displaystyle z=\frac{-1}{y}, then \displaystyle\frac {1}{z^3}(z^3+z^2-2z-1)=0, obviously z\neq 0, thus \displaystyle z=\frac{-1}{r+1} is a root of f(x)=0.
f(x) is monic and has no root in \mathbb{Z}, so it is irreducible in \mathbb{Q}. f(x) is minimal polynomial of r. So [E:\mathbb{Q}]=3. \text{Gal } E/\mathbb{Q} must be the cyclic group of order 3. By theorem 4.7 chapter 4 in Jacoboson’s book, E is normal over \mathbb{Q}.
\mathbf{Remark:} Jacobson, Algebra I. p243.