## 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 cos$nx$ 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.

$\mathbf{Proof:}$ 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.