## 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 $p$th 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 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.

### 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 $E(\xi)=E$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

### 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