Galois extension and quotient group of multiplicative groups

Question

I am struggling with the problem related to the Galois extension of fields.

Let $E/F$ be a finite Galois extension. Show that if the quotient group $E^\times / F^\times$ has an element of order $n$, then $E^\times$ has an element of order $n$.

There are several facts that I tried to use: (1) every finite subgroup of the multiplicative group of a field is cyclic; (2) by the primitive element theorem, $E/F$ is a simple extension. However, I have no idea how to approach this problem. Any ideas or comments are welcome.

Answer 1

Suppose $E,F$ are fields with $F\subseteq E$ such that

• $E/F$ is a Galois extension.$\\[4pt]$
• The quotient group $E^\times / F^\times$ has an element of order $n$.

Our goal is to show that $\;E^\times$ has an element of order $n$.

Proof:

By assumption there is a nonzero element $w\in E$ such that

• $w^n\in F$.$\\[4pt]$
• There is no positive integer $m < n$ for which $w^m\in F$.

Let $a=w^n$, and let $f\in F[x]$ be given by $f=x^n-a$.

Let $g\in F[x]$ be the product of the distinct monic irreducible factors of $f$ in $F[x]$ having a root in $E$.

Since $E/F$ is a Galois extension, it follows that $$ g=(x-r_1)\cdots (x-r_m) $$ where

• $r_1,...,r_m\in E$.$\\[4pt]$
• $r_1,...,r_m$ are distinct.$\\[4pt]$
• $\{r_1,...,r_m\}=\{r\in E{\,\mid\,}f(r)=0\}$.

Since $g\in F[x]$, it follows that $ {\displaystyle{ \prod_{i=1}^m r_i }} \in F$.

Since $g{\,\mid\,}f$, it follows that $m\le n$.

Claim:$\;m=n$.

Suppose instead that $m < n$.

For all $i\in\{1,...,m\}$ we have ${\large{\frac{w^2}{r_i}}}\in E$, and $$ \Bigl( \frac{w^2}{r_i} \Bigr) ^n = \frac{w^{2n}}{r_i^n} = \frac{a^2}{a} = a $$ hence ${\large{\frac{w^2}{r_i}}}\in\{r_1,...,r_m\}$.

Then the map from the set $\{r_1,...,r_m\}$ to itself defined by $$ r_i \mapsto \frac{w^2}{r_i} $$ is injective, hence bijective.

It follows that \begin{align*} & \prod_{i=1}^m r_i = \prod_{i=1}^m \frac{w^2}{r_i} \\[4pt] \implies\;& w^{2m} = \left( \prod_{i=1}^m r_i \right)^{\!2} \\[4pt] \implies\;& w^m = \pm \prod_{i=1}^m r_i \\[4pt] \implies\;& w^m\in F \\[4pt] \end{align*} contradiction, since $m < n$.

Hence $m=n$, as claimed.

Returning now to the main proof . . .

Let $h\in F[x]$ be given by $h=x^n-1$.

For each $i\in\{1,...,n\}$, let $s_i=r_i/w$.

Since $r_1,...,r_n\in E$, and $w\in E$, it follows that $s_1,...,s_n\in E$, and since $r_1,...,r_n$ are distinct, it follows that $s_1,...,s_n$ are distinct.

Since $r_i^n=a$ for all $i$, and $w^n=a$, it follows that $s_i^n=1$ for all $i$.

Thus each $s_i$ is a root of $h$, and since $s_1,...,s_n$ are distinct, it follows that $s_1,...,s_n$ are all the roots of $h$.

Let $S=\{s_1,...,s_n\}$.

For all $i,j\in\{1,...,n\}$, we have $$ (s_is_j)^n=s_i^ns_j^n=1 $$ so $S$ is closed under multiplication, hence $S$ is a finite subgroup of $E^\times$.

It follows that $S$ is a cyclic group, hence since $|S|=n$, there must be some $s_i\in S$ such that $s_i$ has order $n$.

This completes the proof.