Proof of : If $K$ is a finite extension of $F$, then $o(G(K,F)) \leq [K:F]$

Question

$G(K,F)$ is the group of all automorphisms of $K$ that leave $F$ fixed. $F$ is a field of characteristic zero. $[K:F]$ is the dimension of $K$ as a vector space over $F$.

Herstein proves this theorem (5.6.2) in a round about way using the result : If $\sigma_1,...,\sigma_n$ are distinct automorphisms of $K$, then is impossible to find elements $a_1,...,a_n$ not all $0$, in $K$ such that $a_1\sigma_1(u) + ... + a_n\sigma_n(u)=0$ for all $u \in K$.

Here is my attempt at the proof. Please let me know if it is valid.

Let $[K:F]=n$. Let $u_1,...,u_n$ be the basis.

We want to prove the claim using contradiction. Let $\sigma_1, ..., \sigma_{n+1} \in G(K,F)$ be distinct elements of $G(K,F)$.

Consider the $n+1$ equations :

$x_1\sigma_1(u_1) +x_2\sigma_1(u_2) + \ldots + x_n\sigma_1(u_n) =0 $

$\vdots$

$x_1\sigma_{n+1}(u_1) +x_2\sigma_{n+1}(u_2) + \ldots + x_n\sigma_{n+1}(u_n) =0 $

Since there are $n+1$ equations and $n$ unknowns there is a solution for $x_i's$ which is not all zeros. The equations are distinct because we assume $\sigma_i's$ are distinct, for if any two equations are same then their corresponding values for all $u_i's$ are same but since $u_i's$ form a basis, this can only happen if they are equal on all of $K$.

Let a nonzero solution be $a_1,...,a_n$. Let $b_i=\sigma_1^{-1}(a_i)$. If $a_i \neq 0 $ then $b_i \neq 0$ as $\sigma_1$ is an automorphism. Substitute the solution in the first equation to get

$\sigma_1(\Sigma b_iu_i )=0 \iff \Sigma b_iu_i=0$

But since $u_i's$ form a basis this is a contradiction because not all $b_i's$ are zero.

Answer 1

Lemma: Let $\sigma_1,...,\sigma_n$ be distinct nonzero semigroup homomorphism from S to K. If $\sum_{i=1}^n\beta_i\sigma_i(x)=0\; \forall x\in S$, where $\beta_i\in K$, then $\beta_i=0$ for each $i$.

Proof: Let $x,y\in S$, then $\beta_1\sigma_1(xy)+...+\beta_n\sigma_n(xy)=\beta_1\sigma_1(x)\sigma_1(y)+...+\beta_n\sigma_n(x)\sigma_n(y)=0$ and $\sigma_1(x)(\beta_1\sigma_1(y)+...+\beta_n\sigma_n(y))=0$. Therefore, $\beta_2(\sigma_2(x)-\sigma_1(x))\sigma_2(y)+...+\beta_n(\sigma_n(x)-\sigma_1(x))\sigma_n(y)=0$. Notice that $\beta_j(\sigma_j(x)-\sigma_1(x))\in K$, so by induction on n, it equals $0$ for each $x\in S, j=1,...,n$.

Now, for $j\ne 1, \exists x\in S$ such that $\sigma_j(x)\ne \sigma_1(x)$; for $j=1, \beta_1\sigma_1(y)=0 \;\forall y\in S$, so $\beta_i=0$ for $i=1,...,n$.

Theorem: Let E be a field, $\sigma_1,...,\sigma_n\in Aut(E)$ all distinct, and $F=\{r\in E\mid \sigma_i(r)=r\}$, then $[E:F]\ge n$.

Suppose $[E:F]=r\lt n$. Take a basis of E over F, which is $\{u_1,...,u_r\}$. Then $x_1\sigma_1(u_i) +x_2\sigma_2(u_i) + \ldots + x_n\sigma_n(u_i) =0$ for $i=1,..., r$ have solutions in E which are not all zero. Hence $\sum_{i=1}^nx_i\sigma_i(u_j)=0 \;\forall j=1,...,r$. Since $\sigma_i(\beta u_j)=\beta\sigma_i(u_j)$ for each $\beta\in F, \sum_{i=1}^nx_i\sigma_i(z)=0 \;\forall z\in E.$ This is in contradiction with the lemma, and thus $[E:F]\ge n$.

Corollary: $[E:F]\ge o(Aut(E/F))$.

Proof: Let $[E:F]=l $ and suppose $l\lt o(Aut(E/F))$. Choose $l+1$ distinct $\sigma_1,...\sigma_{l+1}\in Aut(E/F)$ and let $F'=\{x\in E\mid \sigma_i(x)=x\}$,then $[K:F]\ge [K:F']\ge l+1\gt l$, which is a contradiction.