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Let $L$ be a field extension of $K$. Consider the set $\operatorname{End}_KL$ of all functions from $L$ to $L$ which are linear over $K$. A subset of $\operatorname{End}_KL$ is $\Gamma(L:K)$, the group (under composition) of all automorphisms on $L$ which fix $K$.

So, we can consider $\operatorname{End}_KL$ to be a vector space over $K$. We have addition of functions in the natural sense, and multiplication of functions by scalars, and all the other properties of a vector space.

I have been working for over a couple days on trying to show that $\Gamma(L:K)$ is linearly independent over $\operatorname{End}_KL$.

In the usual way, we let $\{\sigma_i\} \subset \Gamma(L:K)$ and $\{k_i\} \subset K$, both for $i = 1 \ldots n$ and suppose that $\sum_1^n k_i\sigma_i = 0$. This is an equation in $\operatorname{End}_KL$. Without loss of generality, suppose that $k_i \neq 0$ for each $i = 1 \ldots n$.

I have been wanting to find some contradiction, though unsuccessfully in the general case.

If $n=1$, $k_1 \sigma_1=0 \Rightarrow \sigma_1 = 0$ and this is an immediate contradiction as $\sigma_1$ is not surjective, and therefore not an automorphism.

If $n=2$, $k_1\sigma_1 + k_2\sigma_2=0$. Applying $\sigma_1^{-1}$ to both sides yields $k_1\tau + k_2 \sigma_1^{-1}\sigma_2 = 0$, where $\tau$ is the trivial automorphism. Solving for $\tau$, we have $\tau = \frac{-k_2}{k_1}\sigma_1^{-1}\sigma_2$. Now, evaluating both sides at $-\dfrac{k_2}{k_1}$ gives the equation $-\frac{k_2}{k_1}= (-\frac{k_2}{k_1})^2$. This must imply that $-\frac{k_2}{k_1} = 1$, since $k_2 \neq 0 $. Thus we have that $\sigma_1=\sigma_2$, a contradiction.

Again, any help with solving this in general would be much appreciated.

Thanks.

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    @DylanMoreland: Artin's theorem is the one which bounds $[F : F^H]$, where $H \le \operatorname{Aut} F$.2012-05-06

1 Answers 1

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Let us prove something slightly more general--called Dedekind's lemma.

Let $M$ be a monoid and $k$ a field. A character is then a monoid homomorphism $\chi:M\to k^\times$. The general version of what you are saying then is that the set $C$ of all characters is linearly independent as a subset of the vector space $k^M$ of all functions $M\to k$.

To prove this it suffices to show that if $\chi_1,\cdots,\chi_n$ are distinct characters and

$\sum a_i \chi_i=0\quad\mathbf{(1)}$

for $a_i\in k$ then $a_i=0$ for all $i$. The basic idea is to show that we can always reduce ourselves to a smaller subset of $\{\chi_1,\cdots,\chi_n\}$ that is also l.d. and so we get a contradiction by taking the smallest one.

In particular, note that since $\chi_1\ne\chi_2$ there exists $m\in M$ such that $\chi_1(m)\ne \chi_2(m)$. We see then from $\mathbf{(1)}$ that

$\sum \chi_1(m)a_i \chi_i(x)=0$

for all $x\in M$. But, we also must have that

$\sum a_i \chi_i(mx)=\sum a_i\chi_i(m)\chi_i(x)=0$

Subtracting the two gives

$\sum_{i>1}\chi_i(m)a_i\chi_i(x)=0$

Since this was true for all $x$ we see that $\{\chi_2,\cdots,\chi_n\}$ is l.d. Continuing in this way gives {\chi_n}$ is l.d. which is absurd. Thus, we have reached a contradiction and the conclusion follows.