Suppose that $L:K$ is finite. Show that every K-monomorphism $L\to L$ is an automorphism. Does this result hold if the extension is not finite?

Question

I think I want to show that a $K$-monomorphism, $\sigma$ is onto. That is, every element of $L$ is mapped to by at least one element of $L$. I write $L$ as $L=K(\alpha_1, \dots \alpha_n)$ since the extension is finite.

We know that $\sigma$ fixes $K$, so $\sigma(k)=K$ for all $k\in K$. Now I need to show that $\alpha$'s are sent to $\alpha$s?

Answer 1

This follows directly from the rank-nulity theorem in the finite case because a $K$-monomorphism is also a $K$-linear transformation: $$ \dim_K L = \dim_K \ker \sigma + \dim_K \sigma(L) = \dim_K \sigma(L) $$

Answer 2

One can prove that this is true if $K\subset L$ is merely algebraic, not necessarily finite:

1. If $K\subset L$ is finite and $\sigma : L\to L$ is a $K$-monomorphism, then $\sigma$ is a $K$-linear map if you think of $L$ as a $K$-vector space. Since $L$ is a finite dimensional $K$-vector space, it follows that $\sigma$ is surjective as well.

2. If $K\subset L$ is algebraic (but not necessarily finite), then choose $\beta \in L$, and let $R$ denote the set of all roots of the minimal polynomial of $\beta$ over $K$ which lie in $L$. Now $R$ is a finite set, and $\sigma$ maps $R$ to $R$. Since $\sigma$ is injective, it must must map $R$ onto $R$, and hence $\exists \alpha \in R$ such that $\sigma(\alpha) = \beta$. This is true for any $\beta \in L$, so $\sigma$ is surjective.

However, note that the result is not true if the extension is not algebraic. A simple example is that both $\pi$ and $\pi^2$ are transcendental over $\mathbb{Q}$, and so there is a $\mathbb{Q}$-isomorphism $$ \mathbb{Q}(\pi) \to \mathbb{Q}(\pi^2) $$ which gives a $\mathbb{Q}$-monomorphism from $\mathbb{Q}(\pi)$ to itself which is not surjective.