Finitely many embeddings of a finite extension in an algebraic closure

Question

So I'm reading through Lang's Algebra, and he keeps saying something along the following lines:

"Let $K$ be a finite extension of a field $k$ and let $\sigma_1,\ldots,\sigma_r$ be the distinct embeddings of $K$ in an algebraic closure $k^\text{a}$ of $k$."

But the thing is that it doesn't seem clear to me that there need be only finitely many distinct embeddings, and I can't find anywhere where he proves it or figure out why it should be true or find its proof anywhere else. I feel like it should be obvious but it's not. So can anyone explain this to me?

EDIT: This all certainly makes more sense to me now. The only problem is that all of the answers make use of the minimal polynomial, which (as far as I know) only applies to algebraic finite extensions. I'm willing to believe that Lang is talking about algebraic extensions, but I do wonder if this is true of finite extensions in general.

EDIT 2: Nevermind Edit 1, I forgot what an algebraic closure is.

Answer 1

The key is that if you have a finite extension $K/F$, adding elements to $F$ will always raise the degree of the extension by some factor, so you get a tower of simple extensions $F \subseteq F(\alpha_1) \subseteq F(\alpha_1)(\alpha_2) \subseteq \cdots$. By finiteness of $K/F$, this process has to stop. Then you should recall the following :

If $\varphi : F \to F'$ is an isomorphism of fields, $f(X) \in F[X]$ and $f'(X) \overset{def}=\varphi(f)(X) \in F'[X]$ obtained by applying $\varphi$ on the coefficients of $f$, take $\alpha$ to be a root of $F$ and $\beta$ to be a root of $F'$ (in some extensions). Then there exists a unique isomorphism $\psi : F(\alpha) \to F'(\beta)$ such that $\psi|_F = \varphi$.

This will allow you to do the inductive argument. The main idea is that if you have a map $\psi : F(\alpha) \to K$ where $K$ is a field containing (an isomorphic copy of) $F$ and $\psi$ maps $F$ to this copy, then $\psi$ is entirely determined by where $\alpha$ maps to, and $\alpha$ must map to a root of its minimal polynomial, so it has finitely many options. I leave the details to you.

Hope that helps,

Answer 2

Try proving it for simple extensions of the form $k(a)/k$ first, then induct.

Hint: $k(a)\cong k[x]/(f)$ for a minpoly $f$, and embeddings $\hookrightarrow\overline{k}$ correspond to roots of $f(x)$.

Answer 3

If $K = k(a_1,\dots,a_n)$ is a finite field extension, then an embedding $\sigma: K \rightarrow k^{a}$ must send each $a_i$ to another root of its minimal polynomial over $k$. (Indeed, if $p(a_i)=0$ for $p \in k[x]$, then $p(\sigma(a_i))=0$. ) This shows there are only finitely many possibilities for the image of each $a_i$ under $\sigma$, and these determine $\sigma$. Hence there are only finitely many embeddings $K \hookrightarrow k^a$ over $k$.