Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Re your edit: Every finite extension is algebraic. – Prism Aug 12 '14 at 5:46
Following up on @Prism's comment, if $K$ is a finite extension of $k$, then by definition it is an $n$-dimensional $k$-vector space for some finite $n$. Given any $a \in K$, what can you say about the elements $1,a,a^2,\dots,a^n$? – vociferous_rutabaga Aug 12 '14 at 5:51
Oh haha I forgot that algebraic closures have to be algebraic – desi Aug 12 '14 at 6:50
up vote 3 down vote accepted

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,

share|cite|improve this answer

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 $k$.

share|cite|improve this answer

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 $f(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.