3
$\begingroup$

Let $k$ be a field and let $f(x)\in k[x]$ be irreducible. Let $K$ be the algebraic closure of $k$, and say among the roots of $f(x)$ are $\alpha,\beta\in K$. Then there exists an automorphism of $K$ sending $\alpha$ to $\beta$.

I'm studying basic field/Galois theory (or trying to), and this fact seems used often, and seems as though it should be obvious, but I can't figure it out.

The closest I am able to get is that if $k\subset F$ is the splitting field of $f$, then I know there is an automorphism of $F$ sending $\alpha$ to $\beta$, but I do not know how or if this can be extended to an automorphism of $K$. I found a related question here, but the answer assumes what I want to know.

$\endgroup$
2
  • $\begingroup$ Since in general the algebraic closure has infinite degree, you need Zorn's lemma to extend an isomorphism to it. $\endgroup$
    – fkraiem
    Jul 23, 2015 at 3:43
  • $\begingroup$ Do you know the general approach here, or a source I might find it? I'm at a loss, but if you happen to know the rough picture that would be a great help (I'm familiar enough with Zorn's lemma). $\endgroup$
    – neth
    Jul 23, 2015 at 3:52

1 Answer 1

3
$\begingroup$

The proof should be available in any abstract algebra textbook, the one I have at hand right now is Fraleigh (where it is Theorem 48.3).

Since you are familiar with Zorn's lemma, we apply it to the set $S$ of pairs $(L,\lambda)$, where $L$ is an intermediate field between $k(\alpha)$ and $K$ (inclusive) and $\lambda$ is an injective morphism of $L$ into $K$ (i.e., an isomorphism from $L$ to a subfield of $K$) which extends the given isomorphism $\sigma$ from $k(\alpha)$ to $k(\beta)$. This set is not empty because it contains $(k(\alpha),\sigma)$.

We define an ordering as follows: $(L,\lambda) \le (L',\lambda')$ if and only if $L$ is a subfield of $L'$ and the restriction of $\lambda'$ to $L$ equals $\lambda$. We show that every chain has an upper bound, so $S$ has a maximal element, which can be shown to be the pair $(K,\kappa)$, where $\kappa$ is an isomorphism from $K$ to a subfield of $K$ which extends $\sigma$. $\kappa$ is surjective (if not, we could non-trivially extend $\kappa^{-1}$ from $\kappa(K)$ to $K$, but that's not possible because $\kappa^{-1}$ is already onto $K$, which is algebraically closed), so it is an automorphism of $K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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