# Extension of field automorphism to automorphism of algebraic closure

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.

• Since in general the algebraic closure has infinite degree, you need Zorn's lemma to extend an isomorphism to it. Jul 23, 2015 at 3:43
• 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).
– neth
Jul 23, 2015 at 3:52

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 $K$.