Embedding of $K(\alpha)\hookrightarrow \bar{K}$ fixing $K$

Question

Let $K(\alpha)/K$ be a finite separable extension of fields and let $\bar K$ be the alegbric closure of $K$.

Which embeddings $\beta:K(\alpha)\hookrightarrow \bar K$ fix $K$ (that is $\beta(k)=k,\forall k\in K$)?

Answer 1

As in the comments, if $\beta$ is such a morphism, then it's fixed on $K$ and only the value of $\beta(\alpha)$ determines the morphism, but $\beta(\alpha)$ has to be a root of the minimal polynomial of $\alpha$ over $K$.

First of all, for the definitions, remember that since $K(\alpha)$ is by assumption a finite extension, $\alpha$ is algebraic over $K$, and so the ideal $\{P \in K[X] \mid P(\alpha)=0\}$ is non zero, and since $K[X]$ is principal, it has a unique unitary generator $\pi_\alpha$, called the minimal polynomial of $\alpha$ over $K$. For any morphism $\beta: K \to L$ where $L$ is any field containing $K$, $\pi_\alpha(\beta(\alpha))=0$.

Note also that $\pi_\alpha$ is irreducible: indeed if $\pi_\alpha= PQ$, then $P(\alpha)=0$ or $Q(\alpha)=0$ since $K$ is a field, and thus if both $deg(Q), deg(P)\geq 1$, this contradicts the minimality of $\pi_\alpha$.

Now let $\gamma$ be any root of $\pi_\alpha$. We first note that $\pi_\gamma= \pi_\alpha$. Indeed since $\pi_\alpha$ cancels $\gamma$, $\pi_\gamma \mid \pi_\alpha$ by definition of minimal polynomial. However $\pi_\alpha$ is irreducible and both these polynomials are unitary so $\pi_\alpha= \pi_\gamma$. This will be helpful for what follows.

Define $f: K[X] \to \overline{K}$ by $f(P)= P(\gamma)$. Then by definition $Ker(f) = (\pi_\gamma)$ (the ideal generated by $\pi_\gamma$), and so by the first isomorphism theorem, $f$ factors through the canonical surjection $h: K[X] \to K[X]/(\pi_\gamma)$, which gives an injective morphism $g: K[X]/(\pi_\gamma) \to \overline{K}$. This morphism obviously fixes $K$.

Moreover $\pi_\gamma = \pi_\alpha$ so that $K[X]/(\pi_\gamma) = K[X]/(\pi_\alpha)$. But $s: K[X]/(\pi_\alpha) \to K(\alpha)$ defined by $s(k + (\pi_\alpha))= k$ for $k\in K$, and $s(X+(\pi_\alpha))= \alpha$ is an isomorphism that fixes $K$.

So in the end, putting $\beta := g\circ s^{-1}$ we get an injective morphism $K(\alpha) \to \overline{K}$ that fixes $K$, and such that $\beta(\alpha)=\gamma$.

To sum up, embeddings $\beta: K(\alpha) \to \overline{K}$ that fix $K$ are uniquely determined by the value of $\beta(\alpha)$, and there is such an embedding where $\beta(\alpha)$ is any of the roots of $\pi_\alpha$, the minimal polynomial of $\alpha$ over $K$, and these are the only such embeddings.