An isomorphism of $F$ onto a subfield of $\bar{F}$ has the same number of extensions to each simple algebraic extension of $F$?.

Question

An isomorphism of $F$ onto a subfield of $\bar{F}$ has the same number of extensions to each simple algebraic extension of $F$? ($\bar{F}$ denote an algebraic closure of $F$)

I know this is false but I have not been able to find a counterexample, could someone please give me one? Thank you very much.

Answer 1

Let $F(\alpha)/F$ be a simple extension field and $\varphi:F\to\bar F$ an embedding of $F$ into an algebraically closed field $F$. Let $f\in F[X]$ be the minimum polynomial of $\alpha$ over $F$ and $f^\varphi\in\bar F[X]$ be its image under $\varphi$. There exists a one-to-one corrispondence between embeddings $\varepsilon:F(\alpha)\to\bar F$ extending $\varphi$ and distinct roots of $f^\varphi$ in $\bar F$.

More precisely, let $\text{Alg}_F(F(\alpha),\bar F)$ be the set of $F$-algebra homomorphisms $F(\alpha)\to\bar F$ and let $Z_\alpha\subseteq\bar F$ be the set of roots of $f^\varphi$. Then $$\sigma\mapsto\sigma(\alpha):\text{Alg}_F(F(\alpha),\bar F)\to Z_\alpha$$ is a bijection.

Thus the number of such extensions equals the separable degree of the simple extension.