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Let F be a field, E a finite field extension of F, K the field of separable elements of E over F, C an algebrically closed field containing F. Is it true that every F-homomorphism from K to C extends uniquely to E? If yes, why would it be true? The point should be that the minimum polynomial over K of an inseparable element of E has only a root, but I can't prove this fact. Thank you for help.

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I’ll essay this. Let’s hope I don’t mess it up.

Let $\lambda$ be an element of $E$ not in $K$, and $f(X)\in K[X]$ its minimal polynomial over $K$. This may be written $f_0(X^{p^r})$ for maximal $r\ge0$. Now, $f_0$ is still irreducible over $K$, with root $\lambda^{p^r}$. Furthermore, $f_0'$ is nonzero and thus has no common roots with $f$, so that the roots of $f_0$ are of multiplicity one, i.e. $f_0$ is separable. Thus $\lambda^{p^r}$ is an element of $E$ separable over $K$, thus separable over $F$, thus in $K$, and so the degree of $f_0$ is $1$. So $f(X)=X^{p^r}-\lambda^{p^r}\in K[X]$, and has the single root $\lambda$.

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