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Let $k$ be a field and $k(\alpha)$ a finite extension. Let $f(X) = Irr(\alpha, k, X)$ and suppose that $f$ is separable. Let $k'$ be any extension of $k$. Show that $k(\alpha) \otimes k'$ is a direct sum of fields. If $k'$ is algebraically closed, show that these fields correspond to the embeddings of $k(\alpha)$ in $k'$.

I know that $f$ is separable implies that $f$ has no multiple roots. I was thinking that if $k(\alpha)$ is the direct sum of fields then maybe we can use the result $(a\oplus b)\otimes c \approx (a\otimes c) \oplus (a \otimes b)$ to give the desired result. But I'm not sure if this is the way to go or how to proceed.

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    $\begingroup$ $k(a)$ is $k[X]/(f)$, so $k'\otimes k(a)$ is (show this) $k'[X]/(f)$. Factor $f$ over $k'$, show there are no repeated factors, use the Chinese remainder theorem. $\endgroup$ Sep 23 at 1:43
  • $\begingroup$ $k(a)$ is most certainly a direct product of fields — itself — but that is not going to be of any help. $\endgroup$ Sep 23 at 1:43
  • $\begingroup$ @MarianoSuárez-Álvarez Thank you for the outline. Is the reason that $k' \otimes k(\alpha) = k' \otimes k[X]/(f) = k'[X]/(f)$ because the map $a\otimes g(x) = a g(x)$ is multilinear? How do we know there are no other multilinear maps? I'm also not sure how the chinse remainder theorem and no repeated factors can be used to give a direct sum of fields. $\endgroup$
    – Math_Day
    Sep 25 at 15:59

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