# For separable field $k(\alpha)$ and extension $k'$ show that $k(\alpha) \otimes k'$ is a direct sum of fields

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.

• $k(a)$ 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. Sep 23 at 1:43
• $k(a)$ is most certainly a direct product of fields — itself — but that is not going to be of any help. Sep 23 at 1:43
• @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. Sep 25 at 15:59