I am trying to prove the following assertion (, which is an exercise from a section on tensor products in the book Algebra: Chapter 0 by P.Aluffi):
Let $F=k(\alpha)\supset k$ be a finite simple extension such that $F\otimes_kF\cong F^{[F:k]}$ as a ring. Then the extension is Galois.
A related question has been asked several times on this site, for example in here. Reading the answers to these questions, I learned that the above assertion may be proved as follows:
- There is a canonical isomorphism $F\otimes_k F\cong F[x]/(m(x))$ of $k$-algebras. Now $F^n$ is obviously reduced, so $F[x]/(m(x))$ is also reduced. It follows that $\alpha$ is separable over $k$. (This step requires a little more argument, but I am OK with this step.)
- Therefore, to prove that $F/k$ is Galois, it suffices to show that the $\alpha$ splits over $F$.
- Let $m(x)\in k[x]$ be the minimal polynomial of $\alpha$ over $k$. But $F[x]/(m(x))\cong F^n$ as a ring by hypothesis. By the Chinese remainder theorem, this implies that $m(x)$ factors into linear factors over $F$.
I do not understand the bold faced part in step 3. Sure, the Chinese remainder theorem implies that if $m(x)$ splits over $F$, then $F[x]/(m(x))\cong F^n$ as rings. However, we want to go in the other direction. How do we do this?
Thanks in advance.
In addition to the above question, I would greatly appreciate the answers to the following additional questions. (But you do not have to answer these unless you want to.)
- The question which I cited above actually assumed that the isomorphism $F\otimes_k F\cong F^{[F:k]}$ to be an $F$-algebra isomorphism. In my book, this isomorphism is assumed to be merely a ring isomorphism. Maybe the author took it for granted that the isomorphism to be $F$-linear? Or maybe we can do without the $F$-linearity.
- In general, which quotient of $F[x]$ isomorphic to the direct product of $F$ as a ring (or as an $F$-algebra)?
from Hot Weekly Questions - Mathematics Stack Exchange
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