Galois Extension, Chinese Remainder Theorem, and Quotients of Polynomial Rings 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)?

 A: Let $m(x) = \prod_{i = 1}^l f_i$ be an irreducible factorization in $F[x]$. We want to show that if $F \otimes_k F \cong F^n$, then each $f_i$ is linear. As $\deg m(x) = n$, it suffices to show that $l = n$. By the separability of this extension, which you cited a proof of, the $f_i$ are pairwise coprime. Hence, by the Chinese Remainder Theorem (and the isomorphism $F \otimes_K F \cong F[x] / (m(x))$ you cited), $F \otimes_k F \cong \prod_{i = 1}^l F[x]/(f_i)$. By assumption, this is isomorphic (as a ring) to $F^n$. Furthermore, each $F[x]/(f_i)$ is a field. Hence, $\prod_{i = 1}^l F[x]/(f_i)$ is a product of $l$ fields. Hence, it has precisely $l$ prime ideals $(*)$. Furthermore, as it is isomorphic to the ring $F^n$, which has $n$ prime ideals, $n = l$ and we are done.
$(*)$ To show this, you can first show the following.
Lemma. For $R_1, \dots, R_n$ commutative rings, every ideal of $R_1 \times \dots \times R_n$ looks like $I_1 \times \dots \times I_n$, for $I_j \subseteq R_j$ an ideal. Hint: finiteness is absolutely essential.
Given this, you can refine this to say that the prime ideals of $R_1 \times \dots \times R_n$ look like $R_1 \times \dots \mathfrak{p}_j \times \dots \times R_n$ for $\mathfrak{p}_j \subseteq R_j$ a prime ideal. Hint: an ideal is prime iff the quotient by it is a domain.
For your first additional question, you can see that my proof did not require the isomorphism to be of $F$-algebras, only rings. Furthermore, the proof of separability only used that $\prod_{i = 1}^l F[x]/(f_i)$ is reduced, which is a property of rings. Hence, this also only needed the isomorphism to be of rings.
For your second question, let $F[x] / (f) \cong F^n$ for some arbitrary ideal $f$. We can show, similarly to the separability case, that $F[x] / (f)$ is reduced, so $f$ is squarefree. Following the proof I wrote above, this implies that $f$ must factor as a product of $n$ many irreducible polynomials. If this were an isomorphism of $F$-algebras, then by linear algebra $deg(f)$ would therefore have to be $n$, so $f$ would factor as a product of linear polynomials. I'll think more about the case where this is only an isomorphism of rings.
EDIT: If $F[x] / (f) \cong F^n$ as rings, the above still implies that $f$ splits as a product of $n$ many irreducible factors. However, we will not be able to show that $f$ is a product of linear factors. Indeed, the issue is that there are nontrivial finite extensions $E/F$ with $E \cong F$ as rings. More precisely, let $F = \mathbb Q(t^2)$. Let $f = x^2 - t^2 \in F[x]$. $f$ is irreducible in $F$ and $F[x] / (f) \cong \mathbb Q(t)$. We claim that $\mathbb Q(t) \cong \mathbb Q(t^2)$. Indeed, we can simply take the map sending $t \mapsto t^2$. More rigorously, the map $\mathbb Z[t] \longrightarrow \mathbb Z[t^2]$ via $t \mapsto t^2$ is a well defined isomorphism. Take quotient fields to get the above map and show that it is an isomorphism.
Hence, we have an example of a field $F$ such that $F[x] / (f)$ is a direct product of $F$ as rings, but $f$ is not a product of linear polynomials. To summarize, my answer to your second question is as follows:
Let $F[x] / (f) \cong F^n$ as rings. This occurs if $f$ is a product of $n$ pairwise coprime irreducible polynomials in $F[x]$. If this is, in addition, an isomorphism of $F$-algebras, then each irreducible factor of $f$ is linear.
