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Let $K/F$ be a field extension with characteristic $0$, and $\alpha, \beta \in K$ be algebraic over $F$. Show that $F(\alpha) \otimes_F F(\beta)$ is isomorphic (as rings) to a finite product of fields.

Let $f_1, f_2 \in F[x]$ be the minimal polynomial of $\alpha, \beta$, respectively. Then $F(\alpha) \cong K[x]/(f_1)$, $F(\beta) \cong K[x]/(f_2)$ as $F$-algebras.
Pick $a_1, ..., a_m \in \overline{F}$ so that $f_1(x) = (x-a_1)\cdots(x-a_m)$. Since $F$ is a perfect field and $f_1$ is irreducible, $a_1,..., a_m$ must be pairwise distinct. Therefore, by the Chinese remainder theorem, $K[x]/(f_1) \cong K[x]/(x-a_1)\oplus \cdots \oplus K[x]/(x-a_m)$ as $F$-algebras.
Likewise, $K[x]/(f_2) \cong K[x]/(x-b_1)\oplus \cdots \oplus K[x]/(x-b_n)$ for some $b_1, ..., b_n \in \overline{F}$.
As tensor product distributes over direct sum, $F(\alpha) \otimes_F F(\beta) \cong \bigoplus_{i, j} A_i \otimes_F B_j$, where $A_i := K[x]/(x-a_i)$, $B_j := K[x]/(x-b_j)$.

I would appreciate if you could verify whether the proof is correct so far, and give some hints/reference for how to prove that $A_i \otimes_F B_j$ is a field.

Luke
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    Problems early on $F(\alpha)\simeq F[x]/(f_1)$, same with $F(\beta)$. – Jyrki Lahtonen Sep 03 '22 at 04:11
  • But you are right about using the Chinese Remainder Theroem. Because the components are extensions of $F$, you cannot really use $K[x]$ at all. Instead, factorize $f_1$ over $F(\beta)$ like here. Apologies for using a not perfectly fitting answer of my own. Try as I may I could not easily find the same logic in a more general setting. Equally well you can factor $f_2$ over $F(\alpha)$. – Jyrki Lahtonen Sep 03 '22 at 04:18

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Your use of the Chinese remainder theorem here is incorrect; you only have that factorization into linear factors over $\bar{F}$ so you only get those isomorphisms over $\bar{F}$, not over $F$ (which you should write instead of $K$ here, and in any case $K$ is irrelevant).

The correct way to use the Chinese remainder theorem here is to write $F[\beta] \cong F[x]/f(x)$ for $f$ the minimal polynomial but then leave $F[\alpha]$ alone; it will be easier notationally to write $L = F[\alpha]$. Then

$$F[\alpha] \otimes_F F[\beta] \cong L \otimes_F F[\beta] \cong L[\beta] \cong L[x]/f(x)$$

and now it remains to consider the factorization of $f(x)$ over $L$. You can just as well do the same argument but with the roles of $\alpha$ and $\beta$ swapped. It's a funny feature of this argument that it requires you to break the symmetry between the two.

Qiaochu Yuan
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