## When the tensor product of finite field extensions is a field

Let $K_1$ and $K_2$ be finite extensions of a field $F$ contained in a field $K$. Prove that the $F$-algebra $K_1 \otimes_F K_2$ is a field if and only if $[K_1K_2 : F] = [K_1:F][K_2:F]$.

First, define $\psi : K_1 \times K_2 \rightarrow K_1K_2$ by $(a,b) \mapsto ab$. Clearly $\psi$ is $F$-bilinear, and so induces an $F$-module homomorphism $\Psi : K_1 \otimes_F K_2 \rightarrow K_1K_2$. In fact $\Psi$ is an $F$-algebra homomorphism. Using Proposition 21 in D&F, if $A = \{\alpha_i\}$ and $B = \{\beta_j\}$ are bases of $K_1$ and $K_2$ over $F$, then $AB = \{\alpha_i\beta_j\}$ spans $K_1K_2$ over $F$. In particular, $\Psi$ is surjective.

Suppose $K_1 \otimes_F K_2$ is a field. Now $\mathsf{ker}\ \Psi$ is an ideal of $K_1 \otimes_F K_2$, and so must be trivial- so $\Psi$ is an isomorphism of $F$-algebras, and thus an isomorphism of fields. Using Proposition 21 on page 421 of D&F, $K_1 \otimes_F K_2$ has dimension $[K_1:F][K_2:F]$, and so $[K_1K_2 : F] = [K_1:F][K_2:F]$ as desired.

Conversely, suppose $[K_1K_2 : F] = [K_1:F][K_2:F]$. That is, $K_1 \otimes_F K_2$ and $K_1K_2$ have the same dimension as $F$-algebras. By Corollary 9 on page 413 of D&F, $\Psi$ is injective, and so $K_1 \otimes_F K_2$ and $K_1K_2$ are isomorphic as $F$-algebras, hence as rings, and so $K_1 \otimes_F K_2$ is a field.