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.

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