Tensor Products and Field Extensions (13.3.22)

Dummit and Foote Abstract Algebra, section 12.3, exercise 22:

Let $K/F$ be a finite field extension, and let $K_1,K_2⊆K$ be field extensions of $F$. Show the $F$-algebra $K_1 ⊗_F K_2$ is a field if and only if $[K_1K_2~:~F] = [K_1~:~F][K_2~:~F]$.

Proof: Let $A$ be the set of finite sums of elements of the form $k_1k_2$ for $k_1∈K_1,k_2∈K_2$, let $φ : K_1 × K_2 → A$ be the bilinear map defined by $φ(k_1,k_2)=k_1k_2$, and let $Φ : K_1 ⊗ K_2 → A$ be the corresponding $F$-linear transformation. We observe$$Φ(k_1⊗k_2)Φ(k_1'⊗k_2') = k_1k_1'k_2k_2' = Φ((k_1⊗k_2)(k_1'⊗k_2'))$$allowing us to show$$Φ(\sum_i k_{i1} ⊗ k_{i2})Φ(\sum_j k_{j1}'⊗k_{j2}')=\sum_{i,j} Φ(k_{i1} ⊗ k_{i2})Φ(k_{j1}' ⊗ k_{j2}') =$$$$\sum_{i,j} Φ((k_{i1} ⊗ k_{i2})(k_{j1}' ⊗ k_{j2}')) = Φ((\sum_i k_{i1} ⊗ k_{i2})(\sum_j k_{j1}' ⊗ k_{j2}'))$$so that $Φ$ is an $F$-algebra homomorphism. Note $A = \text{img Φ}$. As well, let $K_1$ have for basis over $F$ $\{n_i\}$ and let $K_2$ have $\{m_j\}$.

($⇒$) We thus have $Φ$ is a nonzero field homomorphism, and is thus an isomorphism. Now $A$ is a field and by definition we have $K_1K_2 ⊆ A$ and by construction we observe $A ⊆ K_1K_2$ so that $K_1K_2 ≅ K_1 ⊗ K_2$ as $F$-algebras, the latter of which has for basis $\{n_i ⊗ m_j\}$ of order $[K_1~:~F][K_2~:~F]$. ($⇐$) We still have $A ⊆ K_1K_2$, and since we observe $[K_1~:~F][K_2~:~F]$ elements $n_im_j$ of $A$ linearly independent over $F$ by Proposition 13.2.21, we must have this is a basis for $K_1K_2$ and thus again $A = K_1K_2$. The $F$-algebra homomorphism $Φ$ above sends basis to basis and is thus an isomorphism, and now $K_1 ⊗ K_2$ is a field.$~\square$