Prove that the tensor product of non algebraic extensions is not a field

Solution 1:

Use Sharp-Grothendieck's miraculous equality computing the Krull dimension of the tensor product of two completely arbitrary field extensions as a function of the transcendence degrees of the extensions : $$\operatorname {dim_{Krull}} (E\otimes _K F) =\min\: ( \operatorname {trdeg}_K E, \operatorname {trdeg}_KF) $$ From this equality the contraposition of your statement easily follws:
If $E,F$ are both non algebraic, then since $ \operatorname {trdeg}_K E \geq 1$ and $ \operatorname {trdeg}_K F\geq 1$ we have $\operatorname{dim_{Krull}} (E\otimes _K F) \geq 1$.
Thus $E\otimes _K F$ is not a field because a field has Krull dimension zero.

Solution 2:

Here's a more hands-on proof. We'll prove the contrapositive: if $E$ and $F$ are both transcendental over $K$ we'll show that $E \otimes_K F$ is not a field by exhibiting a nontrivial ideal. Let $e_i, i \in I$ and $f_j, j \in J$ be transcendence bases of $E$ and $F$ respectively over $K$, so that $E$ is an algebraic extension of $K(\{ e_i \})$ and $F$ is an algebraic extension of $K(\{f_j\})$.

Assume WLOG that $|I| \le |J|$ and pick an embedding $g : I \hookrightarrow J$. This induces an embedding of both $E$ and $F$ into the algebraic closure $\overline{K(\{f_j\})}$, which induces a map

$$E \otimes_K F \to \overline{K(\{f_j\})}.$$

The kernel of this map is nontrivial because it contains, for example, elements of the form $e_i \otimes 1 - 1 \otimes f_{g(i)}$ (and these elements are nonzero because, for example, instead of using the embedding $g$ above we can write down a map from $E \otimes_K F$ to $\overline{K(\{e_i\} \cup \{ f_j\})}$, and the image of $e_i \otimes 1 - 1 \otimes f_{g(i)}$ here is $e_i - f_{g(i)}$ where $e_i$ and $f_{g(i)}$ are algebraically independent). So $E \otimes_K F$ can't be a field.