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

Suppose that $E,F$ are two extensions of $K$ and $E\otimes _K F$ is a field. Prove that $E$ or $F$ are algebraic over $K$.

Is there any hint to prove this? Thanks.

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.

• @GeorgesElencwajg thanks for your answer, but I want a simpler proof. – user115608 Oct 22 '16 at 9:54
• Dear Georges: what a beautiful equality, and a great proof. Thank you for sharing! – Alex Wertheim Nov 12 '16 at 8:16
• Glad you appreciated the equality and proof, dear @Alex. – Georges Elencwajg Nov 12 '16 at 9:13
• @user115608: youtube.com/watch?v=3dfYcQ_r_x8 – Georges Elencwajg May 31 '20 at 7:55

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.