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.
 A: 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.
A: 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.
