The problem comes from a statement in Algebraic Geometry. Let $k$ be a field and $E/k$, $F/k$ are field extensions. Then the statement says there is an bijection between
$$ \operatorname{Spec}(E \otimes_k F) $$ and the set of classes of composite fields of $E$ and $F$.
I know that if $L$ is the composite field of $E$ and $F$. Then there is a natural homomorphism $E \otimes_k F \longrightarrow L$ (Note this does not need to be surjective). The kernel is always a prime ideal. (See Bourbaki's Albgera, V, 2.4)
From my perspective, this is far away from $\operatorname{Spec}(E \otimes_k F)$. Are the prime ideals of tensor product of fields maximal ideals?
Lemma 1. For a commutative ring $A$, $\dim A$ = $\dim (A/rad(A))$.
Lemma 2. [Chapter 5, GTM 256] $\dim k[X_1,\dots, X_n] = n$.
Lemma 3. Assume there are $k$-algebras $A,B$ and $A$ is a field. $A/A'$ is an algebraic extension. Then $\dim A' \otimes_k B = \dim A \otimes_k B$.
Proof: Using Zorn's Lemma, we can reduce to the case where $A = A'[u]$, where $u$ is algebraic over $A'$. $$ A \otimes_k B = A \otimes_{A'} (A' \otimes_k B) $$ Essentially, $A \cong A'[X]/(m(X))$, where $m(X)$ is an irreducible polynomial of $A'[X]$. So $$ A \otimes_k B = (A' \otimes_k B)[X]/((A' \otimes_k B)m(X)) $$ Obviously, $\dim A' \otimes_k B \leq \dim (A' \otimes_k B)[X]/((A' \otimes_k B)m(X))$.
To prove the converse inequality, note that the constant term of $m(X)$ is not zero. So $X \in (A' \otimes_k B)[X]/((A' \otimes_k B)m(X))$ is invertible. Every prime ideal in $(A' \otimes_k B)[X]/((A' \otimes_k B)m(X))$ can be contracted to a prime ideal in $A' \otimes_k B$. If prime ideals $P_1 \subset P_2$ of $(A' \otimes_k B)[X]/((A' \otimes_k B)m(X))$ are not equal, then $P_1 \cap (A' \otimes_k B ) \subset P_2 \cap (A' \otimes_k B)$ is a strict inclusion. Otherwise, if $P_1 \cap (A' \otimes_k B ) = P_2 \cap (A' \otimes_k B )$, then for $f_2(X) \in P_2 \backslash P_1$, $\exists f_1(X) \in P_1(X)$ with the same constant term. Then $f_2(X) - f_1(X) \not= 0$ is invertible, contrdicting the fact $P_2$ is a prime ideal.
Therefore, $\dim A' \otimes_k B = \dim A \otimes_k B$.
Finally, I prove the main theorem.
Theorem. $\dim E\otimes_k F = \min(\operatorname{trdeg} E/k, \operatorname{trdeg} F/k)$.
Proof: This follows from the three lemmas above.