Here is a classical theorem from algebraic geometry which might motivate what I'm about to ask.
Let $k$ be an algebraically closed field. If $A$ and $B$ are finitely generated $k$-algebras, then
$$(\mathfrak m, \mathfrak n) \mapsto \mathfrak m \otimes B + A \otimes \mathfrak n$$
defines a bijection $\operatorname{m-spec} A \times \operatorname{m-spec} B \rightarrow \operatorname{m-spec} A \otimes_k B$.
Let $R$ be a $\mathbb Q$-algebra. Suppose that every prime ideal $\mathfrak p$ of $R$ is maximal, and moreover that $R/\mathfrak p \cong \mathbb Q$.
Does the tensor product $R \otimes_{\mathbb Q} R$ then have the same properties?
For an example, consider a totally disconnected Hausdorff space $X$, and let $R$ be the ring of locally constant functions $X \rightarrow \mathbb Q$, i.e. which are continuous when $\mathbb Q$ is given the discrete topology. Then this question of mine shows that $R$ satisfies the above hypothesis. And I believe it can be shown that $R \otimes_{\mathbb Q} R$ identifies with the ring of locally constant functions $X \times X \rightarrow X$.
Oh, I see that the proof of the classical result I mentioned carries over here:
Let $\mathfrak P$ be a prime of $R \otimes_{\mathbb Q}R$. Let $\mathfrak p$ (resp. $\mathfrak q$) be the contraction of $\mathfrak P$ under $r \mapsto r \otimes 1$ (resp. $r \mapsto 1 \otimes r$). Then $I :=\mathfrak p \otimes R + R \otimes \mathfrak q$ is contained in $\mathfrak P$, with
$$(R \otimes_{\mathbb Q} R) /I \cong R/\mathfrak p \otimes_{\mathbb Q} R/\mathfrak q \cong \mathbb Q \otimes_{\mathbb Q} \mathbb Q \cong \mathbb Q$$
and therefore $I$ is a maximal ideal of $R \otimes_{\mathbb Q} R$. Hence $I = \mathfrak P$. This shows that every prime of $R \otimes_{\mathbb Q} R$ is maximal with residue field $\mathbb Q$.