Consider the following statement
If $A,B$ are both $\overline{k}$-algebras and integral domains, then $A \otimes_\overline{k} B$ is also an integral domain
I am trying to understand the proof of this statement in this Stack Project Blog. I'll copy the first part of the proof here to lead to the question (with $k$ replaced by $\overline{k}$ to better match with Vakil's notation). Let $(\sum_i a_i \otimes b_i)(\sum_j c_j \otimes d_j) = 0$ with the $b_i$'s and $d_j$'s $\overline{k}$-linearly independent (in particular, they are nonzero). Assume, in contrary, that not all the $a_i$ (resp. $c_j$) are zero. Let $A'$ be the subalgebra of $A$ generated by $\{a_i\}_i$ and $\{c_j\}_j$. Then since $A'$ is an integral domain, there is a maximal ideal $\mathfrak{m} \subset A$ that does not contain $a_1, c_1 \neq 0$.
Here is the thing I can't understand no matter how I think about it: why does such an $\mathfrak{m}$ exist? I don't see (1) how the integral-domainness of $A'$ is used here, and (2) why, in particular, can't all maximal ideals of $A$ contain $A'$. Can someone provide me with a more elaborate version of the argument for the existence of $\mathfrak{m}$?
There is actually one more question: in the proof, it states that $A/\mathfrak{m}=\overline{k}$ by the Nullstellensatz, but don't we need $A$ to be finitely generated? (Here is what I'm thinking: By the Zariski lemma (Vakil, 3.2.5), any finitely generated field extension as a ring is a finite extension, but $\overline{k}$ is algebraically closed and hence has no nontrivial algebraic extension, so $A/\mathfrak{m}=\overline{k}$.)
Best Answer
You're mixing up $A$ and $A'$ and that's causing your issue. The point is that you can reduce checking whether $A\otimes B$ is an integral domain to checking whether $A'\otimes B'$ is an integral domain where $A'$ and $B'$ are finitely generated over $\overline{k}$, and now you should work with $A'$ and $B'$ by themselves and put $A$ and $B$ out of your mind while you work on this argument. (This is what Johan is doing - he's implicitly assuming $A$ and $B$ finitely generated without saying anything about it in that blog post.)
Once you're reduced to working with finitely generated $A'$ and $B'$, everything is easy enough: for finitely generated algebras over a field, the Jacobson and nil radicals coincide, so if $a_1,c_1$ are in every maximal ideal they are nilpotent and therefore zero since $A'$ is a domain. But they're not, so there's some maximal ideal not containing them. The second statement you're having trouble with is similar: once you know $A'$ is finitely generated over $\overline{k}$, Zariski's lemma shows that $A'/\mathfrak{m}=\overline{k}$ as you discuss in your parenthetical.