Bijection between maximal ideals of $A\otimes_kB$ and maximal ideals of $A$ and $B$ for finitely generated $k$ algebras.

Question

I am trying to show that there is a bijection between $|\operatorname{Spec}(A\otimes_kB)|$ and $|\operatorname{Spec}A|\times|\operatorname{Spec}B|$ for finitely generated $k$ algebras $A$ and $B$. Here $|\cdot|$ means the closed points/maximal ideals of $\operatorname{Spec}$ of a ring.

I have shown that if $i_A:a\mapsto a\otimes 1$ and $i_B:b\mapsto 1\otimes b$ are the inclusions (though not necesarilly injective) maps into the tensor product, and $\mathfrak m$ is a maximal ideal of $A\otimes_kB$ then $i_A^{-1}(\mathfrak m)$ and $i_B^{-1}(\mathfrak m)$ are maximal ideals. This gives us a set map in one direction: \begin{align} \phi:|\operatorname{Spec}(A\otimes_kB)|&\longrightarrow |\operatorname{Spec}A|\times|\operatorname{Spec}B|\\ \mathfrak m&\longmapsto (i_A^{-1}(\mathfrak m),i_B^{-1}(\mathfrak m)) \end{align} I want to define an inverse to this map by taking $(\mathfrak a,\mathfrak b)\in |\operatorname{Spec}A|\times|\operatorname{Spec}B|$ to the ideal $I=\left\langle i_A(\mathfrak a),i_B(\mathfrak b)\right\rangle$, but I am struggling in two ways.

First, I think this is the right map because we have that if $a\in \mathfrak a$, then $a\otimes 1\in i_A(\mathfrak a)\subset I$ , so $a\in i_A^{-1}(I)$ implying that $\mathfrak a=i_A^{-1}(I)$ as $\mathfrak a$ is maximal. This suggests that $\phi\circ \phi^{-1}$ is the identity map (supposing first that $\phi^{-1}$ takes the pair of maximal ideals to a maximal ideal).

Now onto the problems: I cannot seems to show that $\phi^{-1}\circ \phi$ is the identity. In particular, I need to show that $\mathfrak m\subset \left\langle i_A(i_A^{-1}(\mathfrak m)), i_B(i_B^{-1}(\mathfrak m))\right\rangle$, but I have no idea how to show this to be honest. Secondly, I am struggling to show that $\phi^{-1}$ take $(\mathfrak a,\mathfrak b)$ to a maximal ideal. I know that there is an isomorphism: $$A\otimes_kB/I\cong A/\mathfrak a\otimes_k B/\mathfrak b$$ and that both $A/\mathfrak a$ and $B/\mathfrak b$ are finite field extensions of $k$ by Zariski's lemma, but how can I deduce from that that this tensor product is now again a field?

Edit: if $k$ is algebraically closed then this the second part is easy as both $A/\mathfrak a$ and $B/\mathfrak b$ must be isomorphic to $k$ so $A\otimes_kB/I$ must be isomorphic to $k$ as well...but even in this simpler case I can't figure out the identity part.

Answer 1

I'm afraid this is not true unless we make the additional assumption that $k$ is algebraically closed. Consider $k=\Bbb R$, $A=B=\Bbb C$, then $\mathrm{Spec}(A)$ and $\mathrm{Spec}(B)$ have exactly one (closed) point, whereas $\mathrm{Spec}(A\otimes_k B)=\mathrm{Spec}(\Bbb C \times \Bbb C)$ has two. This is exactly the issue you have identified that the tensor product of fields may not be a field again. So let's put in this extra assumption that $k$ is algebraically closed.

This solution uses a bit of category theory, but not much. Note that every maximal ideal of a f.g. $k$-algebra has residue field $k$ (due to Zariski's lemma and the assumption that $k$ is algebraically closed.) Thus the maximal ideals of f.g. $k$-algebras $A$ may be identified with $k$-algebra homomorphisms $A \to k$. So the functor we're trying to investigate is naturally isomorphic to the contravariant Hom-functor $\mathbf{Hom}_{k\textrm{-}\mathbf{Alg}}(-,k)$. Now a contravariant Hom-functor sends coproducts to products. But the tensor product over $k$ is the coproduct in the category of commutative $k$-algebras! This finishes the proof. This argument can be phrased without category theory. The category theory part is just applying the universal property of the tensor product of algebras to the particular case when the target is $k$.