Closed points of a fibred product of k-schemes

Question

This question comes from Shafarevich, Chapter V.4,

Let $X$ and $Y$ be schemes over an algebraically closed field $k$. Show that the correspondence $ u \to (p_x(u),p_y(u)) $ establishes a 1-1 map between closed points of $ X \times_k Y$ and the pairs $(x,y)$, where $x,y$ are closed points of $X$ and $Y$ respectively.

I've been trying to tackle this in the case of affine schemes (I'm assuming the general case reduces to this), where $X = \mathrm{Spec }A$ and $Y = \mathrm{Spec }B$. In this case the fibred product is $\mathrm{Spec }A \otimes_k B$, where $A$ and $B$ have the structure of $k$-algebras.

I can see how this works in the case of affine rings (i.e finitely generated, reduced k-algebras), but I don't really see why it should work in general. Why, for example, does the projection map (induced by $ a \to a \otimes 1$) send closed points to closed points?

Answer 1

This is not true. If it were, then $\mathbf C\otimes_\overline{\mathbf{Q}}\mathbf C$ would have only one maximal ideal, hence would be a local ring. However, it follows from the main theorem in this paper of Sweedler that $\mathbf C\otimes_\overline{\mathbf{Q}}\mathbf C$ is not a local ring, because $\mathbf C$ is transcendental over $\overline{\mathbf Q}$.

Shafarevich definitely meant to say that $X$ and $Y$ were of finite type over $k$. I had previously misread your question as being about that case, and I wrote the following answer. Even though I see now that you have already worked this case out, perhaps someone can benefit, so I will leave my old answer here:

---

This is one of those things that is completely obvious from the point of view of classical geometry: the image of a point is a point. What else could it be! For schemes, with "point" meaning "closed point", this is not true in general, and the fact that it is true for varieties reflects a special property of affine algebras:

If $f: A \to B$ is a morphism of affine $k$-algebras and $\mathfrak m$ is a maximal ideal of $B$, then $f^{-1}(\mathfrak m)$ is a maximal ideal of $A$.

To prove it, consider the composite $A \to B \to k$ where $B\to k$ is the evaluation at $\mathfrak m$ (quotient) map. It is obviously surjective with kernel $f^{-1}(\mathfrak m)$, hence $f^{-1}(\mathfrak m)$ is maximal.

Remarks:

This is still true if $k$ is not algebraically closed, but the composite $A \to B/\mathfrak m = k'$ need no longer be surjective, because $k'$ might be a nontrivial extension of $k$. However, its image is still a subfield of $k'$; to prove it, one uses Zariski's lemma, which I used implicitly in the algebraically closed case to identify $B/\mathfrak m$ with $k$.

More generally, this is true if $k$ is replaced by a Jacobson ring $R$. In fact, the statement that this property holds for all morphisms of affine $R$-algebras is equivalent to $R$ being Jacobson.