6

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?

Bruno Joyal
  • 55,975
Gabriel_s_syme
  • 187

1 Answers1

4

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.

Bruno Joyal
  • 55,975
  • Thanks, the result above is very useful, though I fear your proof of it is not correct. You state that the composite map $A \to k$ is surjective, this is not true in general. However, you can say that $A/f^{-1}(m)$ is a finitely generated k subalgebra of $k$, which is algebraic (and a domain) over $k$, this implies that it is a field. All of these results depending on the assumption of being of finite type.

    At least, this is what I have gathered from here: http://math.stackexchange.com/questions/107291/the-preimage-of-a-maximal-ideal-is-maximal

     – Gabriel_s_syme May 16 '14 at 13:56
  • Dear @Gabriel: Do the constants of $A$ not map surjectively to the elements of $k$? – Bruno Joyal May 16 '14 at 16:33
  • That is certainly an example of "a" surjective map from $A$ to $k$, but the map in our argument is the composition of the projection induced by $m$ and the map $f: A \to B$, where $f$ can be any morphism. If $f$ maps $A$ to a single point, then the composition is certainly not surjective.

    Anywho as stated above, the map doesn't need to be surjective anyway.

    Also I still haven't quite figured out why the map is a bijection of closed points.

     – Gabriel_s_syme May 18 '14 at 20:47
  • @Gabriel I think that you are confused. How can $f$ map A to a single element of B? It's a unital ring homomorphism, it sends 0 to 0 and 1 to 1. I stand by what I have written; there is nothing incorrect about it. – Bruno Joyal May 18 '14 at 20:56
  • In fact, in the category of k-algebras, any map $A\to k$ is surjective, for any A and k any field. This follows from the very definition of the morphisms in that category. – Bruno Joyal May 18 '14 at 21:11
  • Ah, ok. I see what you are getting at. – Gabriel_s_syme May 19 '14 at 09:23
  • @BrunoJoyal Did you assert at the end of your statement that the 'the closed points of the fiber product $X\times_{\operatorname{Spec} R} Y$ of schemes of finite type over a scheme $\operatorname{Spec} R$ are in bijection with the Cartesian product of the set of closed points of $X$ with the set of closed points of $Y$ if and only if $R$ is a Jacobson ring? – Tomo Feb 03 '16 at 00:25