1
$\begingroup$

Let $K,X,Y$ be $S-$schemes(i.e. $K\to S, X\to S, Y\to S$ are given and fixed.) Consider $Z=X\times_S Y$ and $\phi: Hom(K,Z)\to Hom(K/S,X/S)\times Hom(K/S,Y/S)$ where $Hom(K/S,X/S)$ is the set of scheme morphism $f:K\to X$ s.t. $f$ is $S-$scheme morphism.

From fibre product definition, it is clear that $\phi$ is bijection. In particular, this indicates $K-$valued points of $X\times_S Y$ is decomposed into product of sets of $K-$valued points. Suppose $K-$valued points form a subscheme of $X\times_S Y$. This decomposition of $Hom(K,X\times_S Y)$ is indicating $K-$valued points become products(set theoretical sense).

$\textbf{Q1:}$ Do I always know $K-$valued points form subscheme?

$\textbf{Q2:}$ Should I say $K-$valued points is really product $K-$valued points in scheme sense?(i.e. Denote $X_K$ set of $K-$ valued points. Then $(X\times_S Y)_K=(X/S)_K\times (Y/S)_K$ where $(X/S)_K$ is defined as $K-$valued points of scheme over $S$(i.e. $K,X$ are $S-$scheme and $K\to X$ is $S-$scheme morphism.) Do I know $(X/S)_K$ is scheme in general?

$\endgroup$

1 Answer 1

1
$\begingroup$

Before getting into specific arguments, I think there's a bit of a misuse of language in your question that is leading you astray. Namely, if $S$ is a set, it doesn't make sense to say that $S$ "is" a scheme. You could put multiple incompatible scheme structures on $S$.

This is no different from the situation for simpler categories like groups and sets: a group is a set together with extra structure, so if you just have a set, it doesn't make sense to ask if it "is" a group without specifying the extra structure. Now, admittedly, sometimes in these easy contexts, when the extra structure is clear, we abuse language and say "is" when we mean "is, together with this structure." For example, if someone says, "the set of matrices with determinant one is a group," it would be unreasonable and pedantic to argue with that person, because from context it is clear that the intended operation is matrix multiplication.

In your case, though, the distinction really isn't pedantic. If $K$ is a scheme, then the $K$-valued points of some other scheme $S$ is just a set and it's not at all clear where it should get a scheme structure. Nonetheless, I'd argue that the best possible formalization of your question Q1 is "no."

Namely, if $K$ happens to be the spectrum of a field, then the $K$-valued points of $S$ do have a bit more of geometric flavor to them; in particular, if $S$ is separated and of finite type over an algebraically closed closed field $k$ and $K = \text{Spec} \ k$, then by the Nullstellensatz, the $K$-valued points of $S$ are exactly the closed points of $S$. This isn't a subscheme, though, in the sense that there is no structure of locally-ringed-space on it such that its inclusion into $S$ is a morphism of schemes.

I think Q2 assumes that the answer to Q1 is yes, but let me know if you'd like more clarifications there.

$\endgroup$
3
  • $\begingroup$ So I do not expect getting scheme structure if $K$ is not a point in general and there is only a single $K-$ point? I see your point for algebraically closed field case for $K=spec(k)$ case. This statement is made in purely set theoretical sense then. $\endgroup$
    – user45765
    Feb 9, 2019 at 18:47
  • 1
    $\begingroup$ Yes, I think ordinarily there is no good choice of scheme structure. Maybe a concrete example -- $\mathbb{Q}$-points of a variety over $\mathbb{Q}$ correspond to rational solutions to the equations -- these can be empty, or Zariski dense, or anywhere in between, depending on the equations, and don't inherit the scheme structure of the ambient scheme. $\endgroup$
    – hunter
    Feb 9, 2019 at 18:56
  • $\begingroup$ Thanks a lot for clarification. $\endgroup$
    – user45765
    Feb 9, 2019 at 18:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .