5
$\begingroup$

This question was posted here on StackExchange.

Let $A$ be a commutative ring and $B,C$ be two commutative $A$-algebras. Consider the pushout square of ring homomorphism $\require{AMScd}$ \begin{CD} A@>\beta>>B\\ @V\gamma VV@VVV\\ C@>>>B\otimes_AC \end{CD} In this answer it is proved that the corresponding commutative square of spectrum $\DeclareMathOperator\Spec{Spec}$ \begin{CD}\tag 1 \Spec(B\otimes_AC)@>>>\Spec(B)\\ @VVV@VV\Spec(\beta)V\\ \Spec(C)@>>\Spec(\gamma)>\Spec(A) \end{CD} is a pullback in the category of schemes.

I'm looking for sufficient conditions on $A\to C$ which make $(1)$ a pullback square in category of sets for every $A\to B$.

Two such conditions are:

  • if $C=A/\mathfrak a$ for some ideal $\mathfrak a$ of $A$;
  • if $C=S^{-1}A$ for some mutliplicative system $S$ of $A$.

Note that in both case, $A\to C$ is an epimorphism of commutative rings; so my question is:

If $A\to C$ is an epimorphism of commutative rings, then $(1)$ is a pullback of sets for every ring homomorphisms $A\to B$?

My try. Let $\beta:A\to B$, $\gamma:A\to C$ and $\tau:A\to B\otimes_AC$ and consider the pushout square of commutative rings above. If $\gamma:A\to C$ is a ring epimorphism, then the right-handed $B\to B\otimes_AC$ is a ring epimorphism as well. Moreover, it's know that the functions $\Spec(B\otimes_AC)\to\Spec(B)$ and $\Spec(C)\to\Spec(A)$ are injective. By [Atiyah & MacDonald - ex. 25 pag. 48] we have $$\operatorname{Im}\Spec(\tau)=\operatorname{Im}\Spec(\beta)\cap\operatorname{Im}\Spec(\gamma)$$ enter image description here

This proves that for every set $X$ there exists a function $h:X\to\Spec(B\otimes_AC)$ and the injectivity of $\Spec(C)\to\Spec(A)$ implies that the left-handed triangle commutes. I've troubles in showing that the upper triangle commutes as well.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Your question is unclear: first you seem to ask the general question (viz.: is it always true that Spec sends tensor products to pullbacks of topological spaces?), to which the answer is a resounding “NO” ($\mathbb{A}^2_{\mathbb{C}}$ as a topological space is not the product of two copies of $\mathbb{A}^1_{\mathbb{C}}$), then you ask about a specific equivalence: so, which is your question? $\endgroup$
    – Gro-Tsen
    Commented Sep 22, 2019 at 8:00
  • $\begingroup$ If the second question is your real question, you can't seriously hope to do a characterization with $A\to C$ only, since the situation is symmetric and $A\to B$ plays exactly the same rôle! If $A\to B$ satisfies one of the sufficient conditions you listed and $A\to C$ does not, then the diagram will still be a pullback of topological spaces! $\endgroup$
    – Gro-Tsen
    Commented Sep 22, 2019 at 8:03
  • 3
    $\begingroup$ So maybe your real question should be: under what conditions on $A\to C$ is it true that Spec takes $B\otimes_A C$ to the fiber product of topological spaces for all $A\to B$. But whatever the case, you should clarify your question because, so far, it's not clear what you're asking. $\endgroup$
    – Gro-Tsen
    Commented Sep 22, 2019 at 8:05
  • $\begingroup$ What is $X$ in your diagram? $\endgroup$
    – Martin Brandenburg
    Commented Jan 10, 2020 at 22:03
  • $\begingroup$ @MartinBrandenburg: Here $X$ is a generic set (or topological space); I'm trying to prove that the square of spectrum has the universal property of a pullback square in the category of sets (or topological spaces). $\endgroup$
    – Fabio Lucchini
    Commented Jan 10, 2020 at 22:06

1 Answer 1

Reset to default
3
$\begingroup$

Let me work directly with schemes (you can of course restrict to affine schemes if necessary).

For $S$-schemes $X,Y$ there is a natural map $$|X \times_S Y| \to |X| \times_{|S|} |Y|,$$ which is surjective, and the fiber of $(x,y,s)$ can be identified with $$\mathrm{Spec}(k(x) \otimes_{k(s)} k(y)\bigr).$$ So the condition is equivalent to saying that these tensor products have exactly one prime ideal.

If $X \to S$ is a monomorphism of schemes, then $k(s) \to k(x)$ is an epimorphism between fields, hence an isomorphism. Hence, $ k(x) \otimes_{k(s)} k(y) \cong k(y)$, and the result follows.

Improve this answer
$\endgroup$

You must log in to answer this question.

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