5
$\begingroup$

The spectrum-functor $$ \operatorname{Spec}: \mathbf{cRng}^{op}\to \mathbf{Set} $$ sends a (commutative unital) ring $R$ to the set $\operatorname{Spec}(R)=\{\mathfrak{p}\mid \mathfrak{p} \mbox{ is a prime ideal of R}\}$ and a morpshim $f:S\to R$ to the map $\operatorname{Spec}(R)\to \operatorname{Spec}(S)$ with $\mathfrak{p}\mapsto f^{-1}(\mathfrak{p})$. Does this functor send pullback squares \begin{eqnarray} S\times_R T&\to& T\\ \downarrow && \downarrow\\ S&\to& R \end{eqnarray} of (commutative unital) rings to pushout squares \begin{eqnarray} \operatorname{Spec}(R)&\to& \operatorname{Spec}(T)\\ \downarrow && \downarrow\\ \operatorname{Spec}(S)&\to& \operatorname{Spec}(S\times_R T) \end{eqnarray} of sets? Put in other words, does the functor $\operatorname{Spec}$ from above preserve pushouts?

$\endgroup$
  • $\begingroup$ I'm not sure about this, but if your claim is true shouldn't Spec send products to coproducts? Isn't that easier to check? $\endgroup$ – Abellan Feb 6 '16 at 15:22
3
$\begingroup$

It will be true if both maps youre pushing out along are closed embeddings. It will also be true if one of the maps is an infinitesimal thickening (= a closed embedding which induces an isomorphism on the reduced closed subschemes). This type of thing is useful in defromation theory. In general, you should be able to find counterexamples pretty easily.

$\endgroup$
  • $\begingroup$ Thank you for the answer. Do you know if this is true if only one of the maps (e.g. $T\to R$) is surjective on the level of rings? $\endgroup$ – user8463524 Feb 6 '16 at 16:58
  • $\begingroup$ not that i know of (again, unless e.g. the ideal is square zero - $I^2=0$) $\endgroup$ – user304022 Feb 6 '16 at 17:03
3
$\begingroup$

Pullbacks in $\mathbf{CRing}$ do not necessarily go to pushouts in $\mathbf{Sch}$ or $\mathbf{Set}$. Consider the construction of $\mathbb{P}^1_k$: in $\mathbf{Sch}$ (resp. $\mathbf{Set}$), we have the following pushout square, $$\require{AMScd} \begin{CD} \mathbb{A}^1_k \setminus \{ 0 \} @>>> \mathbb{A}^1_k \\ @VVV @VVV \\ \mathbb{A}^1_k @>>> \mathbb{P}^1_k \end{CD}$$ but if pullbacks in $\mathbf{CRing}$ go to pushouts in $\mathbf{Sch}$ (resp. $\mathbf{Set}$), that would imply that $\mathbb{P}^1_k \cong \operatorname{Spec} k$, which is nonsense.

$\endgroup$
  • $\begingroup$ I think this is not the OP's question, though. $\endgroup$ – Qiaochu Yuan Feb 6 '16 at 18:02
  • $\begingroup$ This answers the OP's question: the corresponding cospan in $\mathbf{CRing}$ has a pullback, and $\operatorname{Spec}$ does not send it to a pushout. $\endgroup$ – Zhen Lin Feb 6 '16 at 18:05
  • $\begingroup$ But the OP's question is about Zariski Spec regarded as a functor landing in $\text{Set}$, not in $\text{Sch}$. $\endgroup$ – Qiaochu Yuan Feb 6 '16 at 18:07
  • $\begingroup$ The same example works. $\endgroup$ – Zhen Lin Feb 6 '16 at 18:09
  • $\begingroup$ Dear Zhen Lin, thank you for your answer. Does the functor ''underlying set of the underlying space of a scheme'' $\mathbf{Sch}\to\mathbf{Set}$ preserve pushouts? If not, I cannot see why your example answers my question. Thank you for helping. $\endgroup$ – user8463524 Feb 7 '16 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.