# Does $\operatorname{Spec}$ preserve pushouts?

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?

• 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? – Abellan Feb 6 '16 at 15:22

• 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? – user8463524 Feb 6 '16 at 16:58
• not that i know of (again, unless e.g. the ideal is square zero - $I^2=0$) – user304022 Feb 6 '16 at 17:03
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.
• 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. – Zhen Lin Feb 6 '16 at 18:05
• But the OP's question is about Zariski Spec regarded as a functor landing in $\text{Set}$, not in $\text{Sch}$. – Qiaochu Yuan Feb 6 '16 at 18:07
• 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. – user8463524 Feb 7 '16 at 18:56