Does $\operatorname{Spec}$ preserve pushouts?

Question

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?

Answer 1

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.

Answer 2

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.