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Let $K$ be a field.

What is an example of two $K$-algebra morphisms $R\to T$ and $S\to T$ such that $\operatorname{Spec}(R\times_T S)$ is not the pushout of the diagram $$ \operatorname{Spec}(S)\leftarrow\operatorname{Spec}(T)\rightarrow\operatorname{Spec}(R) $$ in the category of schemes? Does an example exist with $R$, $S$, $T$ finitely generated? Does an example exist with $R$, $S$, $T$ and $R\times_T S$ finitely generated?

In particular, the exisitence of such an example would prove that the inclusion functor from affine schemes into schemes does not preserve pushouts.

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3 Answers 3

up vote 4 down vote accepted

The pushout of schemes even doesn't have to exist at all. There is an example which I have learnt from Anton Geraschenko and Brian Conrad.

If $X$ is an integral scheme with a non-closed generic point $\eta$ such that the closed points are dense in $X$ (for example a nontrivial affine variety), then the coequalizer of $\eta \rightrightarrows X \sqcup X$ does not exist in the category of schemes. And every coequalizer in a category with coproducts can be described as a pushout, here it is the pushout of $\eta \sqcup \eta \to X \sqcup X$ and $\eta \sqcup \eta \to \eta$. Hence, for example, $\mathrm{Spec}(k(x)) \leftarrow \mathrm{Spec}(k(x) \times k(x)) \to \mathrm{Spec}(k[x] \times k[x])$ has no pushout at all.

A sufficient condition (but not necessary) for $\mathrm{Spec}(R \times_T S) = \mathrm{Spec}(R) \cup_{\mathrm{Spec}(T)} \mathrm{Spec}(S)$ is that $R \to T$ is surjective, see the paper Gluing Schemes and a Scheme without Closed points by Karl Schwede.

In general, pushouts of schemes are quite delicate and it is quite hard to say something about them. A common mistake (even in published papers) is to assume that the forgetful functor to ringed spaces preserves pushouts. Although this may be true for some reason, it is not clear a priori. This means that it is quite hard to check if a pushout, or more generally colimits, exists or not. For example, it is quite easy to see that the colimit of $\mathbb{A}^0 \hookrightarrow \mathbb{A}^1 \hookrightarrow \mathbb{A}^2 \hookrightarrow \dotsc$ in the category of ringed spaces is not a scheme, but of course this does not prove that there is no colimit in the category of schemes. I only know the following: If a colimit of schemes $X = \mathrm{colim~}_i X_i$ exists, then $\Gamma(X,\mathcal{O}_X)=\mathrm{lim~}_i \Gamma(X_i,X_i)$. The reason is that $\Gamma : \mathsf{Sch} \to \mathsf{CRing}^{\mathrm{op}}$ is left adjoint to the functor $\mathrm{Spec}$, therefore preserves all colimits. But a priori we know nothing about the whole structure sheaf of $X$ or the topology of $X$.

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Thank you for the answer, Martin. Is it true that if the pushout of the diagram of the question exists in schemes, then it has to be $\operatorname{Spec}(R\times_T S)$? –  Ronald Bernard Mar 14 '13 at 11:04
I don't know. This is really a quite difficult problem. You can get a feeling for this by looking at the literature about pushouts of schemes (Li, Ferrand, Schwede). –  Martin Brandenburg Mar 14 '13 at 11:19
Do you refer with your last sentence in the answer to Theorem 3.4 of Schwede's paper? There, he requires $R\to T\cong R/I$ to be the quotient morphism and concludes that the pushout $\operatorname{Spec}(S)\cup_{\operatorname{Spec}(T)}\operatorname{Spec}(R)$ exists in the category of schemes. Does he say that it has to be $\operatorname{Spec}(R\times_T S)$ or is this clear? I cannot find it. Thanks. –  Ronald Bernard Mar 14 '13 at 11:34
The paper explicitly says that the pushout is $\mathrm{Spec}(R \times_T S)$. Read more carefully, you will find it. –  Martin Brandenburg Mar 14 '13 at 11:34
Ah yes, you are right and I am blind. Thank you so much. This is a nice answer of yours. –  Ronald Bernard Mar 14 '13 at 11:37

The functor $\textrm{Spec} : \textbf{CRing}^\textrm{op} \to \textbf{Sch}$ does not preserve pushouts, and this is intentional. Indeed, consider the projective line $\mathbb{P}^1_k$ over a field $k$. This is obtained by gluing two copies of $\mathbb{A}^1_k$ along $X = \mathbb{A}^1_k \setminus \{ 0 \}$, and in fact we have a pushout diagram in $\textbf{Sch}$: $$\tag{1} \begin{array}{ccc} X & \to & \mathbb{A}^1_k \\ \downarrow & & \downarrow \\ \mathbb{A}^1_k & \to & \mathbb{P}^1_k \end{array}$$ Applying the global sections functor to this diagram, we get a pullback diagram in $\textbf{CRing}$: $$\tag{2} \begin{array}{ccc} k & \to & k[t^{-1}] \\ \downarrow & & \downarrow \\ k[t] & \to & k[t, t^{-1}] \end{array}$$ However, if we apply $\textrm{Spec}$ to diagram (2) we do not get back to diagram (1) because $\mathbb{P}^1_k$ is not affine. So we have the desired counterexample.

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Nice and basic example. And here the algebras are also of finite type over $k$. –  Martin Brandenburg Mar 14 '13 at 14:01

For a simple example where the pushout of affine schemes is not preserved by the inclusion to schemes, see my answer to http://mathoverflow.net/questions/29311 .

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