Why is the fibered coproduct of affine schemes not affine?

Question

I am confused about the following issue:

Let $X=SpecS$, $U_1=SpecR_1$, $U_2=SpecR_2$. and suppose we have maps $S \rightarrow R_1$, $S \rightarrow R_2$. Let $U_3=Spec (R_1 \otimes_S R_2)$. We have scheme maps $U_1 \rightarrow X$, $U_2 \rightarrow X$, $U_3 \rightarrow U_1$, $U_3 \rightarrow U_2$. The particular situation I have in mind is when $U_1$ and $U_2$ are distinguished (corresponding to localization of S at some element) open subschemes of $X$. Intersection of $U_1$ and $U_2$ is $U_3$, and the inclusion of $U_3$ in $X$ corresponds to $S$-algebra structure on $(R_1 \otimes_S R_2)$.

The category of affine schemes (ASch) is the opposite category of commutative rings (CRing). In CRing kernels (equalisers) of pairs of maps and products exist, so by a lemma from category theory limits should exist, in particular fibered coproducts should exist, so union of two affine schemes $U_1$ and $U_2$ over $U_3$ should be affine scheme $U_4$! But we know that in general it is not so! Maybe the problem is that abstractly it is an affine scheme but what is it's inclusion map into X? Actually there exists an obvious map on the ring side from $S$ to kernel of a pair of maps $R_1 \rightarrow (R_1 \otimes_S R_2)$, $R_2 \rightarrow (R_1 \otimes_S R_2)$.

Thank you!

Answer 1

The short answer is that the category of affine schemes does have pushouts, but these are not the same as pushouts of affine schemes calculated in the category of all schemes.

For a longer answer, consider an example that's small enough to compute: The projective line (over the complex numbers, say) is not affine but it is gotten by gluing two copies of the affine line along the punctured affine line, so it is the pushout (in the category of schemes) of a diagram of affine schemes. Now what's the pushout of that same diagram in the category of affine schemes? Well, the rings involved are two copies of $C[x]$ and a copy of $C[x,x^{-1}]$. The two maps are the two injections of $C[x]$ into $C[x,x^{-1}]$, one sending $x$ to $x$, and the other sending $x$ to $x^{-1}$. The pullback of these, in the category of commutative rings, is just $C$, because the only way a polynomial in $x$ can equal a polynomial in $x^{-1}$ is for both of them to be constant. Therefore, the affine-scheme pushout is not the projective line but a point.

Intuitively, if you glue together two copies of the line along the punctured line "gently," allowing the result to be non-affine, you get the projective line, but if you demand that the result be affine then your projective line is forced to collapse to a point.

Answer 2

From the categorical point of view the situation is the following. We have the category of affine schemes $AffSch$, the category of all schemes $Sch$, and the inclusion functor $i:AffSch \to Sch$. The functor $i$ has a left adjoint functor $i^*:Sch \to AffSch$, $S \mapsto Spec \Gamma(S,{\mathcal{O}}_S)$, which is sometimes called the affine envelope.

Now, the coequalizer (the pushout) by definition is the object which corepresents a certain contravariant functor to $Sets$. Note that whenever we have a functor $i:C \to D$ having a left adjoint and a contravariant functor $F:D \to Sets$, if $F$ is corepresentable by an object $X$, then $F\circ i$ is corepresentable as well and the corepresenting object is $i^{\ast}(X)$. Indeed, if $F(Y) = Hom(X,Y)$ then $$F(i(Z)) = Hom(X,i(Z)) = Hom(i^{\ast}(X),Z).$$ Applying this to the situation of the first paragraph we see that the restriction of a corepresentable (by an scheme $S$) functor from $Sch$ to $AffSch$ is correspresentable by $i^{\ast}(S)$. In particular, the coequalizer in the category of affine schemes is the affine envelope of a coequalizer in the category of all schemes.