# Todd Trimble Dippy disproof of infinitary extensivity of affine schemes

###### Proposition

The category of affine schemes is lextensive, but is not infinitary extensive.

The fact that the category of affine schemes is lextensive is well-known. The dual statement for the category of commutative rings comes down to some easily verified claims:

• The category $CRing$ of commutative rings is of course finitely cocomplete.

• Product projections $\pi_1: R \times S \to R$, $\pi_2: R \times S \to S$ are epic, and their pushout is terminal.

• The pushout in $CRing$ of two ring maps $R \to S$, $R \to T$ is given by the tensor product $S \otimes_R T$, with the evident commutative ring structure.

• The pushing-out functor along a map $R \to S$, namely $S \otimes_R -: CAlg_R \to CAlg_R$, preserves finite products (since finite products are reflected and preserved by the forgetful functor to $Mod_R$, where they are biproducts and therefore coproducts that are preserved by tensoring).

However, the category of affine schemes is not infinitary extensive. Again, we consider what infinitary extensivity would mean for the dual category:

Take $R$ to be Noetherian. Suppose that a commutative ring $S$ over $R$ is flat as an $R$-module, and that $S \otimes_R -: CAlg_R \to CAlg_R$ preserves products. Since the forgetful functor $CAlg_R \to Mod_R$ preserves and reflects equalizers, it must be that $S \otimes_R -: CAlg_R \to CAlg_R$ also preserves equalizers and therefore all limits. We will apply this observation to certain wide pullbacks in $CAlg_R$ to deduce that $S$ is finitely generated as an $R$-module.

For any $R$-module $M$, let $\hat{M} = R \times M$ be the commutative $R$-algebra where the identity lives in the summand $R$ and the product of any two elements in the summand $M$ is defined to be zero. Then for any collection of $R$-modules $M_i$, the wide pullback in $CAlg_R$ of the wide cospan consisting of projection maps $\hat{M} = R \times M \to R$ is just $\widehat{\prod_i M_i}$. Given that $S \otimes_R -$ preserves this wide pullback, it means that the canonical arrow

$S \otimes_R \widehat{\prod_i M_i} \to S \times \prod_i S \otimes_R M_i$

is an isomorphism, and this forces the functor $S \otimes_R -: Mod_R \to Mod_R$, now regarding $S$ simply as an $R$-module, to preserve the product $\prod_i M_i$. So assuming $S$ is flat and $S \otimes_R -: CAlg_R \to CAlg_R$ preserves products, we see that $S \otimes_R -: Mod_R \to Mod_R$ preserves products and therefore limits (again by flatness).

Now $S \otimes_R -: Mod_R \to Mod_R$ is an accessible (in fact, a finitary) functor between locally presentable categories, so that under the assumption that it preserves limits, it must be a right adjoint. Its left adjoint is, as any left adjoint $Mod_R \to Mod_R$ must be, given by a functor of the form $N \otimes_R -$. Hence we have

$(N \otimes_R -) \dashv (S \otimes_R -)$

and it follows that $S \otimes_R - \cong \hom_R(N, -)$. Now the left side of this isomorphism preserves colimits, but $\hom_R(N, -)$ can preserve colimits only if $N$ is finitely generated and projective, i.e., a direct summand of some finite free $R$-module $R^n$. Hence

$S \cong S \otimes_R R \cong \hom_R(N, R)$

is also a direct summand of $R^n$, and in particular is finitely generated.

The upshot is that if a commutative algebra $S$ over a Noetherian $R$ is flat but not finitely generated as an $R$-module, then products of commutative $R$-algebras are not stable under pushing out along the ring map $R \to S$. For the opposite category of affine schemes, it means that coproducts are not stable under pullback.

Created on August 7, 2013 at 05:45:30 by Todd Trimble