What to do with my proof that LRS is not cartesian closed?

[dschepler]

Hmm... I was planning to submit a proof I've found that LRS is not cartesian closed. The basic idea was to prove that if k is a finite field, then LRS_k is not cartesian closed. The idea of the proof is to examine the functor $\mathbf{Top} \to \mathbf{LRS}_k$ of equipping a topological space with the constant sheaf of $k$. That has a left adjoint given by the forgetful functor, and it has a right adjoint taking a locally ringed space over k to the subspace of points where the canonical morphism from k to the residue field is an isomorphism. The proof of the right adjoint part, in particular, requires the finiteness assumption on k in order to construct locally constant functions. Then, I would use this general lemma to get a contradiction: if D is a coreflective subcategory of C which is essentially closed under binary products (or equivalently D has binary products and the inclusion functor preserves them), and C is cartesian closed, then D is cartesian closed.

Then, to pass to LRS not being cartesian closed, I would use a lemma: if C is a cartesian closed category, and P is a subterminal object, then the slice C/P is also cartesian closed. So, if LRS were cartesian closed, then LRS/F_p would be cartesian closed also for each prime p, giving a contradiction.

But then, I noticed you just pushed changes replacing LRS with LRS_R and Sch with Sch_R. So now, I'm not sure how to proceed since my proof (at least at the moment) is limited to rings R which have a finite field as a quotient.

[ScriptRaccoon]

Great to hear that you have found a proof. Let us specialize R then. We can generalize later. When your proof works for every R that has a finite quotient, let's add this to the definition of the category - for now.

[ScriptRaccoon]

Alternatively, you can add back LRS (over Z) but also keep LRS_R. I don't have a strong opinion.

[dschepler]

I think I've found a proof that the functor has a right adjoint more generally if $R$ is any field $k$. We start with the same construction of letting $X_0$ be the subset of points of $X$ with residue field $k$. But in the more general case, we need to strengthen the subspace topology by declaring a neighborhood subbasis at $x$ to be sets of the form $X_0 \cap Z(f)$, where $f \in O_X(U)$ for some open neighborhood $U$ of $x$ in $X$, such that $f_x \in m_{X,x}$. That strengthened topology appears to be exactly what's needed to be able to take a continuous function into that space and a section of $O_X$, and define a pullback as a locally constant function from the domain to $k$. (My proof in the finite field case amounted to the observation that $X_0 \cap Z(f) = \bigcup_{a\ne 0} X_0 \cap D(f-a)$ so it's already open in the subspace topology.)