When does the categorical definition of a module work?

Question

$\DeclareMathOperator{\ab}{Ab}\DeclareMathOperator{\qcoh}{QCoh}$ This entry in the nlab shows that for $A$ a (commutative unital) ring, the category $\mathsf{Mod}_A$ of $A$-modules is equivalent to the category $\ab(\mathsf{CRing}/A)$ of abelian group objects in the slice category of rings over $A$. A module $M$ is associated with the square-zero extension $A\oplus M$, with the familiar $$ (a,m) (b,n) = (a b, a n + b m) $$ Kahler differentials, and the cotangent complex are easily defined in this setting, and this question asks about geometric intuition.

Of course, the functor $M\mapsto \mathcal{O}_X \oplus M$ works for quasi-coherent modules on an arbitrary scheme $X$, so we get an embedding $\qcoh(X)\hookrightarrow \ab(\mathsf{Sch}^{op}/X)$.

My question is this: is this an equivalence of categories? If so, does the definition of Kahler differentials at the nlab recover $\Omega^1$? More generally, does this work for stacks?

Answer 1

It is not true, but something similar is true. If $X$ is a scheme, the functor you describe is actually $\mathsf{Qcoh}(X) \to \mathsf{Ab}(\mathsf{QAlg}(X)/\mathcal{O}_X)$, where $\mathsf{QAlg}(X)$ denotes the category of quasi-coherent algebras on $X$, and $\mathsf{QAlg}(X)/\mathcal{O}_X$ is the slice category consisting of homomorphisms $A \to \mathcal{O}_X$ (which is anti-equivalent to the category of affine morphisms $Y \to X$ together with a section $X \to Y$; it is not just $\mathsf{Sch}^{\mathrm{op}}/X$!). It is equivalent to the category of non-unital quasi-coherent $\mathcal{O}_X$-algebras (i.e. semigroup objects in $\mathsf{Qcoh}(X)$), by taking the kernel of $A \to \mathcal{O}_X$. Those non-unital qc algebras $B$ for which the addition map $B \times B \to B, (a,b) \mapsto a+b$ is a homomorphism are precisely those with trivial multiplication, i.e. which are just qc modules. And this is the only map which could make $B$ an abelian group object.

Thus, the same proof as in the affine case works. Besides, we don't really use schemes or quasi-coherence here: If $X$ is an arbitrary ringed space, then $\mathsf{Mod}(X) \cong \mathsf{Ab}(\mathsf{Alg}(X)/\mathcal{O}_X)$.

The connection with Kahler differentials is fine: If $M$ is some module on a ringed space $X$, then homomorphisms $\mathcal{O}_X \to \mathcal{O}_X \oplus M$ in $\mathsf{Alg}(X)/\mathcal{O}_X$ correspond 1:1 to derivations $\mathcal{O}_X \to M$.

Finally a remark about the nlab, which I couldn't resist to include here. In my opinion, the statements at the nlab about "The correct definition of the notion of module ..." and "... the correct definition of derivations and Kähler modules" should not be taken too seriously. This is a quite subjective point of view, which may explain some aspects for modules quite elegantly, but not all of them. Besides, there are lots of notions of modules, let alone modules for monads.

Answer 2

I think it should work (modulo perhaps the usual caveat with working with non quasi-compact and quasi-separated things) but I cannot seem to find a reference.

If you don't mind going derived, this should be done in Section 3.2 of Lurie's thesis http://www.math.harvard.edu/~lurie/papers/DAG.pdf, where this approach is used to define the cotangent complex of a morphism of derived stacks. (This construction should also work for geometry over $E_n, E_\infty$ or other more exotic operads)