# Olsson's Algebraic Spaces and Stacks Paragraph 2.3.1

The paragraph reads: Let $$T$$ be a topos, and let $$\Lambda$$ be a ring in $$T$$. Denote by $$\operatorname{Mod}_{\Lambda}$$ the category of $$\Lambda$$-modules in $$T$$. My question is how is the category of $$\Lambda$$-modules defined?

The text defined a sheaf of abelian group(rings) in $$T$$ as a sheaf $$A\in T$$, together with morphisms of sheaves of sets $$m: A\times A\to A$$ and $$e: \{*\}\to A$$, satisfying the corresponding axioms. I assume a $$\Lambda$$-module $$M$$ would be a sheaf of abelian groups, along with an action of $$\Lambda$$ on it, but I don't see how to define the action of $$\Lambda$$ on $$M$$ in this general setting.

The definition works in every category $$\mathcal{C}$$ with finite products. We define a $$\Lambda$$-module in $$\mathcal{C}$$ to be an abelian group $$A$$ in $$\mathcal{C}$$ together with a morphism $$\cdot : \Lambda \times A \to A$$ such that the diagrams commute which correspond to the module axioms. For example, $$1 \cdot a = a$$ means that $$1 \times A \xrightarrow{1 \times A} \Lambda \times A \xrightarrow{\cdot} A$$ is equal to $$p_2$$. Every module axiom (and also every ring axiom) can also be written down more conveniently with generalized elements. The previous axiom really just says that $$1 \cdot a = a$$ holds for all generalized elements $$a \in A$$. If $$\mathcal{C}$$ is the topos of sheaves on a space, this means that $$1 \cdot a = a$$ holds for all local sections $$a$$ of $$A$$.
• Another way of putting the construction using generalized elements would be in terms of Yoneda's lemma: $\cdot : \Lambda \times A \to A$ corresponds to a morphism of functors $\operatorname{Hom}({-},\Lambda) \times \operatorname{Hom}({-},A) \to \operatorname{Hom}({-},A)$, and similarly for the ring operations on $\Lambda$ and group operations on $A$. So then for example, the two sides of distributivity $(a + b) x = ax + bx$ give morphisms $\operatorname{Hom}({-},\Lambda) \times \operatorname{Hom}({-},\Lambda) \times \operatorname{Hom}({-},A)\to\operatorname{Hom}({-},A)$, which... Jan 13, 2023 at 21:47
• then correspond to two morphisms $\Lambda\times\Lambda\times A\to A$, and you then assert the equality of those two morphisms. Jan 13, 2023 at 21:48
• Or, given that we're working in a topos, we could just cast the identities as statements in the internal language of a topos, e.g. $(\forall a:\Lambda)(\forall x,y:A) a(x+y)=ax+ay$, and asserting that the interpretation of that statement is equal to $\top : \operatorname{Hom}(1, \Omega)$ would turn out to be exactly what we need. Jan 13, 2023 at 21:51