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In his book Global Calculus, S. Ramanan defines modules over a sheaf of algebras. He demands the abelian group $\mathcal{M}(U)$ to be provided with the structure of $\mathcal{A}(U)$-module for every open $U$. Note that here $\mathcal{A}$ is a sheaf of algebras, not rings.

Strangely enough, I could not find the definition of ‘module over algebra’. Everything I came up with was the following:

If algebra $A$ over $R$ is a homomorphism $R\longrightarrow A$ of commutative rings, and module $M$ over ring $A$ is a homomorphism $A \longrightarrow \mathop{\mathrm{End}} M$ in the same category, then module over algebra should be an arrow from $R\longrightarrow A$ to $R\longrightarrow \mathop{\mathrm{End}} M$ (an arrow between objects under $R$),

but the homomorphism $R\longrightarrow \mathop{\mathrm{End}} M$ is not given, unless $M$ is already an $R$-module (mostly a real vector space, for $\mathcal{A}$ is usually a sheaf of $\mathbb{R}$-algebras of differentiable functions).

How should I think of modules over algebras in this context? Is there a common definition?

akater
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2 Answers2

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A module over an $R$-algebra $A$ is just an $A$-module.

(When one is dealing with a bimodule $M$ over an $R$-algebra $A$, in general one only considers those such that the action of $R$ on the left and on the right on $M$ coincide, but there is no such condition for one-sided modules)

Mariano Suárez-Álvarez
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I'm not sure to understand your question: what do you mean by "the homomorphism $R \longrightarrow \mathrm{End} M$ is not given"? -If you have an $A$-module structure on $M$, that is, a morphism $A \longrightarrow \mathrm{End}M$, and $A$ is an $R$-algebra, that is, a morphism, $R \longrightarrow A$, of course you are given a morphism $R \longrightarrow \mathrm{End} M$: it's just the composition of the previous ones.

Agustí Roig
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