Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Isn't it the same thing as a module over $A$, where you think of $A$ as a ring? – Eric O. Korman Jul 28 '11 at 0:30
up vote 4 down vote accepted

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)

share|cite|improve this answer

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.

share|cite|improve this answer
I don't have an $A$-module structure on $M$; I only have an abelian group and an algebra. – Akater Jul 28 '11 at 0:55
If you say so... – a.r. Jul 28 '11 at 6:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.