# Seeing modules as ring extensions

In a paper about Poisson modules that I can't seem to find at the moment, I saw a description of Poisson modules that was rather different from most descriptions I'd seen; however, after considering it, it seemed to naturally extend to many other situations. The description was approximately:

Let $A$ be a Poisson algebra. Then a Poisson module over $A$ is a vector space $E$ and a Poisson algebra structure on $A \oplus E$ such that 1) it agrees with the original structure on $A$, and 2) $\{e_1, e_2\} = e_1 e_2 = 0$ for any $e_1, e_2 \in E$.

The general idea does seem to work for other situations than Poisson; it clearly works for commutative algebras, and works for Lie algebras in the same way as for Poisson algebras.

This leads to a few questions, then. 1) Is this a "good" way to think about algebras (of various forms) and modules? Are there any situations where it doesn't work? 2) The apparent thing that "separates" the "base" algebra structure from the module structure is that the module part is nilpotent (in fact, it "squares" to 0). This made me think back to non-reduced schemes, and made me think that the "right way" to think of a nonreduced scheme is a reduced scheme with extra module structures on top "localized at" the nonreduced points? 3) If so, is there a natural way to think about the "higher modules", since nilpotence can happen at higher powers than 2? 4) Are there any other natural ideas that result from this?

Edited to give a partial answer I thought of afterwards: This really isn't correct; the appropriate idea here is when $E$ is a bimodule. It just happens that in the cases I was thinking of, a module is naturally a bimodule. So if the type of algebra was "arbitrary not-necessarily-commutative algebra", then the above structure would not be equivalent to a module, but a bimodule.