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So a module over a commutative ring $ R$ is an abelian group $G$ equipped with an action given by the product $R\times G\rightarrow G$ that satisfies a few conditions. What if $G$ itself is a ring? Is there a notion of a "ring acting on another ring?"

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    $\begingroup$ Look up "$R$-algebra." $\endgroup$ May 29, 2016 at 4:34

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There are plenty of cases where this happens. It's really only interesting when the module action interacts in an interesting way with the multiplication

The most straightforward example, which works when $R$ is commutative, is the concept of an $R$-algebra, which I'll let you look up yourself. In this case it's easy to see what the action does.

If you look in my profile you can find my dissertation which has other examples. For an easier non-$R$-algebra example, let's make the polynomial ring $\Bbb R[y]$ act on $C^\infty(\mathbb R)$ by the rule $$(yf)(x)=f'(x)$$ Then the interaction with the multiplication is given by the Leibniz formula, $$y^n(fg)=\sum_{k=0}^n{\binom{n}{k}(y^kf)(y^{n-k}g)}$$

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This is called an $R$-algebra. An $R$-algebra is a ring $M$ with a unital map $R\to M$ whose image is contained in the center of $M$. Alternatively, an $R$-algebra is an $R$-module $M$ such that the multiplication $M\times M\to M$ is $R$-bilinear.

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