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I've looked up numerous definitions, and all of them talk of a new operator that is added to the $R$-module that is the $R$-algebra.

One definition says

An $R$-algebra, where $R$ is a commutative ring, is a ring with identity together with a ring homomorphism $f\colon R \to A$ such that the subring $f(R)$ of $A$ is contained within the center of $A$."

I don't see how the fact that an $R$-algebra is an $R$-module with a bilinear operator follows from this definition. Thanks in advance.

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1 Answer

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An $R$-algebra is an $R$-module, which also has a ring structure. That's probably the definition you have in mind.

The definition above does it the other way round. It says an $R$-algebra is a ring which also has the described additional structure. In fact the morphism $f$ gives $A$ an $R$-module structure:

$A$ is an $R$-module via $ra:=f(r)a$ for $r\in R$ and $a\in A$. Now you want the multiplication in $A$ to be bilinear wrt $R$, which amounts to saying that $f(A)$ lies in the centre.

Conversely if $A$ is an algebra in the original sense, define $f(r)=r\cdot1$, where the multiplication comes from the module structure and $1\in A$ is the identity in the ring $A$.

Edit: Just checked the definition on wikipedia. I understand under an $R$-algebra what Wiki calls associative $R$-algebra. The definition you gave in your question also gives an associative $R$-algebra.

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