When trying to prove some $M$ is an $A$-module, with $A$ being an algebra, how do I deal with the set of generators of $A$?

Question

Suppose I want to find out if $M$ is an $A$-module, with $A$ being an algebra with a set of generators $S$ and $R$ a set of equations satisfied by the generators of $A$.

For the sake of simplicity, suppose there are only two elements in the generating set $S$, say $x, y$, and assume further that $R=\{x^2=y\}$. I am now interested in showing that $M$ is an $A$-module.

I have defined multiplication only between the generators $x, y$ of $A$ and the elements $m$ of $M$. Our teacher said this does not suffice to give $M$ an $A$-module structure. He said that we need to also make sure that $x^2*m=y*m$ for all $m\in M$, after defining the multiplication. Why is this required?

Answer 1

Recall that an abelian group $M$ is a left $R$-module if the scalar product $$ R \times M \to M, (r,m) \mapsto rm $$ satisfies the usual axioms.

We need this scalar product to be well-defined!

Answer 2

By defining what the generators do to elements of $M$, you are giving a putative action of $A$ on $M$. What remains to check is that the relations act like zero.

So in your example, if $M = \mathbb{Z}^2$ and $x$ acts by flip-flopping the factors and $y$ acts as the identity, you're all set. But if $x$ acts by multiplication by $0$ and $y$ acts by multiplication by $-1$, you haven't defined the structure of an $A$-module on $M$.