If $\mathfrak{m}M = o$, then the $R$-module $M$ is equivalently an $R/\mathfrak{m}$-module.

Question

While I was reading a book, I encountered this proposition:

Suppose that $\mathfrak{m}$ is an ideal of a ring $R$ and that $M$ is an $R$-module. If $\mathfrak{m}M = o$, then the $R$-module $M$ is equivalently an $R/\mathfrak{m}$-module.

I just can't come to grips with this proposition. Can anyone explain this statement? Thanks.

Answer 1

Recall that the elements of $R/\mathfrak m$ are cosets $a + \mathfrak m$. The representative element $a$ is not unique, you can adjust by any element of $\mathfrak m$ and still identify the same coset.

The obvious choice for turning $M$ into an $R/\mathfrak m$-module is to define $(a + \mathfrak m)\cdot x = ax$. But if $m \in \mathfrak m$ then $a + \mathfrak m = (a + m) + \mathfrak m$ therefore we need $(a + \mathfrak m)\cdot x = ((a + m) + \mathfrak m)\cdot x$ to hold. This simplifys to $ax = (a + m)x$ or equivalently $mx = 0$. Thus it must be that elements of $\mathfrak m$ annihilate $M$, otherwise the action $(a + \mathfrak m)\cdot m = am$ would not be well defined.

Answer 2

For the $R/\mathfrak m$ module structure given by $[r] \cdot m := rm$ on $M$ to be well defined, we need $\mathfrak mM=0$.

Or in the more algebraic way: The Ringhomomorphism $R \to End(M)$ factors over $R/\mathfrak m$ if and only if $\mathfrak mM = 0$.