Suppose M is a left R-module, and I is an ideal of R. Let IM be the set of all sums of elements of M of the form rx, where r is in I and x is in M. Then, first of all, IM is an R-submodule of M. You should convince yourself of this; it is straightforward. Then the quotient abelian group M/IM is an R-module, as we showed last semester.

Moreover, there is a natural, well-defined, R/I-module structure on M/IM. If r+I is an element of R/I, and x+IM is an element of M/IM, define the product (r+I)(x+IM) to equal rx+IM. It is slightly more involved, but still straightforward, to show that this product is well-defined.

In particular, if R/I is a field (e.g., if R is a commutative ring and I is a maximal ideal), then M/IM is a vector space over the field R/I.

This is well-illustrated by the group of integers, considered as a Z-module, and its finite cyclic quotients. That is, let R=Z and M=Z, and let I be the principal ideal generated by 28, for instance. Then IM is 28Z, and M/IM is Z/28Z, or Z₂₈. At the same time, R/I is also Z₂₈. Of course, Z₂₈ is naturally a Z-module. But in addition, being a ring, Z₂₈ is naturally a module over itself. Thus M/IM is a module over R/I.