# A module problem

1. May 5, 2007

### Hurkyl

Staff Emeritus
Let R be a commutative ring, with subring S.

Let M be an R-module.

Does there exist a S-module N such that $N \otimes_S R \cong M$ as R-modules? Preferably with N a sub-S-module of M?

Even better, can we choose such modules N so that if we have an R-module homomorphism
f:M --> M'​
it yields an S-module homomorphism
g:N --> N'​
so that

Code (Text):
N (x) R ---> M
|            |
g (x) R |            | f
V            V
N' (x) R --> M'

(with the horizontal arrows the aforementioned isomorphisms)

Last edited: May 5, 2007
2. May 5, 2007

### matt grime

No. Is the simple answer. Such a module, M, I would be tempted to call S-free.

For a well understood (by me) counter example, I like to think of group algebras. The functor $?\otimes_{kH}kG$ is the induction functor, and it is certainly not essentially surjective - though it is true, that every module is a summand of a kH-free object for some H (called the vertex subgroup, and this is a lovely theory of Green's from the 50s.

In general, you're probably asking for an equivalence of categories here, given by restriction from R to S.

3. May 5, 2007

### Hurkyl

Staff Emeritus
I didn't think it'd be that easy, but I couldn't see how to go about constructing a counterexample.

I'm hoping to find circumstances for which, if I had two maps R --> T with S their equalizer, that S-mod --> R-mod would resemble an equalizer of the induced functors R --> T. And if not in Cat, then in what sort of 2-category it does.