# Module equivalence

1. Aug 25, 2010

### Bleys

Module "equivalence"

There is a problem in a book I'm not quite understanding.
Let M be an R-module and let I=Ann(M). Show that M can be regarder as an R/I-Module where scalar multiplication is given by the rule m(I+r)=mr

I don't understand what they mean by "regarded as". Am I suppose to show there is an isomorphism? (I don't know how to do that between modules over different rings), or just a bijection? Should I create a one-to-one mapping from the scalars R to R/I instead?

Any help is appreciated

2. Aug 25, 2010

### Hurkyl

Staff Emeritus
Re: Module "equivalence"

Really, they probably just want you to show that the underlying additive group of M, together with that scalar multiplication, is an R/I-module.

Bonus points for explicitly showing that, if $\pi$ is the projection from R to R/I, that $r \cdot m = \pi(r) \cdot m$. (where the two $\cdot$ symbols refer to the appropriate scalar products)

One main point is that, in the expression $r \cdot m$ where r is in R and m is in M, it doesn't matter if we mean this product as the scalar product on M, or if we are using r to name an element of R/I, and the product is the scalar product on the R/I-module defined in the problem; the result of the arithmetic expression is the same.

3. Aug 26, 2010

### Bleys

Re: Module "equivalence"

Ok, thank you Hurkyl!
Just out of curiosity, what exactly does this result mean, the fact that the set M can be regarded as both an R-module and an R/I-module? Is this mostly a consequence of the properties of the quotient ring R/I (for example of, like you pointed out, the canonical homomorphism from R to R/I)?
I'm finding it hard to wrap my head around it because R and R/I are not isomorphic (possibly not even of the same order if finite) yet behave the same as scalars over M?

4. Aug 26, 2010

### Hurkyl

Staff Emeritus
Re: Module "equivalence"

The more opportunities you have to use this fact, the more it will make sense.

You already know that all of the elements of I behave exactly the same as scalar multipliers on M. So what do you get when you impose that they are all congruent?

Incidentally, there are two well-behaved constructions related to a ring homomorphism f:R --> S.
• (The underlying group of) any S-module M can be given an R-module structure defined by $r\cdot m = f(r)\cdot m$.
• By the above, S itself can be viewed as an R-module. For any R-module M, the (underlying group of) the tensor product $S \otimes_R M$ can be given an S-module structure defined by $s \cdot (s' \otimes m) = (ss') \otimes m$.
(These constructions extend to transform S-module homomorphisms into R-module ones and R-module into S-module ones respectively)

When S is the quotient ring R/I, the constructions are even more well-behaved. The underlying group of the second construction is just (isomorphic to) the quotient $M / IM$. Furthermore, the two constructions essentially become one-sided inverses -- applying the first construction and then the second construction gives you a module naturally isomorphic to the one you started with.

Roughly speaking, this means that you can view the entire category of R/I-modules as if it was just a subcategory of R-modules. Another way of saying that is that it's unusually simple to use the module theory of R to study the module theory of R/I. How do you identify which R modules are also R/I modules under this correspondence? They are the ones with $I \subseteq Ann(M)$.