# [SOLVED]R-modules and Homomorphism of Rings

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, I find it difficult to get the exact meaning of the following question. What does "in a natural way" means? Is it that we have to show that there exist a isomorphism between $$S$$ and a sub-module of $$R$$? Any ideas are greatly appreciated. Question:

Given a ring homomorphism $$f:R\rightarrow S$$, show that every $$S$$-module can be considered as an $$R$$-module in a natural way.

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, I find it difficult to get the exact meaning of the following question. What does "in a natural way" means? Is it that we have to show that there exist a isomorphism between $$S$$ and a sub-module of $$R$$? Any ideas are greatly appreciated. Question:

Given a ring homomorphism $$f:R\rightarrow S$$, show that every $$S$$-module can be considered as an $$R$$-module in a natural way.
I think I am getting some understanding. Suppose if we have a $$S$$-module called $$M$$. Then we can define the operation $$R\times M\rightarrow M$$ as, $$r.m=f( r)\,m$$. Under this operation it's clear that $$M$$ becomes a $$R$$-module. This can be verified by showing that $$M$$ satisfies the $$R$$-module properties under the operation defined above.

This I believe is the natural way of defining an $$R$$-module from a given $$S$$-module where $$R$$ and $$S$$ are homomorphic. Correct me if I am wrong. #### Deveno

##### Well-known member
MHB Math Scholar
Yes.

It is "natural" because it's pretty much the only way you have available to define

$r.m$.

This theorem tells us, for example, that any $\Bbb Z_n$-module can simply be regarded as a $\Bbb Z$-module, that is: an abelian group, and explains why reducing a polynomial (mod p) can be so useful in investigating polynomials with integer coefficients, because often information in the $S$-module can be "lifted" to an $R$-module.

#### Sudharaka

##### Well-known member
MHB Math Helper
Yes.

It is "natural" because it's pretty much the only way you have available to define

$r.m$.

This theorem tells us, for example, that any $\Bbb Z_n$-module can simply be regarded as a $\Bbb Z$-module, that is: an abelian group, and explains why reducing a polynomial (mod p) can be so useful in investigating polynomials with integer coefficients, because often information in the $S$-module can be "lifted" to an $R$-module.
Thank you so much. I am sure your immense knowledge about these things will be of great help to me this semester. 