R-modules and Homomorphism of Rings

[Sudharaka]

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]

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]

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]

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. :)

 

[blue_raver22]

I understand "in a natural way" to mean that there is a clear and intuitive way to establish a relationship between the two structures, in this case between the \(S\)-module and the \(R\)-module. In other words, there exists a natural correspondence between the elements of the \(S\)-module and the elements of the \(R\)-module.

In this context, it means that we need to show that there exists an isomorphism between the \(S\)-module and a sub-module of \(R\), where the isomorphism preserves the module structure. This means that the operations of addition and scalar multiplication in the \(S\)-module are preserved when we consider it as an \(R\)-module.

One way to show this is to define a map from the elements of the \(S\)-module to the elements of the \(R\)-module in such a way that it preserves the module structure. This map should be bijective and the inverse map should also preserve the module structure. This will establish the isomorphism between the two structures.

In summary, to show that every \(S\)-module can be considered as an \(R\)-module in a natural way, we need to find an isomorphism that preserves the module structure between the two structures. This will allow us to view the elements of the \(S\)-module as elements of the \(R\)-module and vice versa, in a way that is intuitive and consistent with the operations defined in both structures.