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[SOLVED] R-modules and Homomorphism of Rings

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 5, 2012
1,621
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
Feb 15, 2012
1,967
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
Feb 5, 2012
1,621
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. :)