# localization of modules

• Oct 17th 2011, 03:58 PM
Joolz
localization of modules
Hi there,

I would like some help with the following:

Let f be a homomorphism between two commutative rings A and B. S is a multiplicatively closed subset of A. Then f(S) is a multiplicatively closed subset of B. Now the book says that S(^-1)B and (f(S)^-1)B are isomorphic as S(^-1)A-modules.

I get that we can see B as an A-module and from there that S(^-1)B is an S(^-1)A-module. But why is (f(S)^-1)B an S(^-1)A-module?

J.
• Oct 19th 2011, 11:45 PM
NonCommAlg
Re: localization of modules
Quote:

Originally Posted by Joolz
Hi there,

I would like some help with the following:

Let f be a homomorphism between two commutative rings A and B. S is a multiplicatively closed subset of A. Then f(S) is a multiplicatively closed subset of B. Now the book says that S(^-1)B and (f(S)^-1)B are isomorphic as S(^-1)A-modules.

I get that we can see B as an A-module and from there that S(^-1)B is an S(^-1)A-module. But why is (f(S)^-1)B an S(^-1)A-module?

J.

why can we see B as an A-module? because we can define $\displaystyle a \cdot b = f(a)b$, for all $\displaystyle a \in A$ and $\displaystyle b \in B$. so if $\displaystyle s,t \in S$ and $\displaystyle a \in A, b \in B$, it is natural to define $\displaystyle (s^{-1}a) \cdot (f(t))^{-1}b =(f(st))^{-1}f(a)b,$ which makes $\displaystyle (f(S))^{-1}B$ an $\displaystyle S^{-1}A$-module.