The following question was asked in my assignment of Introduction to commutative algebra and I was not able to solve this particular problem.
Actually, I am not very good in localization of Modules. So, I am asking for help here.
Question: Let A be a commutative ring and $S \subset A$ is a multiplicative set.
(a) Let S has no zero divisors. Show that if $S^{-1}A$ is an A-module of finite type, then $S^{-1} A=A$ ie S contains invertible elements.
(b) Let $A_1= A/I$ and $S_1$ be the image of S in $A_1$ . Show that there exists an isomorphism $S^{-1} A \to {S_1}^{-1} A_1$.
(c)$S^{-1} A$ is an A-module of finite type iff $S_1\subset A^{\times}_1$ and in this case $S^{-1} A $ is isomorphic to A/I.
Attempt: (a) Any element of $S^{-1} A$ is of the form a/s and as $S^{-1} A$ is an A -module of finite type, so, $x\in S^{-1} A$ implies $x= a/s = {a_1}' \frac{a_1} {s_1} +... + \frac{a_n} {s_n} {a_n}' $ ${a_i}' \in A$ and $\frac{a_i} {s_n} \in A \S$( Definition of finite type).
Now, I have to show that for any $s \in S$ , If inverse of s exists ( call it s') then $s'\in S$.
But I am not able to make any progress.
(b) So,$S_1= S/I$ . Now an element of $S^{-1} A$ is given by a/s , where $a\in A$ and $s\in S$ and an element of ${S_1 }= s+I $ and an element of $A_1$ is given by $a+I$. $ f : \frac{a} {s} \to \frac{a'+ I} {s' +I}$. f is onto. I am not able to prove f to be 1-1. Can you please help me with this?
(c) I don't have much to show as attempt in this as well.
Please help me.
Let me do the first one for you.
So, assume that $S^{-1}A$ is a finitely generated module over $A$. So pick a set of generators $y_i=a_i/s$ for some $a_i\in A, s\in S$. (Do you see why we can choose a common denominator $s$?) Let $t\in S$. Then $1/ts\in S^{-1} A$, so we can write it as $$1/ts= \sum b_iy_i$$ for some $b_i\in A$. Multiply by $ts$ to get $$1= t(\sum a_ib_i)$$ showing $t$ is a unit.