# A finitely generated $S^{-1}R$-module isomorphic to the localization of a finitely generated $R$-module

I am trying to prove the following statement:

Every finitely generated $$S^{-1}R$$-module, say $$M$$, is isomorphic to $$S^{-1}N$$, where $$N$$ is a finitely generated $$R$$-module ($$S$$ is an arbitrary multiplicative set).

My thoughts: the localization $$S^{-1}N$$ is a finitely generated $$S^{-1}R$$ module since $$N$$ is finitely generated.$$\,$$ Now I am looking for an isomorphism, say $$\alpha$$, between $$S^{-1}N$$ and $$M$$ such that for each generator of $$S^{-1}N$$,$$\,\,$$say $$n_i$$,$$\,$$ $$\alpha(n_i)$$ is a generator of $$M$$. But how can we tell that the cardinality of the generators set of $$S^{-1}N$$ is equal to the cardinality of the generators set of $$M$$. Any ideas?

• But you haven't even found an $N$ ! How about taking a generating set $x_1,...,x_n$ of $M$ and taking $N$ to be the sub-$R$-module of $M$ generated by these ? – Maxime Ramzi Jul 2 '19 at 10:24
• @Max Let $x_1,...,x_n$ be the generator set of $M$ and define $N$ as a sub-$R$-module of $M$ generated by these. So, the cardinality of the generator set of $S^{-1}N$ is also n. Hence we can construct an isomorphism between $S^{-1}N$ and M, say $\alpha$, such that if $n_i$ is a generator of $S^{-1}N$ then $\alpha(n_i)$ is a generator of $M$. Am I correct? by the way, is it just a speical case of my question? – Ariel Jul 2 '19 at 10:51
• No that's not correct in general. $\mathbb Z$ and $\mathbb Z/2 \mathbb Z$ have the same number of generators over $\mathbb Z$, they're not isomorphic. You have to use that $N$ is not any module, it is related to $M$. That's why what you explained in your question made no sense : you had not picked any specific $N$ ! – Maxime Ramzi Jul 2 '19 at 10:58
• @Max we know that there is an isomorphism from $S^{-1}N$ to $N$. If you will take $N$ to be a sub module of $M$ you will get an injective homomorphism to $M$. – Ariel Jul 2 '19 at 11:48
• Maybe you know that but unluckily for you it's not true – Maxime Ramzi Jul 2 '19 at 11:57

Let $$M$$ be an $$S^{-1}R$$-module, generated by, say, $$x_1,...,x_n$$. Let $$N$$ be the sub-$$R$$-module of $$M$$ generated by $$x_1,...,x_n$$.
Clearly $$N$$ is finitely generated as an $$R$$-module, and there is a natural map of $$S^{-1}R$$-modules $$S^{-1}N\to M$$.
It will be surjective because its image contains $$x_1,...,x_n$$.
It will be injective because it factors as $$S^{-1}N\to S^{-1}M\to M$$, the first one is injective as $$S^{-1}$$ preserves injections, the second one is an isomorphism.
• Why does the map $S^{-1}N\rightarrow M$ factors through $S^{-1}N\to S^{-1}M\to M$? – Babai Sep 14 '20 at 19:18
• @Babai : Apply $S^{-1}$ to the nap $S^{-1}N \to M$ – Maxime Ramzi Sep 14 '20 at 19:52