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?

  • $\begingroup$ 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 ? $\endgroup$ – Maxime Ramzi Jul 2 '19 at 10:24
  • $\begingroup$ @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? $\endgroup$ – Ariel Jul 2 '19 at 10:51
  • $\begingroup$ 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 $N$ ! $\endgroup$ – Maxime Ramzi Jul 2 '19 at 10:58
  • $\begingroup$ @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 $M$. $\endgroup$ – Ariel Jul 2 '19 at 11:48
  • $\begingroup$ Maybe you know that but unluckily for you it's not true $\endgroup$ – 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.

Therefore it is an isomorphism.

  • $\begingroup$ Why does the map $S^{-1}N\rightarrow M$ factors through $S^{-1}N\to S^{-1}M\to M$? $\endgroup$ – Babai Sep 14 '20 at 19:18
  • $\begingroup$ @Babai : Apply $S^{-1}$ to the nap $S^{-1}N \to M$ $\endgroup$ – Maxime Ramzi Sep 14 '20 at 19:52

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