# Localization of a finitely generated module.

Let $M$ be a finitely generated $R$-module, then for some multiplicative subset $D \subseteq R$ is $D^{-1}M$ finitely generated as a $D^{-1}R$-module?

Let $m_1, \dots, m_n$ be a set of generators of $M$, with some possible relations. Then every element $m \in M$ can be written as

$$m=a_1m_1 + \dots + a_n m_n$$

Passing to the localization, each element $\frac{m}{d} \in D^{-1}M$ can then be written as $$\frac{m}{d}= \frac{a_1m_1 + \dots + a_n m_n}{d} = \frac{a_1}{d}\frac{m_1}{1} + \dots + \frac{a_2}{d} \frac{m_n}{1}$$

This make me believe that $\pi: M \to D^{-1}M$ which maps $m \to \frac{m}{1}$ would map generators to generators. This would give that $M$ as an $R$-module is isomorphic to $D^{-1}M$ as a $D^{-1}R$-module.

What is wrong with my thinking?

• Someone should tag this question with #algebraic-geometry because of its relation to finite morphisms of (affine) schemes. – Tanner Strunk Mar 30 '17 at 16:22

$1)$ What is an isomorphism between modules over different rings?
$2)$ Anyway you did not prove the map $M\to D^{-1}M$ is surjective (in general it is neither surjective not injective).
Example: suppose $R$ is an integral domain, $D=R-\{0 \}$, and $D^{-1}R=K$ its quotient field. If $M$ is a non-zero torsion module, then $D^{-1}M=0$, so $M$ and $D^{-1}M$ are certainly not isomorphic in whatever sense.