Localization of a module ($S^{-1}M$)

For a multiplicative subset $S\subseteq R$ and an $R$-module $M$, the localization $S^{-1}M$ consists of fractions $m/s$, with $m/s=m^{\prime }/s^{\prime }$ when some $u\in S$ satisfies $u(s^{\prime }m-s{m}^{\prime })=0$. It is an $S^{-1}R$-module, and the canonical map is $m\mapsto m/1$. If $M$ is finitely generated over a Noetherian ring, then $S^{-1}M$ is a Noetherian module over $S^{-1}R$.

The fraction relation defining localization of a module gives this criterion. For localization at powers of a single element $f$, a fraction vanishes exactly when some power of $f$ annihilates its numerator. Finitely many such vanishing relations can be cleared with one common exponent. This is distinct from detecting global zero by localizing at every prime ideal.

For a multiplicative subset $S$ of a commutative ring $A$, localization preserves short exact sequences of $A$-modules. Surjectivity follows by lifting a numerator. If a fraction maps to zero, some element of $S$ annihilates its image, so multiplying its numerator by that element puts it into the original kernel. For an injective original map, the same criterion shows its localized kernel is zero. Thus $S^{-1}A$ is a flat module over $A$.

For a module homomorphism $f:M\to N$ over a commutative ring $A$, $f$ is injective if and only if every $f_{\mathfrak{m}}$ is injective, with $\mathfrak{m}$ ranging over maximal ideals. In a localization of a module, $x/s$ maps to zero exactly when some denominator $t$ satisfies $tf(x)=0$. If $f$ is injective this gives $tx=0$, so the original fraction is zero. In general it identifies $\ker (f_{\mathfrak{m}})$ with $(\ker f{)}_{\mathfrak{m}}$. A nonzero element of $\ker f$ has a proper annihilator, contained in some maximal ideal; it remains nonzero in that localization. Thus vanishing of all localized kernels forces $\ker f=0$. No finiteness or Noetherian assumption is required.

Let $R$ be Noetherian, $I\subseteq R$ an ideal, $S=1+I$, and $M$ a finitely generated $R$-module. ThenIf $(1+a)m=0$ with $a\in I$, then $m=(-a{)}^{j}m\in I^{j}M$ for every $j$. Conversely, apply the Artin-Rees lemma to $Rm\subseteq M$. For sufficiently large $c$ it givesso $m=am$ for some $a\in I$ and $(1-a)m=0$.

Letbe the semigroup algebra inside $k[t^{\mathbb{Q}}]$, and let $I$ be generated by the $t^q$ with $q>0$. This integral domain is not Noetherian, sinceMoreover $I^2=I$, because $t^q=(t^{q/2}{)}^2$. Thus ${\bigcap }_{j\ge 1}{I}^j=I\ne 0$, whereas the map $R\to (1+I{)}^{-1}R$ is injective because $R$ is a domain.