Exactness of localization

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.