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Localization of a module

For a multiplicative subset S of a commutative ring R, the localization S^{-1}M is the module obtained by making every element of S act invertibly.

Let RR be a , let S⊆RS\subseteq R be a , and let MM be an RR-module. The localization of MM at SS is the S−1RS^{-1}R-module S−1MS^{-1}M of equivalence classes of pairs m,s∈M×Sm,s\in M\times S, with

(m,s)∼(m′,s′)⟺∃t∈S such that t(s′m−sm′)=0 in M.(m,s)\sim(m',s')\Longleftrightarrow\exists t\in S\text{ such that }t(s'm-sm')=0\text{ in }M.

Write the class as m/sm/s. Addition and scalar action are

ms+m′s′=s′m+sm′ss′,(rs)(mt)=rmst.\frac m s+\frac {m'}{s'}=\frac {s'm+sm'}{ss'},\qquad \left(\frac r s\right)\left(\frac m t\right)=\frac {rm}{st}.

Every s∈Ss\in S acts invertibly, and m↦m/1m\mapsto m/1 is RR-linear.

Here S−1RS^{-1}R is the .

Universal property

Let NN be an S−1RS^{-1}R-module, regarded as an RR-module through R→S−1RR\to S^{-1}R. For every RR-linear map f:M→Nf:M\to N, there is a unique S−1RS^{-1}R-linear map f~:S−1M→N\widetilde f:S^{-1}M\to N such that f~∘ιM=f\widetilde f\circ\iota_M=f.

Properties

Localizing at a prime p\mathfrak p means taking S=R∖pS=R\setminus\mathfrak p and writing

Mp:=(R∖p)−1M,M_{\mathfrak p}:=(R\setminus\mathfrak p)^{-1}M,

in parallel with .

Localization interacts well with exact sequences: it is an exact functor on modules (see and compare with the general notion of an ).

Finally, localization can be expressed as a base change: via there is a natural isomorphism

S−1M≅(S−1R)⊗RM.S^{-1}M \cong (S^{-1}R)\otimes_R M.
Examples
  1. Localizing a quotient. If I⊆RI\subseteq R is an ideal, then
    S−1(R/I) ≅ (S−1R)/(S−1I),S^{-1}(R/I)\ \cong\ (S^{-1}R)/(S^{-1}I),
    where S−1IS^{-1}I denotes the image of II in S−1RS^{-1}R.
  1. Torsion killed by localization. Take R=ZR=\mathbb Z, M=Z/nZM=\mathbb Z/n\mathbb Z, and localize at S=Z∖(p)S=\mathbb Z\setminus (p), so S−1Z=Z(p)S^{-1}\mathbb Z=\mathbb Z_{(p)}.
  • If p∤np\nmid n, then n∈Sn\in S becomes a unit, so S−1M=0S^{-1}M=0.
  • If p∣np\mid n, then S−1M≅Z(p)/nZ(p)S^{-1}M\cong \mathbb Z_{(p)}/n\mathbb Z_{(p)}, which is generally nonzero.
  1. An annihilator made invertible. Let R=k[x]R=k[x], M=R/(x)M=R/(x), and S={1,x,x2,… }S=\{1,x,x^2,\dots\}. In S−1RS^{-1}R, the element xx is a unit, but xx annihilates MM, so S−1M=0S^{-1}M=0.