Localization

R. Virk

Department of Mathematics
University of Wisconsin
Madison, WI 53706

virk@math.wisc.edu

Let $A$ be a commutative ring (with $1$). Let $S\subseteq A$ be a multiplicative set, i.e., $1\in S$ and if $x,y\in S$, then $xy\in S$. Suppose $f:A\to B$ is a ring homomorphism satisfying

1. $f(x)$ is a unit of $B$ for all $x\in S$;
2. if $g: A\to C$ is a ring homomorphism taking every element of $S$ to a unit of $C$, then there exist a unique homomorphism $h: B\to C$ such that $g=h\circ f$;

then $B$ is uniquely determined up to isomorphism, and is called the localization of $A$ with respect to $S$. We write $B=S^{-1}A$ and call $f:A\to S^{-1}A$ the canonical map. We prove the existence of $S^{-1}A$ as follows: define a relation $\sim$ on the set $A\times S$ by $(a,s)\sim (b,s')$ if and only if there exists $t\in S$ such that $t(s'a-sb)=0$; it is easy to check that this is an equivalence relation (if we just have $s'a-sb=0$ in the definition, the transitive law fails when $S$ has zero divisors). We write $\frac{a}{s}$ for the class of $(a,s)$ and define sums and products by the usual rules for calculating with fractions, i.e. $\frac{a}{s}+\frac{b}{s'}=\frac{as'+bs}{ss'}$ and $\frac{a}{s}\frac{b}{s'}=\frac{ab}{ss'}$. This makes $B$ a ring and defining $f:A\to B$ by $a\mapsto \frac{a}{1}$ we see that $f$ is a ring homomorphism satisfying the required properties. From this construction we also see that kernel of the canonical map $f:A\to S^{-1}A$ is given by $\ker(f)=\{a\in A\,\vert\, sa=0\,$   for some $s\in S\}$.

Example 0.1   Let $\mathfrak{p}$ be a prime ideal of $A$. Then $S=A-\mathfrak{p}$ is multiplicatively closed. We write $A_{\mathfrak{p}}$ for $S^{-1}A$ in this case. The elements $\frac{a}{s}$, $a\in\mathfrak{p}$ form an ideal $\mathfrak{m}$ in $A_{\mathfrak{p}}$. If $\frac{b}{t} \not\in \mathfrak{m}$, then $b\not\in \mathfrak{p}$, hence $b\in S$ and therefore $\frac{b}{t}$ is a unit in $A_{\mathfrak{p}}$. Consequently, $\mathfrak{m}$ is the unique maximal ideal in $A_{\mathfrak{p}}$.

Remark 0.2   A ring with a unique maximal ideal is called a local ring.

Example 0.3   Let $f\in A$, then $S=\{f^n\}_{n\geq 0}$ is multiplicatively closed. We write $A_f$ for $S^{-1}A$ in this case.

The construction of $S^{-1}A$ can be carried through for an $A$-module $M$ in place of the ring $A$. Define an equivalence relation $\sim$ on $M\times S$ by:

$$(m,s) \sim (m',s') \Leftrightarrow$$   $$\mbox{ there exists $t\in S$\ such that $t(s'm-sm')=0$.}$$$$$$

Let $\frac{m}{s}$ denote the class of the pair $(m,s)$, let $S^{-1}M$ denote the set of such fractions made into a $S^{-1}A$-modile with the obvious definitions of addition and scalar multiplication. (Note that $S^{-1}M$ is also an $A$-module via the canonical map $A\to S^{-1}A$.) As in the examples before, write $M_{\mathfrak{p}}$ when $S=A-\mathfrak{p}$ and $M_f$ when $S=\{f^n\}_{n\geq 0}$.

The module $S^{-1}M$ satisfies the following universal property: suppose we are given a map $\varphi$ from $M$ to an $A$-module $N$ on which the elements of $S$ act by automorphisms. Then there is a unique map $\varphi': S^{-1}M\to N$ such that $\varphi=\varphi'\circ \iota$.

Let $\varphi: M \to N$ be an $A$-module homomorphism. This gives rise to an $S^{-1}A$ module homomorphism $S^{-1}\varphi:S^{-1}M\to S^{-1}N$, namely $S^{-1}\varphi$ maps $\frac{m}{s}$ to $\frac{\varphi(m)}{s}$. We have that $S^{-1}(\varphi\circ\psi)=S^{-1}(\varphi)\circ S^{-1}(\psi)$ and that $S^{-1}\mathrm{id}_M = \mathrm{id}_{S^{-1}M}$. Thus, localization give a functor from the category of $A$-modules to the category of $S^{-1}A$-modules.

Proposition 0.4   Localization is an exact functor. That is, if $\xymatrix{M'\ar[r]^f &M \ar[r]^g& M''}$ is exact at $M$, then

$$\xymatrix{ S^{-1}M' \ar[r]^{S^{-1}f} & S^{-1}M \ar[r]^{S^{-1}g}& M''}$$

is exact at $S^{-1}M$.

Proof. We have that $S^{-1}g \circ S^{-1}f = S^{-1}(f\circ g) = 0$. Hence, $\mathrm{im\,}(S^{-1}f) \subseteq \ker (S^{-1}g)$. Conversely, suppose $\frac{m}{s} \in \ker (S^{-1}g)$. Then $tg(m)=g(tm)=0$ for some $t\in S$. So $tm \in \ker g$, hence $tm=f(m')$ for some $m'\in M'$. Thus, in $S^{-1}M$ we have $\frac{m}{s}=\frac{f(m')}{st}=(S^{-1}f)(\frac{m'}{st})\in \mathrm{im\,}(S^{-1}f)$. Hence $\ker (S^{-1}g) \subseteq \mathrm{im\,}(S^{-1}f)$. $\qedsymbol$

Proposition 0.5   Let $M$ be an $A$-module. Then the $S^{-1}A$ modules $S^{-1}M$ and $S^{-1}A\otimes_A M$ are isomorphic; more precisely this isomorphism is given by the map

$$f: S^{-1}A\otimes_A M \to S^{-1}M,$$

$$\frac{a}{s}\otimes m \mapsto \frac{am}{s}.$$

Proof. The map $S^{-1}A\times M \to S^{-1}M$ is $A$-bilinear induces the map $f$ and is clearly surjective.

Let $\sum_i \frac{a_i}{s_i}\otimes m_i$ be any element of $S^{-1}A\otimes_A M$. Then setting $s=\prod_i s_i$ and $t_i=\prod_{j\neq i}s_i$ we have that

$$\sum_i \frac{a_i}{s_i}\otimes m_i = \sum_i \frac{a_it_i}{s} \otimes m_i = \sum_i \frac{1}{s} \otimes a_it_im_i.$$

Thus, every element of $S^{-1}A\otimes_A M$ can be written in the form $\frac{1}{s}\otimes m$. Suppose $f(\frac{1}{s}\otimes m)=0$, then $\frac{m}{s}=0$, i.e. $tm=0$ for some $t\in S$. Thus,

$$\frac{1}{s}\otimes m = \frac{t}{st}\otimes m = \frac{1}{st} \otimes tm = 0.$$

Hence, $f$ is injective and an isomorphism. $\qedsymbol$

Proposition 0.6   Let $M$ be an $A$-module, and let $m\in M$. Then the following are equivalent:

1. $m=0$;
2. $\frac{m}{1}=0$ in $M_{\mathfrak{p}}$ for each prime ideal $\mathfrak{p}$ of $A$;
3. $\frac{m}{1}=0$ in $M_{\mathfrak{m}}$ for each maximal ideal $\mathfrak{m}$ of $A$;

Proof. It is clear that (i) $\Rightarrow$(ii) $\Rightarrow$(iii). To see that (iii) $\Rightarrow$(i), let $m\in M$ be such that $\frac{m}{1}=0$ in $M_{\mathfrak{m}}$ for each maximal ideal $\mathfrak{m}$ of $A$. Suppose $\mathrm{Ann\,}(m)\neq A$, let $\mathfrak{m}'$ be a maximal ideal containing $\mathrm{Ann\,}(m)$. Then we have that $sm=0$ for some $s \in A - \mathrm{Ann\,}(m)$, which is a contradiction. Thus, $\mathrm{Ann\,}(m)=A$ and $m=0$. $\qedsymbol$

Corollary 0.7   Let $M$ be an $A$-module. Then the following are equivalent:

1. $M=0$;
2. $M_{\mathfrak{p}}=0$ for each prime ideal $\mathfrak{p}$ of $A$;
3. $M_{\mathfrak{m}}=0$ for each maximal ideal $\mathfrak{m}$ of $A$.

Proposition 0.8   Let $\varphi: M \to N$ be an $A$-module homomorphism. Then the following are equivalent:

1. $\varphi$ is injective;
2. $\varphi_{\mathfrak{p}}:M_{\mathfrak{p}}\to N_{\mathfrak{p}}$ is injective for each prime ideal $\mathfrak{p}$ of $A$;
3. $\varphi_{\mathfrak{m}}: M_{\mathfrak{m}} \to N_{\mathfrak{m}}$ is injective for each maximal ideal $\mathfrak{m}$ of $A$.

Similarly with `injective' replaced by `surjective' throughout.

Proof. From 0.4 we have that (i) $\Rightarrow$(ii), (ii) $\Rightarrow$(iii) is clear. To see (iii) $\Rightarrow$(i), let $M'=\ker \varphi$. Then $0 \rightarrow M' \rightarrow M \rightarrow N$ is exact, hence $0 \rightarrow M'_{\mathfrak{m}} \rightarrow M_{\mathfrak{m}} \rightarrow N_{\mathfrak{m}}$ is exact by 0.4. Thus, $M'_{\mathfrak{m}} \simeq \ker \varphi_{\mathfrak{m}} =0$ since $\varphi_{\mathfrak{m}}$ is injective. Hence, $M'=0$ by the previous corollary and $\varphi$ is injective.

The proof with `injective' replaced by `surjective' is similar. $\qedsymbol$