Definition 10.9.1. Let R be a ring, S a subset of R. We say S is a multiplicative subset of R if 1\in S and S is closed under multiplication, i.e., s, s' \in S \Rightarrow ss' \in S.
10.9 Localization
Given a ring A and a multiplicative subset S, we define a relation on A \times S as follows:
(x, s) \sim (y, t) \Leftrightarrow \exists u \in S \text{ such that } (xt-ys)u = 0
It is easily checked that this is an equivalence relation. Let x/s (or \frac{x}{s}) be the equivalence class of (x, s) and S^{-1}A be the set of all equivalence classes. Define addition and multiplication in S^{-1}A as follows:
x/s + y/t = (xt + ys)/st, \quad x/s \cdot y/t = xy/st
One can check that S^{-1}A becomes a ring under these operations.
Definition 10.9.2. This ring is called the localization of A with respect to S.
We have a natural ring map from A to its localization S^{-1}A,
A \longrightarrow S^{-1}A, \quad x \longmapsto x/1
which is sometimes called the localization map. In general the localization map is not injective, unless S contains no zerodivisors. For, if x/1 = 0, then there is a u\in S such that xu = 0 in A and hence x = 0 since there are no zerodivisors in S. The localization of a ring has the following universal property.
Proposition 10.9.3. Let f : A \to B be a ring map that sends every element in S to a unit of B. Then there is a unique homomorphism g : S^{-1}A \to B such that the following diagram commutes.
\xymatrix{ A \ar[rr]^{f} \ar[dr] & & B \\ & S^{-1}A \ar[ur]_ g }
Proof. Existence. We define a map g as follows. For x/s\in S^{-1}A, let g(x/s) = f(x)f(s)^{-1}\in B. It is easily checked from the definition that this is a well-defined ring map. And it is also clear that this makes the diagram commutative.
Uniqueness. We now show that if g' : S^{-1}A \to B satisfies g'(x/1) = f(x), then g = g'. Hence f(s) = g'(s/1) for s \in S by the commutativity of the diagram. But then g'(1/s)f(s) = 1 in B, which implies that g'(1/s) = f(s)^{-1} and hence g'(x/s) = g'(x/1)g'(1/s) = f(x)f(s)^{-1} = g(x/s). \square
Lemma 10.9.4. The localization S^{-1}A is the zero ring if and only if 0\in S.
Proof. If 0\in S, any pair (a, s)\sim (0, 1) by definition. If 0\not\in S, then clearly 1/1 \neq 0/1 in S^{-1}A. \square
Lemma 10.9.5. Let R be a ring. Let S \subset R be a multiplicative subset. The category of S^{-1}R-modules is equivalent to the category of R-modules N with the property that every s \in S acts as an automorphism on N.
Proof. The functor which defines the equivalence associates to an S^{-1}R-module M the same module but now viewed as an R-module via the localization map R \to S^{-1}R. Conversely, if N is an R-module, such that every s \in S acts via an automorphism s_ N, then we can think of N as an S^{-1}R-module by letting x/s act via x_ N \circ s_ N^{-1}. We omit the verification that these two functors are quasi-inverse to each other. \square
The notion of localization of a ring can be generalized to the localization of a module. Let A be a ring, S a multiplicative subset of A and M an A-module. We define a relation on M \times S as follows
(m, s) \sim (n, t) \Leftrightarrow \exists u\in S \text{ such that } (mt-ns)u = 0
This is clearly an equivalence relation. Denote by m/s (or \frac{m}{s}) be the equivalence class of (m, s) and S^{-1}M be the set of all equivalence classes. Define the addition and scalar multiplication as follows
m/s + n/t = (mt + ns)/st,\quad m/s\cdot n/t = mn/st
It is clear that this makes S^{-1}M an S^{-1}A-module.
Definition 10.9.6. The S^{-1}A-module S^{-1}M is called the localization of M at S.
Note that there is an A-module map M \to S^{-1}M, m \mapsto m/1 which is sometimes called the localization map. It satisfies the following universal property.
Lemma 10.9.7. Let R be a ring. Let S \subset R a multiplicative subset. Let M, N be R-modules. Assume all the elements of S act as automorphisms on N. Then the canonical map
\mathop{\mathrm{Hom}}\nolimits _ R(S^{-1}M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(M, N)
induced by the localization map, is an isomorphism.
Proof. It is clear that the map is well-defined and R-linear. Injectivity: Let \alpha \in \mathop{\mathrm{Hom}}\nolimits _ R(S^{-1}M, N) and take an arbitrary element m/s \in S^{-1}M. Then, since s \cdot \alpha (m/s) = \alpha (m/1), we have \alpha (m/s) =s^{-1}(\alpha (m/1)), so \alpha is completely determined by what it does on the image of M in S^{-1}M. Surjectivity: Let \beta : M \rightarrow N be a given R-linear map. We need to show that it can be "extended" to S^{-1}M. Define a map of sets
M \times S \rightarrow N,\quad (m,s) \mapsto s^{-1}\beta (m)
Clearly, this map respects the equivalence relation from above, so it descends to a well-defined map \alpha : S^{-1}M \rightarrow N. It remains to show that this map is R-linear, so take r, r' \in R as well as s, s' \in S and m, m' \in M. Then
\begin{align*} \alpha (r \cdot m/s + r' \cdot m' /s') & = \alpha ((r \cdot s' \cdot m + r' \cdot s \cdot m') /(ss')) \\ & = (ss')^{-1}\beta (r \cdot s' \cdot m + r' \cdot s \cdot m') \\ & = (ss')^{-1} (r \cdot s' \beta (m) + r' \cdot s \beta (m')) \\ & = r \alpha (m/s) + r' \alpha (m' /s') \end{align*}
and we win. \square
Example 10.9.8. Let A be a ring and let M be an A-module. Here are some important examples of localizations.
Given \mathfrak p a prime ideal of A consider S = A\setminus \mathfrak p. It is immediately checked that S is a multiplicative set. In this case we denote A_\mathfrak p and M_\mathfrak p the localization of A and M with respect to S respectively. These are called the localization of A, resp. M at \mathfrak p.
Let f\in A. Consider S = \{ 1, f, f^2, \ldots \} . This is clearly a multiplicative subset of A. In this case we denote A_ f (resp. M_ f) the localization S^{-1}A (resp. S^{-1}M). This is called the localization of A, resp. M with respect to f. Note that A_ f = 0 if and only if f is nilpotent in A.
Let S = \{ f \in A \mid f \text{ is not a zerodivisor in }A\} . This is a multiplicative subset of A. In this case the ring Q(A) = S^{-1}A is called either the total quotient ring, or the total ring of fractions of A.
If A is a domain, then the total quotient ring Q(A) is the field of fractions of A. Please see Fields, Example 9.3.4.
Lemma 10.9.9. Let R be a ring. Let S \subset R be a multiplicative subset. Let M be an R-module. Then
S^{-1}M = \mathop{\mathrm{colim}}\nolimits _{f \in S} M_ f
where the preorder on S is given by f \geq f' \Leftrightarrow f = f'f'' for some f'' \in R in which case the map M_{f'} \to M_ f is given by m/(f')^ e \mapsto m(f'')^ e/f^ e.
Proof. Omitted. Hint: Use the universal property of Lemma 10.9.7. \square
In the following paragraph, let A denote a ring, and M, N denote modules over A.
If S and S' are multiplicative sets of A, then it is clear that
SS' = \{ ss' : s\in S, \ s'\in S'\}
is also a multiplicative set of A. Then the following holds.
Proposition 10.9.10. Let \overline{S} be the image of S in S'^{-1}A, then (SS')^{-1}A is isomorphic to \overline{S}^{-1}(S'^{-1}A).
Proof. The map sending x\in A to x/1\in (SS')^{-1}A induces a map sending x/s\in S'^{-1}A to x/s \in (SS')^{-1}A, by universal property. The image of the elements in \overline{S} are invertible in (SS')^{-1}A. By the universal property we get a map f : \overline{S}^{-1}(S'^{-1}A) \to (SS')^{-1}A which maps (x/s')/(s/1) to x/ss'.
On the other hand, the map from A to \overline{S}^{-1}(S'^{-1}A) sending x\in A to (x/1)/(1/1) also induces a map g : (SS')^{-1}A \to \overline{S}^{-1}(S'^{-1}A) which sends x/ss' to (x/s')/(s/1), by the universal property again. It is immediately checked that f and g are inverse to each other, hence they are both isomorphisms. \square
For the module M we have
Proposition 10.9.11. View S'^{-1}M as an A-module, then S^{-1}(S'^{-1}M) is isomorphic to (SS')^{-1}M.
Proof. Note that given a A-module M, we have not proved any universal property for S^{-1}M. Hence we cannot reason as in the preceding proof; we have to construct the isomorphism explicitly.
We define the maps as follows
\begin{align*} & f : S^{-1}(S'^{-1}M) \longrightarrow (SS')^{-1}M, \quad \frac{x/s'}{s}\mapsto x/ss'\\ & g : (SS')^{-1}M \longrightarrow S^{-1}(S'^{-1}M), \quad x/t\mapsto \frac{x/s'}{s}\ \text{for some }s\in S, s'\in S', \text{ and } t = ss' \end{align*}
We have to check that these homomorphisms are well-defined, that is, independent the choice of the fraction. This is easily checked and it is also straightforward to show that they are inverse to each other. \square
If u : M \to N is an A homomorphism, then the localization indeed induces a well-defined S^{-1}A homomorphism S^{-1}u : S^{-1}M \to S^{-1}N which sends x/s to u(x)/s. It is immediately checked that this construction is functorial, so that S^{-1} is actually a functor from the category of A-modules to the category of S^{-1}A-modules. Moreover this functor is exact, as we show in the following proposition.
Proposition 10.9.12.slogan Let L\xrightarrow {u} M\xrightarrow {v} N be an exact sequence of R-modules. Then S^{-1}L \to S^{-1}M \to S^{-1}N is also exact.
Proof. First it is clear that S^{-1}L \to S^{-1}M \to S^{-1}N is a complex since localization is a functor. Next suppose that x/s maps to zero in S^{-1}N for some x/s \in S^{-1}M. Then by definition there is a t\in S such that v(xt) = v(x)t = 0 in M, which means xt \in \mathop{\mathrm{Ker}}(v). By the exactness of L \to M \to N we have xt = u(y) for some y in L. Then x/s is the image of y/st. This proves the exactness. \square
Lemma 10.9.13. Localization respects quotients, i.e. if N is a submodule of M, then S^{-1}(M/N)\simeq (S^{-1}M)/(S^{-1}N).
Proof. From the exact sequence
0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0
we have
0 \longrightarrow S^{-1}N \longrightarrow S^{-1}M \longrightarrow S^{-1}(M/N) \longrightarrow 0
The corollary then follows. \square
If, in the preceding Corollary, we take N = I and M = A for an ideal I of A, we see that S^{-1}A/S^{-1}I \simeq S^{-1}(A/I) as A-modules. The next proposition shows that they are isomorphic as rings.
Proposition 10.9.14. Let I be an ideal of A, S a multiplicative set of A. Then S^{-1}I is an ideal of S^{-1}A and \overline{S}^{-1}(A/I) is isomorphic to S^{-1}A/S^{-1}I, where \overline{S} is the image of S in A/I.
Proof. The fact that S^{-1}I is an ideal is clear since I itself is an ideal. Define
f : S^{-1}A\longrightarrow \overline{S}^{-1}(A/I), \quad x/s\mapsto \overline{x}/\overline{s}
where \overline{x} and \overline{s} are the images of x and s in A/I. We shall keep similar notations in this proof. This map is well-defined by the universal property of S^{-1}A, and S^{-1}I is contained in the kernel of it, therefore it induces a map
\overline{f} : S^{-1}A/S^{-1}I \longrightarrow \overline{S}^{-1}(A/I), \quad \overline{x/s}\mapsto \overline{x}/\overline{s}
On the other hand, the map A \to S^{-1}A/S^{-1}I sending x to \overline{x/1} induces a map A/I \to S^{-1}A/S^{-1}I sending \overline{x} to \overline{x/1}. The image of \overline{S} is invertible in S^{-1}A/S^{-1}I, thus induces a map
g : \overline{S}^{-1}(A/I) \longrightarrow S^{-1}A/S^{-1}I, \quad \frac{\overline{x}}{\overline{s}}\mapsto \overline{x/s}
by the universal property. It is then clear that \overline{f} and g are inverse to each other, hence are both isomorphisms. \square
We now consider how submodules behave in localization.
Lemma 10.9.15. Any submodule N' of S^{-1}M is of the form S^{-1}N for some N\subset M. Indeed one can take N to be the inverse image of N' in M.
Proof. Let N be the inverse image of N' in M. Then one can see that S^{-1}N\supset N'. To show they are equal, take x/s in S^{-1}N, where s\in S and x\in N. This yields that x/1\in N'. Since N' is an S^{-1}R-submodule we have x/s = x/1\cdot 1/s\in N'. This finishes the proof. \square
Taking M = A and N = I an ideal of A, we have the following corollary, which can be viewed as a converse of the first part of Proposition 10.9.14.
Lemma 10.9.16.slogan Each ideal I' of S^{-1}A takes the form S^{-1}I, where one can take I to be the inverse image of I' in A.
Proof. Immediate from Lemma 10.9.15. \square
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