We next discuss an important construction that is the algebraic analogue of the geometric notion of concentrating attention near a point.
Let \(A\) be an integral domain and \(S \subset A\) a subset not containing \(0\). We can then consider the subring \[ S^{-1}A = A[ \{1/s : s \in S\} ] \subset \mathrm {Frac}(A). \] There is the inclusion homomorphism \(A \to S^{-1}A\), and every element of \(S\) becomes invertible in \(S^{-1}A\). Recall that in the construction of \(\mathrm {Frac}(A)\) we use the equivalence relation which declares \(\frac {a}{s} = \frac {b}{s'}\) if \(as' - bs = 0\). More generally, if \(A\) is not an integral domain but \(S\) is a multiplicative set that does not contain any zero divisors, then the same construction works.
Now let \(A\) be any ring and \(S \subset A\) a multiplicative set, that is, \(1 \in S\) and \(S\) is closed under multiplication, which possibly contains zero divisors. We would like to construct a ring \(B\) and a homomorphism \(\iota : A \to B\) such that \(\iota (s)\) is invertible for all \(s \in S\). If \(as = 0\) for some \(a \in A\) and \(s \in S\), then \(\iota (a)\iota (s) = 0\) forces \(\iota (a) = 0\), so \(\iota \) cannot be injective: such elements \(a\) have to be killed first. It turns out that this can be done directly, by imitating the construction of the field of fractions with a small modification of the equivalence relation.
Definition 6.13. Let \(A\) be a ring and \(S \subset A\) a multiplicative set. Consider the set \(A \times S\) with the equivalence relation \[ (a,s) \sim (b,s') \text {\ if and only if there exists\ } t \in S \text {\ such that\ } t(as'-bs)=0. \] Writing \(\frac {a}{s}\) for the equivalence class of \((a,s)\), we define the ring of fractions \(S^{-1}A \coloneqq \{ \frac {a}{s} : a \in A, s \in S \}\) with the following rules for addition and multiplication: \[ \frac {a}{s} + \frac {b}{s'} \coloneqq \frac {as' + bs}{ss'}, \qquad \frac {a}{s} \cdot \frac {b}{s'} \coloneqq \frac {ab}{ss'}. \]
The ring \(S^{-1}A\) is also called the localisation of \(A\) with respect to \(S\).
\(\sphericalangle \) Verify that \(\sim \) is an equivalence relation and that addition and multiplication are well defined, so that \(S^{-1}A\) is a ring. The transitivity of \(\sim \) is instructive: it is where the element \(t\) in the definition is needed.
\(\blacktriangleright \) There is a homomorphism \(\iota : A \to S^{-1}A\) given by \(a \mapsto \frac {a}{1}\), and it satisfies \(\iota (s) \frac {1}{s}=1\) for all \(s \in S\). One often omits \(\iota \) from the notation and writes \(a \in S^{-1}A\) instead of \(\iota (a) \in S^{-1}A\) for \(a\in A\). The kernel of \(\iota \) is \(\mathrm {ker}(\iota ) = \{ a\in A : \frac {a}{1} = \frac {0}{1} \} = \{ a\in A : \exists \, t \in S,\ at =0 \}\). In particular, \(S^{-1}A = \{0\}\) if and only if \(0 \in S\), and \(\iota \) is injective if \(A\) is an integral domain and \(0 \notin S\).
\(\blacktriangleright \) If \(I \subset A\) is an ideal, we write \(S^{-1}I\) for the ideal of \(S^{-1}A\) generated by \(\iota (I)\). Explicitly, \(S^{-1}I = \{ \frac {a}{s} : a \in I,\ s \in S \}\), since the right-hand side is an ideal containing \(\iota (I)\), and \(\frac {a}{s} = \frac {1}{s}\, \iota (a)\).
\(\blacktriangleright \) Localisation satisfies the following universal property: suppose \(B\) is a ring and \(g: A \to B\) is a ring homomorphism such that \(g(s)\) is invertible for all \(s \in S\). Then there exists a unique homomorphism \(h: S^{-1}A \to B\) such that \(g = h \circ \iota \). Indeed, \(h(\frac {a}{s})\, g(s) = h(\frac {a}{s})\, h(\frac {s}{1}) = h(\frac {a}{1}) = g(a)\) forces \(h(\frac {a}{s}) = g(a) g(s)^{-1}\), and one verifies that this formula gives a well-defined homomorphism of rings.
\(\sphericalangle \) Let \(\mathfrak {n} = \mathrm {ker}(\iota ) = \{ a \in A : \exists \, t \in S,\ at=0 \}\), and let \(\overline {S}\) be the image of \(S\) in \(A/\mathfrak {n}\). Show that the multiplicative set \(\overline {S}\) does not contain any zero divisors of \(A/\mathfrak {n}\), and that \(\overline {S}^{-1}(A/\mathfrak {n})\) is isomorphic to \(S^{-1}A\).
The following two are the main examples of localisation.
(a) Let \(f \in A\) and \(S = \{ 1, f, f^2, \ldots \}\). We write \(A_f \coloneqq S^{-1}A\). We have \(\mathrm {ker}(\iota ) = \{ a \in A : \exists \, n \geq 0, \ af^n =0 \}\), and \(A_f = \{0\}\) if and only if \(f\) is nilpotent.
\(\sphericalangle \) Prove that \(A_f\) is isomorphic to the ring \(A[X]/ (Xf -1)\).
\(\blacktriangleright \) For \(0 \neq b \in \mathbb {Z}\) we have \(\mathbb {Z}_b = \{ \frac {a}{b^s} \in \mathbb {Q} : a \in \mathbb {Z},\ s \geq 0 \} = \mathbb {Z}[1/b]\).
\(\sphericalangle \) Let \(A = k[X,Y]/(XY)\), the coordinate ring of the union of the two axes in \(k^2\). Show that \(A_X\) is isomorphic to the ring \(k[X,X^{-1}]\), see Figure 12.
(b) Let \(\mathfrak {p} \subset A\) be a prime ideal. Then \(S \coloneqq A \setminus \mathfrak {p}\) is a multiplicative set, by the definition of a prime ideal, and we write \(A_\mathfrak {p} \coloneqq S^{-1}A\). This is a local ring with unique maximal ideal \(S^{-1}\mathfrak {p}\): by Lemma 6.14(ii) below, its prime ideals are the ideals \(S^{-1}\mathfrak {q}\) with \(\mathfrak {q} \subset \mathfrak {p}\). Beware of the notation: in \(A_f\) the element \(f\) is inverted, while in \(A_{\mathfrak {p}}\) the elements outside \(\mathfrak {p}\) are inverted.
\(\blacktriangleright \) If \(A\) is an integral domain, then \((0)\) is a prime ideal, and \(A_{(0)}\) is the field of fractions of \(A\).
\(\blacktriangleright \) For the prime ideal \((p) \subset \mathbb {Z}\) we get \(\mathbb {Z}_{(p)} = \{ \frac {a}{b} \in \mathbb {Q} : p \nmid b \}\).
\(\blacktriangleright \) For the maximal ideal \((X) \subset k[X]\) we get \(k[X]_{(X)} = \{ \frac {f}{g} \in k(X): X \nmid g \}\). Since \(X \nmid g\) means \(g(0) \neq 0\), these are the rational functions that are defined at the point \(0\).
Here are some basic properties of localisation.
Lemma 6.14. (i) Let \(J \subset S^{-1}A\) be an ideal and \(I = \iota ^{-1}(J) \subset A\). Then \(J = S^{-1}I \). In particular, \(S^{-1}A\) is Noetherian if \(A\) is Noetherian.
(ii) The map \(\iota ^*: \mathrm {Spec}(S^{-1}A) \to \mathrm {Spec}(A)\), \(\mathfrak {q} \mapsto \iota ^{-1}(\mathfrak {q})\), is an inclusion-preserving bijection onto the subset \(\{ \mathfrak {p} \in \mathrm {Spec}(A) : \mathfrak {p}\cap S = \emptyset \}\), with inverse given by \(\mathfrak {p} \mapsto S^{-1}\mathfrak {p}\).
(iii) (Localisation commutes with quotients) Let \(I\) be an ideal in \(A\), and write \(\overline {S}\) for the image of \(S\) under the canonical ring homomorphism \(A \to A/I\). Then \(\overline {S}^{-1} (A/I) \simeq S^{-1}A / S^{-1}I\).
\(\blacktriangleright \) For \(A = \mathbb {Z}\), part (ii) of the lemma says that \(\mathrm {Spec}(\mathbb {Z}[1/6])\) consists of the points of \(\mathrm {Spec}(\mathbb {Z})\) other than \((2)\) and \((3)\), while \(\mathrm {Spec}(\mathbb {Z}_{(5)})\) consists only of \((5)\) and \((0)\), see Figure 13.
Proof. (i) Since \(J\) is an ideal containing \(\iota (I)\), we have \(S^{-1}I \subset J\). Conversely, let \(x=\frac {a}{s} \in J\). Then \(\frac {a}{1} = \frac {s}{1} \cdot \frac {a}{s} \in J\), hence \(a \in \iota ^{-1}(J) =I\) and \(x \in S^{-1}I\). If \(A\) is Noetherian, then \(I = (a_1,\ldots , a_n)\) for some \(a_i \in A\), and \(J = S^{-1}I\) is generated by \(\frac {a_1}{1},\ldots , \frac {a_n}{1}\); so every ideal of \(S^{-1}A\) is finitely generated.
(ii) The key point is the following: if \(\mathfrak {p} \subset A\) is a prime ideal with \(\mathfrak {p} \cap S = \emptyset \), then for \(a \in A\) and \(s \in S\), \[ \tfrac {a}{s} \in S^{-1}\mathfrak {p} \iff a \in \mathfrak {p}. \] Indeed, if \(\frac {a}{s} = \frac {p}{r}\) with \(p \in \mathfrak {p}\) and \(r \in S\), then there exists \(u \in S\) such that \(u(ar - ps) = 0\), so \(a \cdot ru = p \cdot us \in \mathfrak {p}\). Since \(ru \in S\) does not lie in \(\mathfrak {p}\), we get \(a \in \mathfrak {p}\). It follows that \(S^{-1}\mathfrak {p}\) is a prime ideal of \(S^{-1}A\): it is proper, since \(1 \notin \mathfrak {p}\), and if \(\frac {a}{s} \cdot \frac {b}{t} = \frac {ab}{st} \in S^{-1}\mathfrak {p}\), then \(ab \in \mathfrak {p}\), so \(a \in \mathfrak {p}\) or \(b \in \mathfrak {p}\). It also follows that \(\iota ^{-1}(S^{-1}\mathfrak {p}) = \mathfrak {p}\).
Conversely, let \(\mathfrak {q} \subset S^{-1}A\) be a prime ideal and \(\mathfrak {p}= \iota ^{-1}(\mathfrak {q})\), a prime ideal of \(A\). Then \(\mathfrak {p} \cap S = \emptyset \), since otherwise \(\mathfrak {q}\) would contain a unit of \(S^{-1}A\), and \(S^{-1}\mathfrak {p} = \mathfrak {q}\) by part (i). So the two maps are inverse to each other, and both clearly preserve inclusions.
(iii) Both sides have the universal property for ring homomorphisms \(g : A \to C\) which map the elements of \(S\) to units of \(C\) and the elements of \(I\) to \(0\), so they are isomorphic. More concretely, one checks directly that \(\frac {a}{s} + S^{-1}I \mapsto \frac {a+I}{s+I}\) is a well-defined ring isomorphism from \(S^{-1}A /S^{-1}I\) to \(\overline {S}^{-1}(A/I)\). □