Lecture 4: Localization and regular functions
1
Localization
Definition 1. Let A be a ring, and S ⊆ A. We say S is a multiplicative subset if 1 ∈ S and
st ∈ S whenever s, t ∈ S.
Example 2. The following are multiplicative subsets of the ring A:
(i) for each f ∈ A, the subset S B {1, f , f 2 , . . .} ⊆ A;
(ii) for each prime ideal p ⊆ A, the subset S B A − p.
Definition 3. Let S ⊆ A be a multiplicative subset of the ring A. A localization of A with
respect to S is a ring map λ : A → A0 such that λ(S) ⊆ (A0 )× , and for each ring map φ : A → B
such that φ(S) ⊆ B× , there exists a unique ring map φ̃ : A0 → B such that the diagram
λ
A
A0
φ̃
φ
B
commutes.
Proposition 4. Let S ⊆ A be a multiplicative subset of the ring A.
(i) There exists a localization λ : A → S−1 A of A with respect to S.
(ii) If µ : A → A0 is any other localization of A with respect to S, then there is a unique
∼ A0 such that µ = αλ.
isomorphism α : S−1 A →
Proof. Let S−1 A denote the set of formal fractions a/s, a ∈ A, s ∈ S, modulo the equivalence
relation a/s ∼ b/t if and only if there exists y ∈ S such that u(at − bs) = 0. The binary
operations (a/s) + (b/t) B (at + bs)/st and (a/s) · (b/t) B ab/st are compatible with this
equivalence relation and make S−1 A a ring, and the map λ : a 7→ a/1 : A → S−1 A is a
localization of A with respect to S.
Consider µ as in (ii). Since λ is a localization, we have a unique map µ̃ : S−1 A → A0
such that µ̃λ = µ. Since µ is a localization, we have a unique map λ̃ : A0 → S−1 A such that
λ̃µ = λ. The uniqueness condition in the definition of a localiztion therefore implies that,
since λ̃µ̃ = λ and idS−1 A λ = λ, we must have λ̃µ̃ = idS−1 A . The same argument shows that
µ̃λ̃ = idA0 , so λ̃ = µ̃−1 , which proves (ii).
Example 5. Let A be a ring.
(i) If f ∈ A and S B {1, f , f 2 , . . .}, then we often denote S−1 A by Af or A[f −1 ].
(ii) If p ⊆ A is a prime ideal and S B A − p, then we often denote S−1 A by Ap .
(iii) If A is an integral domain and S = A − {0}, then S−1 A = Frac(A) is the fraction field.
Exercise 6. Let S ⊆ A be a multiplicative subset and λ : A → S−1 A the localization map.
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(i) When is λ injective?
(ii) Find a necessary and sufficient condition on S to guarantee that S−1 A = 0.
(iii) If A = k[x1 , . . . , Xn ]/a with a radical and S = {1, f , f 2 , . . . , }, show that the maximal
ideals of Af = S−1 A correspond bijectively to points of D(f ) ⊆ V(a).
(iv) More generally, for any ring A and any multiplicative subset S ⊆ A, show that the
function p 7→ λ−1 (p) induces a bijection between the set of prime ideals of S−1 A and the set
of prime ideals of A that do not meet S.
2
Directed colimits
Definition 7. Let C be a category.
(i) A directed set (I, ≤) is a nonempty set I equipped with a reflexive, transitive relation
≤ with respect to which each pair of elements has an upper bound. For example, if X is
a topological space and x ∈ X, then the set nbd(x) of open neighborhoods of x, equipped
with the containment relation ⊇ is a directed set.
(ii) Let (I, ≤) be a directed set. A direct system in C indexed by (I, ≤) consists of an Iindexed family {Xi |i ∈ I} ⊆ C of objects of C, and, for each i ≤ j, a morphism fij : Xi → Xj
such that:
(a) for each i ∈ I, fii = idXi ; and
(b) for each i ≤ j ≤ k, fik = fjk fij .
In other words, a direct system indexed by (I, ≤) is a functor (I, ≤) → C.
(iii) Let (Xi , fij ) be a direct system in C. A colimit of (Xi , fij ) consists of the data (X, φi )
of an object X ∈ C and a family of morphisms {φi : Xi → X | i ∈ I} of C such that:
(a) for each i ≤ j, φi = φj fij ; and
(b) given any family of morphisms {gi : Xi → Y | i ∈ I} into a fixed object Y ∈ C such
that gi = gj fij for each i ≤ j, there exists a unique morphism g : X → Y such that
gi = gφi for each i ∈ I.
Proposition 8. Let (I, ≤) be a directed set, and (Xi , fij ) an (I, ≤)-indexed direct system in C.
(i) If a colimit (Xi , fij ) exists, then it is unique up to unique isomorphism.
(ii) If C is the category of rings, then a colimit of (Xi , fij ) exists.
Proof. For (i), suppose (X, φi ) and (Y, ψi ) are both colimits of (Xi , fij ). Then we have unique
morphisms ψ : X → Y and φ : Y → X compatible with the φi and the ψi . The uniqueness of
Definition 7(iii)(b) now implies that φψ = idX and ψφ = idY .
F
For (ii), consider the following equivalence relation on the disjoint union i∈I Xi of the
sets underlying the Xi : if x ∈ Xi and y ∈ Xj , then x ∼ y if there exists k ∈ I such that k ≥ i,
k ≥ j and fik (x) = fjk (y). With this notation, we set x+y B fik (x)+fjk (y), and xy B fik (x)fjk (y).
F
As an exercise, check that these operations make colimi∈I Xi B
i∈I Xi /∼ a ring, and that
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the evident morphisms Xi → colimi∈I Xi realize it as the colimit of (Xi , fij ).
Notation 9. When a colimit of (Xi , fij ) exists, we denote it by colimi∈I Xi . Although colimits
are not unique, any two are canonically isomorphic by Proposition 8(i), so this is only
slightly abusive.
Exercise 10. Let A be a ring, S ⊆ A a multiplicative subset. Consider S as a directed set
with respect to the relation s ≤ t if s divides t. Then the morphisms a/sn 7→ a/sn : As → S−1 A
exhibit S−1 A as the colimit of the direct system of rings given by the As and the transition
maps a/sn 7→ (at n )/(st)n : As → Ast .
3
Regular functions
Notation: We will abusively denote X ∩ V(a) by V(a) and X ∩ D(f ) by D(f ).
Definition 11. Let X be an algebraic set, and U ⊆ X open. A function f : U → k is regular
at p ∈ U if there exist g, h ∈ A(X) and an open neighborhood p ∈ V ⊆ U such that h(p) , 0
and f (q) = g(q)/h(q) for each q ∈ V. We say f : U → k is regular if it is regular at each p ∈ U.
The ring of all regular functions f : U → k is denoted by OX (U).
Lemma 12. Let X be an algebraic set, and g, H, h ∈ A(X) such that H(p) , 0 for each p ∈ D(h).
There exist r ∈ Z≥0 and g 0 ∈ A(X) such that g(p)/H(p) = g 0 (p)/hr (p) for each p ∈ D(h).
Proof. Since H(p) , 0 for each p ∈ D(h), we have
D(h) ⊆ D(H)
⇒
V(H) ⊆ V(h)
⇒
rad(h) ⊆ rad(H)
⇒
∃r ∈ Z>0 [hr ∈ (H)].
Thus, there exist a ∈ A(X) and r ∈ Z>0 such that hr = aH. Setting g 0 B ag, we have
g(p)/H(p) = a(p)g(p)/(a(p)H(p)) = g 0 (p)/hr (p).
Proposition 13. Let X be an algebraic set, H ∈ A(X), and f : D(H) → k a regular function.
There exist g ∈ A(X), n ∈ Z≥0 such that f (p) = g(p)/Hn (p) for each p ∈ D(H).
Proof. By Definition 11 and quasi-compactness of the Zariski topology (Proposition 2.14),
S
there exist a finite open cover D(H) = m
i=1 Ui and Gi , Hi ∈ A(X) such that Hi (p) , 0 and
f (p) = Gi (p)/Hi (p) for each each i and each p ∈ Ui . Distinguished opens form a base
(Proposition 2.4), so, for each i, we may assume Ui = D(hi ) for some hi ∈ A(X). By Lemma
r
12, for each i, there exist gi ∈ A(X) and ri ∈ Z≥0 such that Gi (p)/Hi (p) = gi (p)/hi i (p) for each
p ∈ D(hi ). We may furthermore arrange things so that r = ri for each i.
For each p ∈ D(hi ) ∩ D(hj ) = D(hi hj ), we have
gj (p)
gi (p)
= f (p) = r ,
r
hi (p)
hj (p)
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so gi hrj = gj hri on D(hi hj ). Since hi (p)hj (p) = 0 on V(hi hj ) and X = V(hi hi ) ∪ D(hi hj ), we have
hi hj (gi hrj − gj hri ) = 0 ∈ A(X).
P
r+1
As V(H) = V(h1 , . . . , hm ) = V(h1r+1 , . . . , hr+1
), the Nullstellensatz implies Hn = m
i=1 ai hi
Pmm
n
for some n ∈ Z>0 and ai ∈ A(X). Let g B i=1 ai gi hi ∈ A(X). We claim f (p) = g(p)/H (p) for
each p ∈ D(H). Restricting all domains to D(hi ), we have
hr+1
i
m
X
aj gj hj =
j=1
As D(H) =
m
X
j=1
Sm
i=1 D(hi ),
aj gj hr+1
i hj =
m
X
aj gi hi hr+1
= gi hi Hn
j
j=1
⇒
g
gi hi
g
= r+1
= ri = f .
n
H
hi
hi
this completes the proof.
Corollary 14. For each f ∈ A(X), the map a/f r 7→ (p 7→ a(p)/f r (p)) : A(X)f → OX (D(f )) is an
isomorphism. In particular, A(X) ' OX (X).
Proof. By Proposition 13, the map is surjective. Let a ∈ A(X), r ∈ Z≥0 . If a(p)/f r (p) = 0 for
each p ∈ D(f ), then a(p) = 0 for each p ∈ D(f ), so f (p)a(p) = 0 for each p ∈ X. In that case,
f a = 0 ∈ A(X), so a/f r = 0 ∈ A(X)f .
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