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ring zero ring algebra group of units
@qK
Chapter 1
Rings and Schemes
We study commutative rings and their geometric counterparts, the affine schemes.
The affine scheme associated to a commutative ring A consists of two pieces: a topological space SpecA, whose “points” are the prime ideals ofA, together with a sheaf of rings OA, whose sections are the “functions” of the affine scheme. A general scheme will then be obtained by “gluing” affine schemes, much in the same way a manifold is obtained by gluing open balls of some euclidean space.
1 Basics
Here we fix the notation and recall some basic concepts and results that will be used throughout. The reader is advised to skim through this section and come back as needed.
1.1 Rings
Unless otherwise stated from now on the unadorned term ringwill refer to a com- mutative ring with 1. Thezero ring0 is a legitimate one: it has a single element 0 = 1. We also require that morphisms of rings map 1 into 1, and ifM is a module over some ring, that 1·m=mfor all elementsm∈M.
Now let A be a ring. By an A-algebra we understand any ring morphism φ:A → B, or just the ring B if φ is clear from the context. For example, the polynomial ring A[x1, . . . , xn] is an A-algebra with respect to the inclusion map A ֒→A[x1, . . . , xn]. Given anA-algebraφ:A→B and elements a∈A andb∈B, we will usually writea·b instead ofφ(a)·b(even thoughφmay not be injective).
A morphism of A-algebras between A → B and A → C is a ring morphism f:B→C making the diagram
B f - C
A
6 -
commutative. In other words, using the shorthand notation for multiplication by elements of A, a ring morphismf:B →Cis a morphism ofA-algebras if and only if it isA-linear: f(ab) =af(b) for alla∈Aandb∈B.
We writeA×for the group of units ofA. The ideal generated bya1, . . . , an∈A will be denoted by one of the two forms
(a1, . . . , an) =A·a1+· · ·+A·an
Ideals can be added and multiplied: if {aλ} is a family of ideals of A, we write P
λaλ for the ideal generated by this family, and for idealsa,bofA, we write ab for the ideal generated by the productsab witha∈aandb∈b.
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ideal correspondence theorem
Ifa is an ideal ofAand a, b∈Awe write
a≡b (moda) ⇐⇒ a−b∈a
We will represent the image ofa∈Ain A/a bya+a,amodaor even by aifais clear from the context.
Recall the important
Theorem 1.1.1 (Ideal correspondence theorem) Let Abe a ring and a⊂A be an ideal. Then the quotient map A→A/a gives a bijection:
{idealsb⊂A|b⊃a} ←→ {ideals of A/a} b7→b/a
where b/a is the image ofbunder the quotient map.
Two kinds of ideals will be specially important to us. Recall that an idealp⊂A is primeif it is proper andab∈p⇒a∈p orb∈pfora, b∈A. Equivalently,pis prime if and only ifA/pis a domain. An idealm⊂Aismaximalif it is maximal under inclusion among the proper ideals of A. Equivalently,m is maximal if and only if A/mis a field. In fact, this follows from the ideal correspondence theorem above: m is maximal if and only ifA/m has just one proper ideal, and thus must be a field. Maximal ideals are prime since fields are domains. An easy application of Zorn’s lemma shows that every nonzero ring has a maximal ideal, and hence by the ideal correspondence theorem we have that every proper ideal is contained in a maximal ideal.
Commutative rings abound in nature. Number Theory and Algebraic Geom- etry are two major sources of commutative rings. In Number Theory one studies the ring of integers Zand its friends, such as the ring of Gaußian integersZ[i] or the ring ofp-adic numbersZp. In Algebraic Geometry, one is mainly interested in algebraic sets, which are subsetsV ⊂Cn “cut out” by a finite set of polynomials f1, . . . , fd∈C[x1, . . . , xn]:
V ={(a1, . . . , an)∈Cn|fi(a1, . . . , an) = 0 for alli}
Now if p, q∈C[x1, . . . , xn] are polynomials withp≡q (modf1, . . . , fd), we have that p(a1, . . . , an) = q(a1, . . . , an) for all points (a1, . . . , an) ∈ V. Hence the ele- ments of the quotient ring
A=C[x1, . . . , xn]/(f1, . . . , fd)
give rise to well-defined polynomial functions from V toC. Not surprisingly such ring of functions encodes many geometric properties of the original algebraic set and is a central object of study in Algebraic Geometry. For instance, we have a map
V → {maximal ideals ofA}
(a1, . . . , an)7→(x1−a1, . . . , xn−an)/(f1, . . . , fd)
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Basics 3
neighbourhood topological closure dense set topological basis induced topology
In fact, to check that (x1−a1, . . . , xn−an)/(f1, . . . , fd) is maximal inAfirst observe that
fi(x1, . . . , xn)≡fi(a1, . . . , an)≡0 (modx1−a1, . . . , xn−an) for alli and hence (x1−a1, . . . , xn−an)⊃(f1, . . . , fd). Therefore
C[x1, . . . , xn]/(f1, . . . , fd)
(x1−a1, . . . , xn−an)/(f1, . . . , fd) = C[x1, . . . , xn]
(x1−a1, . . . , xn−an) ∼=C
is a field, showing that (x1−a1, . . . , xn−an)/(f1, . . . , fn) is indeed maximal. Later we will prove Hilbert’s Nullstellensatz, which states that all maximal ideals ofA have this form, and thus the above map between points of V and maximal ideals of Ais a bijection.
1.2 Topology
Let X be a topological space. By a neighbourhoodof a point x∈X we simply mean any open setU ∋x. For any subsetY ⊂X we define itsclosureY to be
Y df=n x∈X
every neighbourhood of x has a nonempty intersection withY
o= \
F⊃Y F closed
F
Hence Y is the set of points of X which are “arbitrarily close” to points ofY, or equivalently the smallest closed set containingY. We say that Y isdenseinX if Y =X.
Recall that a basisof a topological space X is a collectionBof open subsets of X such that every open set ofX can be written as a union of open sets in B. For instance, the open balls form a basis for the standard topology onRn. A basis B ofX satisfies the following two properties:
1. S
U∈BU=X
2. ifU, U′∈ Bandx∈U∩U′, then there is aU′′∈ Bsuch thatU′′⊂U∩U′ andx∈U′′.
Conversely, if X is an arbitrary set then giving a topology on X is equivalent to giving a familyB of subsets ofX satisfying the two axioms above: forBdefines a topology onX whose open sets are precisely those that can be written as unions of elements of B. The advantage of working with a basis is that it can simplify some tasks. For instance in order to check that a map f:X → Y between topological spaces is continuous, it is enough to check that the pre-imagesf−1(W) are open in X for the open setsW in a basis ofY.
IfY ⊂X is an arbitrary subset of the topological spaceX, we will considerY as a topological space with itsinduced topology: the open sets of Y are those of the form U ∩Y, where U is an open set in X. The induced topology is the coarsest topology for which the inclusion mapY ֒→X is continuous. If ∼ is an equivalence relation onX, the quotient space X/∼is also a topological space: a set of equivalence classes inX/∼is open if and only if their union is open in X.
This is the so-calledquotient topology, the finest one for which the projection
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product topology Tychonoff’s theorem Hausdorff space
irreducible topological space
map X → X/ ∼ is continuous. Finally if {Xi}i∈I is a collection of topological spaces its cartesian product X =Q
i∈IXi will be considered a topological space with its product topology, the coarsest topology for which all projection maps pi:X→Xiare continuous. A basis ofX consists of the sets of the formT
i∈I0p−1i Ui
where each Ui ⊂Xi is open in the respective space andI0 is afinite subset of I.
The following result is equivalent to the axiom of choice.
Theorem 1.2.1 (Tychonoff ’s theorem) The product of compact spaces is com- pact.
Finally for a topological spaceX we say that
1. X is Hausdorff if for every pair of distinct points x, y ∈ X there exist two disjoint open setsU andU′ withx∈U andy∈U′;
2. X is quasi-compact if any open cover X = S
λUλ of X has a finite subcoverX=Uλ1∪ · · · ∪Uλn; following the French tradition, we say that X iscompactif it is quasi-compact and Hausdorff;
3. X isdisconnectedif it is the union of two disjoint nonempty closed (and thus also open) sets. Otherwise,X is said to beconnected;
4. X isreducibleif it is the union of two proper closed subsets. Otherwise, X is calledirreducible. Equivalently,X is irreducible if and only if any two nonempty open subsets have a nonempty intersection. In particular, irreducible spaces are connected. This concept is only interesting for non- Hausdorff spaces.
IfY is an arbitrary subset ofX, we say thatY is quasi-compact, irreducible and so on if if has the corresponding property with respect to the induced topology from X.
1.3 Categories
A Grothendieck universeis a nonempty setU such that 1. ifx∈ U andy∈xtheny ∈ U;
2. ifx, y∈ U then{x, y} ∈ U;
3. ifx∈ U then 2x∈ U, where 2xdenotes the set of subsets ofx;
4. ifI∈ U and (xi)i∈I is a family of elements of U thenS
i∈Ixi∈ U.
A Grothendieck universe is a very large set in which one can perform the usual operations of set theory on its elements without leavingU. Theaxiom of universe says that for any set xthere exists a universe U such that x∈ U. We adopt this axiom, which is independent of the other axioms of set theory, and by Sets or Ab we always mean the category of sets or abelian groups that belong to some fixed universe U. This will prevent logical difficulties of the type “sets of all sets”.
If C is a category we denote by C◦itsdualoroppositecategory, the category obtained by reversing the arrows of C. By abuse of language, we usually write X ∈C if X is an object of C. IfX, Y ∈ C, we write HomC(X, Y) for the set of morphisms from X to Y. We say thatf ∈ HomC(X, Y) is an isomorphism if there exists g ∈ HomC(Y, X) such that g◦f = idX and f ◦g = idY. We write X ∼=Y if there is an isomorphism betweenX andY. It may be useful to think of
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Basics 5
a category in terms of its underlying graph (actually a directed multigraph), whose vertices are the objects of the category and whose edges are the morphisms of the category. Each vertex has a distinguished loop, the identity arrow, and we have a composition law on the edges satisfying the usual axioms.
If D is another category, then by a functor F: C → D we always mean a covariant functor: it assigns an object F(X) ∈ D to each X ∈ C and a mor- phism F(φ)∈HomD(F(X), F(Y)) to every morphismφ∈HomC(X, Y) such that F(idX) = idF(X) for allX ∈C andF(φ◦ψ) =F(φ)◦F(ψ) for all morphisms φ andψof C that can be composed (in terms of the underlying graphs, a functor is a morphism of graphs respecting the composition laws on the edges). A contravariant functor from C to D is then just a functorF: C◦→D. Amorphism of functors or natural transformation φ:F → G between F: C → D and G: C → D is a collection of morphisms φX ∈HomD(F(X), G(X)), one for eachX ∈C, such that
F(X) φX- G(X)
F(Y) F(f)
? φY- G(Y) G(f)
?
commutes for all X, Y ∈ C and all f ∈ HomC(X, Y). The set of all morphisms betweenF andGwill be denoted Hom(F, G).
A functorF: C→D gives a set mapFX,Y: HomC(X, Y)→HomD(F(X), F(Y)) for each pair of objectsX, Y ∈C. We say thatFisfull(respectivelyfaithful,fully faithful) ifFX,Y is surjective (respectively injective, bijective) for all X, Y ∈ C.
A faithful functor need not be injective on objects or morphisms: we may have f:X →Y andf′:X′ →Y′ mapping to the same arrow; injectivity is only guaran- teed for morphisms between a fixed pair of objects. Similarly a full functor need not be surjective on objects or morphisms. A subcategoryC′ of C is a category whose objects and morphisms are subsets of those of C and whose composition law on arrows is that of C; a subcategory C′ of C is said to be full if the inclusion functor C′ ֒→C is fully faithful (in terms of the underlying graphs, a subcategory corresponds to a subgraph and a full subcategory, to an induced subgraph, i.e., a subgraph obtained by choosing vertices of a graph and including all edges between them).
We say that two categories C and D are naturally equivalent if there are functorsF: C→D andG: D→C such that we have isomorphismsG◦F ∼= idCand F◦G∼= idD, where id denotes the identity functor. A functorF: C→D gives an equivalence of categories if and only if it is fully faithful andessentially surjective:
every object of D is isomorphic to some F(X) with X ∈ D. In particular, an equivalence of categories does not need to be a bijection: for instance the categories given by the graphs
• and • ⇄ •
are equivalent. A fully faithful functorF: C→D establishes an equivalence between C and its image, a full subcategory of D.
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For each object X ∈ C we have a functor X: C → Sets given by X(−) = HomC(X,−). We say that a functorF: C →Sets isrepresentable by an object X ∈C if there is an isomorphismF ∼=X. A direct but useful consequence of the definitions is the so-called Yoneda’s lemma, which states that for any functor F: C→Sets and for eachX ∈C there is a natural bijection
Hom(X, F) ≈- F(X) φ7→φX(idX)
which isfunctorialinX, i.e., asX runs over the objects of X the above bijections yield an isomorphism between the functors X7→Hom(X, F) andF. In particular, we have that there is a natural isomorphism Hom(X, Y) = HomC(X, Y) for any X, Y ∈ C, showing that an object representing a functor is unique up to isomor- phism.
Let F: C → D and G: D→ C be functors. We say that F is a left adjoint of Gand thatG is aright adjointof F if for eachX ∈C andY ∈D there is a natural bijection
HomD(F(X), Y) ≈- HomC(X, G(Y))
functorial in both X and Y: for a fixed X, asY runs over the objects of D the above bijections yield an isomorphism between the functors HomD(F(X),−) and HomC(X, G(−)), and similarly for a fixed Y.
1.4 Limits
Definition 1.4.1 A category I isfiltered if it satisfies the following axioms:
1. Given two arrows
j ր i
ց k
they “eventually converge”: there existj→l andk→l such that j
ր ց
i l
ց ր
k commutes.
2. Given two arrows
i⇉j
they can be “equalised”: there existsj →ksuch that the compositions i⇉j→k
from itokare equal.
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Basics 7
3. I isconnectedas an undirected graph, i.e., any two objects iand j of I can be connected by a path of arrows, ignoring their orientation:
i←x1→x2←x3→ · · · ←xn−1→xn ←j
We say that a category I iscofilteredif the opposite category I◦is filtered.
The main example of a filtered category is the one given by a partially ordered set (I,≤) satisfying the following axiom: for any two elements i, j∈I, there exists another elementk ∈ I with i ≤ k and j ≤k. To such (I,≤), we can associate a filtered category whose objects are the elements of I and whose arrows are given by
Hom(i, j)=df
{i→j} ifi≤j
∅ otherwise
for any two i, j ∈ I. For instance, if M is a fixed A-module for some ring A, the partially ordered set of all finitely generated submodules of M (ordered by inclusion) gives rise to a filtered category.
Let I be a filtered category and C be any category. LetF: I→C be a functor.
Think of I as an “index set” for the objects F(i) ∈ C. We now define a sort of
“union” of theF(i). For any objectX ∈C, letcX: I→C be the constant functor taking every i∈I toX and every arrow to idX. A director inductive limitof F is any object representing the functorX 7→Hom(F, cX). If it exists, this object is unique up to isomorphism and we denote it by
−→limF or −→lim
i∈I
F(i)
Explicitly the direct limit of F: I→C is characterised by the “formula”
HomC( lim−→F, X)∼= Hom(F, cX)
where these bijections are functorial inX ∈C. To understand this formula, first observe that a morphism betweenF andcX consists of a family of compatible mor- phisms{φi:F(i)→X}i∈I in the sense that the following diagram is commutative for each arrowt:i→j in I:
F(j) φj - X
F(i) F(t)
6 φi
-
Now, using the formula, let{fi:F(i)→ lim
−→F}i∈Ibe the element of Hom(F, clim
−→F) corresponding to id lim
−→F ∈Hom( lim
−→F,lim
−→F). Intuitively, we think of the mor- phisms F(t):F(i)→F(j) as “inclusion maps”; then lim−→F will be the “union” of theF(i), where the fi:F(i)→ lim
−→F are the natural “inclusion maps”. What the
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formula then says is that to give a mapφfrom the “union” lim−→F toX is the same as to give a family{φi:F(i)→X}i∈Iof compatible maps from theF(i) toX. This bijection is given byφ7→ {φ◦fi}i∈I since we have a commutative diagram
id lim
−→F ∈ HomC( lim−→F,−→limF)∼= Hom(F, clim
−→F)∋ {fi}i∈I
φ∈ HomC( lim
−→F, X)
?
∼= Hom(F, cX)∋ {φ◦fi}i∈I
?
where the vertical arrows are induced by φ.
Example 1.4.2 Let M be a fixed A-module for some ring A and let I be the filtered category of its finitely generated submodules. Let F: I → A-mod be the inclusion functor. Then M = lim
−→F since to give a morphism fromM to another A-moduleN is the same as to give a family of compatible morphisms F(i)→ N from the submodules F(i)⊂M to N.
Example 1.4.3
Let I be a filtered category and F: I→ Ab a functor. We now show how to construct the direct limit ofF; the same method applies to other categories other than Ab. We define
−→limF df= L
i∈IF(i) H
whereH is the following subgroup: identifying an element ofF(i) with its image in L
i∈IF(i),H is generated by differencesxi−xj wherexi∈F(i) andxj ∈F(j) are two elements that “eventually agree”, that is, such that there exist arrowsf:i→k andg:j→kin I withF(f)(xi) =F(g)(xj). Then it is easy to show that the object
−→limF together with the natural mapsF(i)→ lim
−→Findeed qualify as a direct limit of F.
Next we briefly discuss the dual concept of direct limits. Let I be a filtered category, C some category andF: I→C be a functor. An inverseor projective limitofF, denoted by
←−limF or lim
←−i∈I F(i),
is an object in C that representsX 7→HomC(cX, F), i.e., such that we have bijec- tions
HomC(X,lim
←−F)∼= Hom(cX, F)
which are functorial in X. As before, we have a family {fi: lim
←−F →F(i)}i∈I of compatible maps such that the above bijection is given byφ7→ {fi◦φ}i∈I. Example 1.4.4 Here is the canonical example of a projective limit. Let k be a field and ks be its separable closure, and consider the filtered category I whose objects are the subfields l of ks which are finite and Galois over k, and where we have an arrow l → l′ if and only if l ⊃ l′. We have a functor F: I → Groups given by F(l) = Gal(l/k) and where the maps F(l) → F(l′), for l ⊃ l′, are the
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Spectrum of a ring 9
spectrum
natural projections Gal(l/k) → Gal(l′/k) taking a k-automorphism φ of l to its restrictionφ|l′. Then lim
←−F = Gal(ks/k) since to give a morphism from a groupH to Gal(ks/k) is the same as to give a family of compatible morphismsH →Gal(l/k) for all finite Galois extensionsl ofk.
Let I be a filtered category and F: I→ Ab a functor. We now show how to construct lim
←−F. Set
←−limF =
(xi)∈Y
i∈I
F(i)|F(f)(xi) =xj for allf ∈HomI(i, j) and alli, j∈I
Then it is not difficult to show that lim
←−F together with the natural projection maps lim
←−F →F(i) have the required property.
2 Spectrum of a ring
As we have already noted, prime ideals play a prominent role in the study of rings.
We now make the following
Definition 2.1 The set of all prime ideals of a ring A is called spectrum of A and is denoted by SpecA.
Example 2.2 If k is a field, Speck = {(0)} has a single element. On the other hand, since every nonzero ring has a maximal ideal, and the ring 0 has no proper ideals, we have that SpecA=∅ ⇐⇒ A= 0.
Example 2.3 (Affine Line) It is easy to check that ifAis a PID then SpecA={(0)} ∪ {(p)|pis irreducible}
All prime ideals but (0) are maximal in this case. For instance, forA =C[t], we have
SpecC[t] ={(0)} ∪ {(t−a)|a∈C}
Notice that we have a bijection a ↔ (t−a) between C and the set of maximal ideals of C[t]. We represent SpecC[t] as the “line” C: while the maximal ideals correspond to points in this line, the prime ideal (0) is represented by the line itself. Later, we will make SpecC[t] into a topological space, and then it will be shown that the closure of{(0)}is the whole space SpecC[t], so that (0) is a “dense point”, one that is “spread out” through the whole line. This explains our choice for the representation of (0).
←(0) (t)
0
(t−1) 1 (t+ 2)
−2
SpecC[t]
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It is important to understand how spectra of different rings relate to each other.
Letφ:A→B be a morphism of rings. Ifq∈SpecB then its pre-image φ−1(q) ={a∈A|φ(a)∈q}
is a prime ideal of A: φ−1(q) is proper since 1∈φ−1(q) ⇐⇒ 1 =φ(1)∈q, and if ab∈φ−1(q), then φ(ab) =φ(a)φ(b)∈q, so we must have, say, φ(a)∈q ⇐⇒ a∈ φ−1(q). Therefore we have an associated morphism between the spectra
Spec(φ): SpecB→SpecA q7→φ−1(q)
As one can easily check, this defines a functor Spec: Rings◦→Sets from the opposite of the category of rings to the category of sets.
Example 2.4 IfAis a subring of a ringBandi:A ֒→B is the inclusion map then Spec(i) takesq∈SpecB to q∩A∈SpecA.
If ais an ideal of a ring Aand q:A։A/ais the quotient map, then by the ideal correspondence theorem Spec(q): SpecA/a ֒→ SpecA is injective and its image is the subset of SpecA
V(a)df={p∈SpecA|p⊃a}
Example 2.5 Let A = C[x, y]/(x−y2) and B = C[t]. The map of C-algebras φ:A →B given by ¯x7→ t2 and ¯y 7→t is an isomorphism; the inverse map is the C-algebra morphism ψ:B →Agiven by t 7→y. Hence Spec(φ): Spec¯ B → SpecA is a bijection. The prime ideal (0)∈SpecB is sent toφ−1(0) =ψ(0) = (0). The maximal ideal (t−a)∈SpecB,a∈C, is sent toφ−1(t−a) =ψ(t−a) = (¯y−a) = (¯x−a2,y¯−a), which is also maximal. Hence
SpecA={(0)} ∪ {(¯x−a2,y¯−a)|a∈C}
We represent SpecAas the parabolax−y2= 0 inC2: the maximal ideal (¯x−a2,y¯− a) corresponds to the point (a2, a) while (0) corresponds to the parabola itself. Then Spec(ψ) can be thought of as the “projection” to the y-axis and Spec(φ), as the inverse map.
SpecA (0)
SpecB (0)
(¯x−a2,y¯−a) (t−a) Spec(ψ)
Spec(φ)
The maps Spec(φ) and Spec(ψ)
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Spectrum of a ring 11
Example 2.6 (Double cover) LetAandBas in the previous example, and con- sider the map of C-algebrasτ:B→Agiven byτ(t) = ¯x. Then Spec(τ): SpecA→ SpecB sends (0) toτ−1(0) = (0) sinceτ is injective. On the other hand, we clearly have (t−a2) ⊂ τ−1(¯x−a2,y¯−a); but since (t−a2) is maximal, we must have equality, and hence the image under Spec(τ) of (¯x−a2,y¯−a) is the maximal ideal (t−a2). Hence Spec(τ) corresponds to the “projection” to the x-axis, and the parabola can be thought of as a “double cover” of the line.
Let σ:A → A be the automorphism of B-algebras (via τ) given by σ(¯x) = ¯x and σ(¯y) = −y. Then Spec(σ) sends (0) to (0) while it sends (¯¯ x−a2,y¯−a) to (¯x−a2,y¯+a). From the fact that the diagram
A σ - A
B τ
6 τ
-
is commutative and Spec is a functor we have that SpecA Spec(σ)
SpecA
SpecB Spec(τ)
? Spec(τ)
is also commutative, and hence Spec(σ) permutes the primes of SpecAover a given prime of SpecB, inducing an action on each “fibre” of Spec(τ). It corresponds to the automorphism of the “double cover” that exchanges the two “sheets”.
SpecA
(0)
SpecB
(0) (¯x−a2,y¯+a)
(¯x−a2,y¯−a)
(t−a2) Spec(τ) Spec(σ)
Double cover of SpecC[t]
Example 2.7 (Arithmetic double cover) Let A = Z[√
5]. We will describe SpecAusing the inclusion map i:Z֒→A. Since Spec(i) takesp∈SpecAto p∩Z, we have two cases: p∩Z= (0) orp∩Z= (p) for some prime numberp.
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multiplicative set
Ifplies over (0), thenp= (0). This follows from the fact that for any nonzero ideal a of A,a∩Zis not zero. In fact, let a+b√
5 be a nonzero element ofa,a, b∈Z.
Then a−b√
5 6= 0 as well, and therefore a2−5b2 = a−b√ 5
· a+b√ 5
is a nonzero element in a∩Z.
For the second case, p contains the principal ideal (p), so we need to describe SpecA/(p). Since A∼=Z[x]/(x2−5), we have that A/(p)∼=Fp[x]/(x2−5). Now there are a few possibilities depending on howx2−5 factors inFp[x].
First, ifx2−5 is irreducible, thenA/(p) is a field isomorphic toFp2, and therefore (p) is a maximal ideal.
Second, if p6= 2 and p6= 5 andx2−5 splits, then x2−5 has two distinct roots
−aanda, whereais an integer such thata2≡5 (modp). Then by the Chinese Remainder Theorem we have an isomorphismFp[x]/(x2−5)∼=Fp×Fpgiven by the mapf(x)7→(f(a), f(−a)). Since Spec(Fp×Fp) consists of two primes, (0)×Fpand Fp×(0), SpecA/(p) also has two primes, namely (x−a)/(x2−5) and (x+a)/(x2−5).
They correspond to the prime ideals (p,√
5−a) and (p,√
5 +a) ofA, respectively.
Third, ifp= 2 orp= 5, thenx2−5 is a square. Let us analyse the casep= 5. We have thatx2−5 =x2, andA/(5)∼=F5[x]/(x2) has a single prime ideal (x)/(x2). In fact, a prime ideal (f(x)) ofF5[x] contains (x2) if and only iff(x) dividesx2, so we must have (f(x)) = (x). The corresponding prime ideal ofAis then (5,√
5) = (√ 5).
Similarly,A/(2)∼=F2[x]/(x−¯1)2and we obtain that the only prime ideal ofAlying over p= 2 is (2,√
5−1).
Since Zis a PID likeC[t], we representZas a line, and Spec(i): SpecA→SpecZ can be thought of as a “double cover” of this “arithmetic line”.
Now ifσ:A→Ais theZ-algebra automorphism given by√
57→ −√
5 we have that Spec(σ) is an automorphism of SpecApermuting the two “sheets” of the cover.
SpecZ
(2) (3) (5) (7) (11)
(2,√
5−1) (3) (√
5) (7)
(11,√ 5−4)
(11,√ 5 + 4)
SpecZ[√ 5]
Double cover of SpecZ
Remark 2.8 Using quadratic reciprocity, we can actually describe how x2 −5 factors for different primes: if p6= 2 andp6= 5 then x2−5 splits inFp[x] if and only if
5 p
= 1 ⇐⇒
p 5
= 1 ⇐⇒ p≡ ±1 (mod 5).
3 Localisation
Definition 3.1 LetAbe a ring. Amultiplicative subsetSofAis a subset such
DRAFT ET, May 08, 2007 (21:24)
Localisation 13
localisation localisation map field of fractions quotient field
that 1∈S, and which is closed under multiplication: t, s∈S ⇒ts∈S. The two most common choices ofS are:
1. Set of powersS={hn |n≥0}of a fixed elementh∈A.
2. The complementS =A−pof a prime idealp∈SpecA.
Given a multiplicative subsetS of a ringA, we want to construct a new ring S−1Awhich is the “closest possible” to Aand in which all elements ofS become units. For instance whenA =Z andS =A− {0} we should obtain S−1A =Q.
The idea is that the elements ofS−1A should be pairs (a, s)∈A×S representing
“fractions”a/s. One just has to be careful about the presence of zero-divisors: since the elements ofS are to become units inS−1Awe would expect thata/1 = 0/1 in S−1Aifsa= 0 inAfor some s∈S. This leads to the following
Definition 3.2 LetS be a multiplicative subset of a ring A. The localisation S−1A ofA with respect to S is the ring
S−1A=
fractions a/s witha ∈A and s∈S, where two fractions a1/s1 and a2/s2 are identified if there exists some t ∈S such thatt·(a1s2−a2s1) = 0 inA
where addition and multiplication are performed in the usual way:
a1
s1
+a2
s2
= a1s2+a2s1
s1s2
a1
s1 ·a2
s2
= a1a2
s1s2
The ringS−1A comes equipped with a mapρ:A→S−1A, defined byρ(a) =a/1 for alla∈A, calledlocalisation map.
It is easy (but tedious, and better done in private when no one is looking) to check that all the operations above do not depend on the choice of the representa- tives of the fractions. When A is a domain, (0) is a prime ideal and applying the above toS =A−(0) we obtain a fieldS−1A, thefield of fractionsorquotient field of A, denoted by FracA. In this case the localisation map A → FracA is injective, hence we can view A as a subring of FracA; more generally, S−1A can be viewed as a subring of FracAfor any multiplicative subsetS ⊂A. But for an arbitrary ringAthe localisation mapρ:A→S−1A may fail to be injective.
If A and B are rings andS ⊂A is multiplicative, we write HomS(A, B) for the subset of Hom(A, B) consisting of those morphisms that send each element of S to a unit in B. Then it is easy to check that S−1A represents the functor A7→HomS(A, B), i.e., there is a natural bijection, functorial inB,
HomS(A, B)∼= Hom(S−1A, B)
In terms of universal properties: if φ∈HomS(A, B) then there is a unique map ψ:S−1A→B such thatφ=ψ◦ρ. In this sense,S−1A is the “closest” ring toA in which all elements of S are units.
One can also localise ideals, or more generally, modules or algebras: given an A-module M, the localisation S−1M of M with respect to S is the S−1A- module
S−1M =
fractionsm/swithm∈M ands∈S, where two fractions m1/s1and m2/s2 are identified if there exists somet∈S such thatt·(s2m1−s1m2) = 0 inM
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where the operations are defined by m1
s1
+m2
s2
= s2m1+s1m2
s1s2
a t · m
s =am ts for alla∈A, s, t∈S, andm, m1, m2∈M.
Localisation is actually a functor from A-Mod to S−1A-Mod. A morphism φ:M →N ofA-modules gives rise to a morphism
S−1φ:S−1M →S−1N m
s 7→ φ(m) s
(form∈M ands∈S)
Again, one can easily check thatS−1φis well-defined, and that localisation respects composition and the identity map, making it into a functor.
Definition 3.3 Let A be a ring and M be an A-module. For h ∈ A we define Ah and Mh to be the localisations of A and M with respect to the powers of h.
Similarly for p∈SpecA we defineAp andMp to be the localisations of AandM with respect toA−p.
Theorem 3.4 (Localisation preserves exactness) Let Abe a ring andS be a multiplicative subset ofA. If
M f- N g- P is an exact sequence of A-modules then
S−1M S
−1-f S−1N S
−1-g S−1P
is an exact sequence ofS−1A-modules. In particular, the localisation of an injective (respectively surjective) map is also injective (respectively surjective).
Proof Notice first that imS−1f ⊂kerS−1g sinceS−1g◦S−1f =S−1(g◦f) = 0.
To show the opposite containment, let n/s∈ kerS−1g where n ∈ N and s ∈ S.
Since g(n)/s= 0/1 inS−1P there exists t∈S such thatt·g(n) =g(t·n) = 0. By exactness, we can findm∈M such thatf(m) =t·nand therefore (S−1f)(m/ts) = tn/ts=n/s, showing that n/s∈imS−1f, as required.
For example, localising the exact sequence ofA-modules 0 - kerf - M f- N
with respect to a multiplicative setSwe obtain an exact sequence ofS−1A-modules 0 - S−1kerf - S−1M S
−1-f S−1N
which shows that kerS−1f =S−1kerf. We say thatkernels commute with locali- sation. One can similarly show that images, cokernels and quotients also commute with localisation.
One has an important partial converse to the above theorem:
DRAFT ET, May 08, 2007 (21:24)
Localisation 15
annihilator
Theorem 3.5 (A local-global principle) Let Abe a ring.
1. LetM be anA-module. Then
M = 0 ⇐⇒ Mm= 0 for all maximal ideals mof A 2. Let
M f- N g- P be a complex of A-modules. If the localisations
Mm
fm- Nm
gm- Pm
are exact for all maximal idealsmofAthen the original complex is exact.
In particular, a morphism of A-modules is injective (respectively surjec- tive) if and only if its localisations with respect to all maximal ideals are injective (respectively surjective).
Proof To obtain the second statement we just have to apply the first one to the A-module kerg/imf and use the fact that (kerg/imf)m = kergm/imfm since localisation is an exact functor.
So we are left to prove 1, and the implication ⇒ is clear. To prove ⇐, let m∈M; we show thatm= 0. For that consider theannihilatorofm:
ann(m)df={a∈A|a·m= 0}
Clearly it is an ideal of A, and we have to show that ann(m) =A. Since Mm = 0 there existss∈A−msuch that s∈ann(m) and therefore ann(m)6⊂m. But then ann(m) is not contained in any maximal ideal, hence ann(m) =A, as required.
Localisation has the effect of “killing” primes, which is what makes it such a useful tool in the study of rings. We have:
Theorem 3.6 (Prime ideals in localisation) Let A be a ring andS be a mul- tiplicative subset ofA. Denote by ρ:A→S−1Athe localisation map, and let
DS df
={p∈SpecA|p∩S =∅}
be the the set of primes ofA disjoint fromS. Then Spec(ρ): SpecS−1A ֒→SpecA
is an injection with imageDS. The pre-image of p∈DS is given by S−1p, viewed as an ideal ofS−1A.
Proof Ifp= Spec(ρ)(q) and there is s∈S∩p then ρ(s)∈q, but sinceρ(s) is a unit inS−1A, this is impossible. Hence the image of Spec(ρ) is contained inDS. Next we prove that p 7→ S−1p is a well-defined map from DS to SpecS−1A and that this map is an inverse to Spec(ρ) restricted toDS.
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local ring residue field local morphism
First we show that forp∈DS,a∈Aands∈S we have that a
s ∈S−1p ⇐⇒ a∈p (∗)
The implication ⇐ is obvious; on the other hand, ifa/s∈S−1p, then there exist p∈pandt∈S such thata/s=p/t, and this in turn implies that there is anr∈S such that r(at−ps) = 0. Hencerta∈p, and sincep∩S=∅,amust be inp.
Now we show thatS−1p∈SpecS−1Awhenp∈DS: first,S−1pis proper since 1/1∈S−1p would imply that 1∈p by (∗); now if as ·as′′ ∈S−1p fora, a′ ∈Aand s, s′∈S thenaa′ ∈pagain by (∗) and eitheraor a′ must be inp, and thus either a/s ora′/s′ must be inS−1p, as was to be shown.
Finally we prove that Spec(ρ) and p 7→ S−1p establish a bijection between SpecS−1A andDS. If p∈DS then Spec(ρ)(S−1p) =p; this follows directly from (∗). Now givenq∈SpecS−1A, letp= Spec(ρ)(q); we have to show thatq=S−1p.
Clearly q ⊃ S−1p. To prove the reverse containment, let a/s ∈ q where a ∈ A and s∈ S. Thena/1 ∈q as well and hence a ∈p, i.e., a/s ∈ S−1p, and we are done.
Example 3.7 (Localisation at a prime) Letp ⊂A be a prime ideal, and let ρ:A → Ap be the localisation map. Let S = A−p. In this case, the imageDS
of Spec(ρ) consists of the prime ideals of A which are contained in p; all other primes are “killed” once we invert the elements of S. In this regard, we have a complementary effect to the map induced by the quotient mapA։A/p, since the image of SpecA/p֒→SpecA consists of primes that containp.
Since Spec(ρ) preserves inclusion of ideals, we have that Ap has a unique maxi- mal ideal, namely S−1p. We will often write pAp for this maximal ideal. Since localisation commutes with quotients we have that the field
Ap/pAp=S−1(A/p) = Frac(A/p) is the field of fractions of the domain A/p.
In view of the local-global principle (theorem 3.5) and the last example, rings with a unique maximal ideal are bound to play a central role in the study of rings and hence deserve a name:
Definition 3.8 Alocal ringAis a ring which has a unique maximal idealm. The quotientk=A/mis calledresidue field. As a shorthand, we write (A,m, k) for a local ring Awith maximal idealmand residue fieldk.
A morphism φ:A → B between two local rings (A,m, k) and (B,n, l) is said to be a local morphism if φ−1n = m or, equivalently, φ(m) ⊂ n. Notice that a local morphism induces an injective morphism of residue fields ¯φ:k →l given by amodm7→φ(a) modnfora∈A.
Let Abe any ring and u∈A. Thenuis a unit if and only if (u) =A, i.e, if and only ifudoes not belong to any maximal ideal. HenceAis a local ring if and only if A−A×is an ideal (in which case it must be maximal).
DRAFT ET, May 08, 2007 (21:24)
Localisation 17
Example 3.9 (Power series) For an arbitrary ring A, the ring of formal power seriesA[[t]] with coefficients inAis the ring
A[[t]]df= lim
←−n≥1
A[t]/(tn)
Its elements can be represented by formal expressionsa0+a1t+a2t2+· · · with ai ∈ A; the projection maps A[[t]] → A[t]/(tn) then send the power series to its
“truncations”a0+a1t+. . .+an−1tn−1modtn. Sums and products are computed in the usual way:
X
n≥0
antn+X
n≥0
bntndf=X
n≥0
(an+bn)tn X
n≥0
antn
·X
n≥0
bntndf
=X
n≥0
X
i+j=n
aibj
tn
To determineA[[t]]×, observe that (P
n≥0antn)(P
n≥0bntn) = 1 is equivalent to a0b0= 1
a0b1+a1b0= 0 a0b2+a1b1+a2b0= 0
... Hence if P
n≥0antn ∈ A[[t]]× then a0 ∈ A×. Conversely, if a0 ∈ A×, we can set b0 =a−10 , b1 =−a−10 a1b0, b2 =−a−10 (a1b1+a2b0) and so on, recursively solving for thebi. Therefore
A[[t]]×=nX
n≥0
antn
a0∈A×o
In particular if k is a field then all elements of k[[t]] that are not in the principal ideal (t) are units, and hence (k[[t]],(t), k) is a local ring.
Whenk =R or k =C, one can also consider the subring k{{t}} ⊂ k[[t]] of con- vergent power series: its elements are power series that converge in some neigh- bourhood of 0. The ringk{{t}}is again a local, with maximal ideal (t) and residue field k.
Example 3.10 (p-adic numbers) Letpbe a prime number. The ring Zp
= limdf
←−n≥1Z/(pn)
is called the ring ofp-adicnumbers. Its elements can be represented in “base p”
by an infinite suma0+a1p+a2p2+· · ·with 0≤ai≤p−1. As before, this infinite sum is a unit if and only ifa06= 0. Notice thatZpis a domain (although theZ/(pn) are not) and thatZis a subring ofZp via the “diagonal map”a7→(amodpn)n≥1. We have that (Zp,(p),Fp) is a local domain.
Local rings are somewhat “simpler” than general rings, so one would expect that they are easier to handle. In this regard, one of the central results about local rings is the apparently innocuous
DRAFT ET, May 08, 2007 (21:24)
minimal basis nilpotent element nilradical
Lemma 3.11 (Nakayama’s lemma) Let(A,m, k)be a local ring,abe a proper ideal of A, andM be a finitely generatedA-module.
1. IfaM =M thenM = 0.
2. IfN is a submodule of M such thatM =N+aM then M =N.
Proof The second item follow from the first applied toM/N. For the first item, suppose M 6= 0 and let ω1, . . . , ωn be a minimal set of generators of M. By hypothesis, we may write ω1=a1ω1+· · ·+anωn for someai ∈a ⊂m. But then ω1∈Aω2+· · ·+Aωn since 1−a1 is a unit (it is not inm), which contradicts the minimality ofω1, . . . , ωn.
Let us illustrate the “niceness” of local rings by showing that every finitely generated module over a local ring has what we can call a minimal basis. In general, the number of elements in different minimal sets of generators of a finite A-moduleM can vary. But ifAis local, we have
Corollary 3.12 (Minimal basis) Let M be a finitely generated module over a local ring (A,m, k). Then every minimal set of generators of M has dimkM/mM elements.
Proof Letm= dimkM/mMand letnbe the smallest number of elements needed to generateM. Suppose thatω1, . . . , ωngenerateM; then their imagesω1, . . . , ωn ∈ M/mM also generate the k-vector spaceM/mM and they are all distinct: if for instanceω1=ω2then the submoduleN=dfAω2+· · ·+Aωn⊂M is such thatM = N+mM and hence by Nakayama’s lemmaM =N contradicting the minimality of n. Hence n≥m. Conversely,M can be generated bym elements: letτ1, . . . , τm∈ M be such that their images form a basis of thek-vector spaceM/mM. Then the submodule N df= A τ1+· · ·+A τm ⊂ M is such that M = N +mM, hence by Nakayama’s lemmaM =N.
4 Some applications
Let us see a couple of examples of how to apply the techniques seen so far. The first one is a characterisation of the nilradical of a ring in terms of prime ideals.
Recall that an elementa∈Ais said to benilpotentif there existsn >0 such that an = 0. The set of all nilpotent elements of Ais callednilradicaland is denoted p(0). It is an ideal: if a∈ p
(0), then clearly ra ∈ p
(0) for any r ∈ A; and if a, b ∈ p
(0), we can choose n large enough so that both an and bn are 0. Then (a+b)2n = 0, since in the expansion of (a+b)2n each of the termsaibn−i is 0.
Hence a+b∈p (0).
Surprisingly there is a very strong connexion between the nilradical and the prime ideals of a ring. If p ∈ SpecA and a ∈A is nilpotent then for some nwe have
an= 0⇒an ∈p⇒a∈p
Therefore a nilpotent element belongs to everyprime ideal and hence p(0)⊂ \
p∈SpecA
p
DRAFT ET, May 08, 2007 (21:24)
Some applications 19
radical
minimal prime ideal
Actually the above containment is an equality! To show the opposite inclusion, we prove that for everyh /∈p
(0) there exists p ∈SpecA such that p 6∋ h. It is enough to show that SpecAh6=∅, for then any primep∈SpecA in the image of Spec(ρ): SpecAh→SpecAwill do, whereρ:A→Ah denotes the localisation map.
In fact, we have thath /∈p
(0) ⇐⇒ Ah 6= 0 since Ah = 0 ⇐⇒ 1/1 = 0/1 in Ah ⇐⇒ hn= 0 inAfor somen≥0. We have just shown:
Theorem 4.1 (Nilradical) For any ringA, p(0) = \
p∈SpecA
p
We can extend the above definitions a bit:
Definition 4.2 LetAbe a ring andabe an ideal ofA. Theradicalofais defined to be the ideal
√adf={a∈A|an∈afor some n≥1} Observe that by the ideal correspondence theorem√
a⊂Acorresponds exactly to the nilradicalp
(0)⊂A/a. Hence from the theorem we obtain Corollary 4.3 For any ideal a of a ringA,
√a= \
p⊃a
p (p∈SpecA)
Example 4.4 Letkbe a field and consider the polynomial ringA=k[x1, . . . , xn].
Letf ∈A be a nonzero element and let
f =u·pe11. . . perr, u∈k×
be the factorisation of f into powers of distinct monic irreducible polynomials p1, . . . , pr∈A. Then, from the fact thatA is a UFD, we obtain
p(f) = (p1. . . pr) = (p1)∩ · · · ∩(pr)
Notice that ifp ∈ SpecA then f ∈p ⇒ pi ∈p for some i, and since each (pi) is already a prime ideal ofA, the (pi) are theminimal prime idealscontaining (f).
As a result, the radical of (f) is just the intersection of the (pi), which agrees with the direct computation above.
Now we come to the second application. Letφ:A→Bbe a morphism of rings and writef: SpecB →SpecA for the associated map of spectra. Forp∈SpecA, we wish to compute thefibref−1(p) ofp with respect tof, i.e., the pre-image of punderf:
f−1(p)df={q∈SpecB |f(q) =p}={q∈SpecB |φ−1(q) =p}
DRAFT ET, May 08, 2007 (21:24)
To this end, letS =A−p. Thenφ(S) is a multiplicative subset ofB and we can consider the composition
B→ B
φ(p)B →φ(S)−1 B φ(p)B
The corresponding maps on spectra are:
Specφ(S)−1 B φ(p)B
֒→Spec B
φ(p)B ֒→SpecB
They are all injections: moreover the image of this composition consists of the primes q∈SpecB such that
q⊃φ(p)B
q∩φ(S) =∅ ⇐⇒
φ−1(q)⊃p
φ−1(q)⊂p ⇐⇒ φ−1(q) =p
which are precisely those primes in f−1(p). We summarise the discussion above in the next theorem where (in accordance with the conventions for A-algebras mentioned in the beginning of the chapter) we write B/pB instead of B/φ(p)B and S−1(B/pB) instead ofφ(S)−1(B/pB).
Theorem 4.5 (Fibres) Let φ:A → B be a morphism of rings and denote by f: SpecB → SpecA the corresponding map on spectra. Let p ∈ SpecA and let S =A−p. Then we have a one-to-one correspondence
primes ideals in SpecS−1(B/pB)↔primes ideals inf−1(p) given by the map
SpecS−1(B/pB)→SpecB corresponding to the composition B→B/pB→S−1(B/pB).
Example 4.6 (Blow-up) LetX = SpecC[x, y] andY = SpecC[x, y, z]/(y−xz), and let p:Y → X defined by the inclusion C[x, y]֒→C[x, y, z]/(y−xz). We can represent the maximal ideals (x−a, y−b) ∈ X with a, b ∈C as points (a, b) of the plane C2, and maximal ideals (¯x−a,y¯−b,z¯−c)∈ Y withb =ac as points (a, b, c) of the helicoid of equation y−xz = 0 in C3. Then pis the “projection”
map from the helicoid onto thexy-plane taking (¯x−a,y¯−b,¯z−c) to (x−a, y−b).
The surfacey−xz = 0 is an affine piece of what is known as theblow-upof the plane C2 at the origin (0,0). This name stems from the fact that it “takes apart”
the lines in C2 going through the origin: the liney=mx,m∈C, is shifted to the
“height” min the surfacey−xz = 0.
Let us compute the fibre of a maximal ideal (x−a, y−b)∈X. LetS=C[x, y]− (x−a, y−b). By the theorem, this fibre is given by
SpecS−1 C[x, y, z]
(y−xz, x−a, y−b)
∼= Spec C[z]
(b−az)
Hence if a6= 0 then C[z]/(b−az)∼=Cand the fibre consists of a single maximal ideal (¯x−a,y¯−b,z¯−b/a)∈Y; ifa= 0 andb6= 0 then C[z]/(b−az)∼= 0 and the fibre is empty; finally if a=b= 0 thenC[z]/(b−az)∼=C[z] and all maximal ideals of the form (¯x,y,¯ z¯−c),c∈C, as well as the prime ideal (¯x,y), belong to the fibre.¯ Looking at fibres can be an effective technique to “compute” spectra. For instance, we have
DRAFT ET, May 08, 2007 (21:24)
Zariski Topology 21
Theorem 4.7 (The Affine Plane) Let k be an algebraically closed field. Any prime ideal of the polynomial ring k[x, y] has one of the following forms:
1. (0) or
2. (f)for some irreducible polynomial f or 3. (x−a, y−b)for somea, b∈k.
Hence we have a one-to-one correspondence
prime ideal (0)↔the whole planek2 prime ideal (f)↔the curvef(x, y) = 0 maximal ideal (x−a, y−b)↔the point(a, b)∈k2
Proof Letp: Speck[x, y]→Speck[x] be the map corresponding to the inclusion k[x]֒→k[x, y]. We begin by computing the fibre of (x−a)∈Speck[x],a∈k. Let S=k[x]−(x−a). Then the fibre is given by
SpecS−1(k[x, y]/(x−a))∼= Speck[y]
Hence the fibre corresponds to the primes of the form (x−a) and (x−a, y−b) withb∈k.
Now we look at the fibre of (0). LetS =k[x]−(0). We have SpecS−1(k[x, y]) = Speck(x)[y]
Sincek(x)[y] is a PID, its prime ideals are of the form (0) or (f) wheref ∈k(x)[y]
is irreducible. Since the localisation mapk[x, y]֒→k(x)[y] is injective, the first case corresponds to the ideal (0) of Speck[x, y]. For the second case, writeq = (f)⊂ k(x)[y]; the corresponding prime ideal ink[x, y] is p =k[x, y]∩q. Multiplying f by a convenient element ofk(x)×, we may assume that it is in k[x, y] and that it is primitive (i.e., the gcd of its coefficients as a polynomial iny is 1). But then f is irreducible ink[x, y], and since k[x, y] is a UFD we have that (f)⊂k[x, y] is a prime ideal. We claim thatp = (f). We already have that (f)⊂p; to prove the reverse inclusion, letg∈p. Sinceg∈q we can writeg=f h/sfor someh∈k[x, y]
and some nonzeros∈k[x]. Thengs∈(f), but sincef has degree at least 1 with respect to the variabley(otherwise we would get the contradictionq=k(x)[y]) we have thats /∈(f) and thusg∈(f), as required.
5 Zariski Topology
In this section we introduce a topology on the spectrum of a ring. LetAbe a ring.
To describe the topology of SpecAwe introduce two special types of subsets. First, for an idealaofA(prime or not) let
V(a)df={p∈SpecA|p⊃a} Second, for an elementh∈Awe let
D(h)df={p∈SpecA|p6∋h}
Here is the geometric interpretation: an element a ∈ A should be thought of as a “function” on SpecA and whose “zeroes” are the prime ideals p ∋ a; after all, these are the primes for whicha≡0 (modp). ThenV(a) should be regarded as the “variety” defined by the zeroes of the “functions”a∈a, while D(h) should be thought of as the “domain” of the “function” 1/h. We have the following identities:
DRAFT ET, May 08, 2007 (21:24)
Zariski topology
Lemma 5.1 LetAbe a ring. Letg, h∈A, leta,b⊂Abe ideals ofA, and let{aλ} be an arbitrary family of ideals of A.
1. V((0)) = SpecAandV(A) =∅; 2. V(a)∪V(b) =V(ab);
3. T
λV(aλ) =V(P
λaλ);
4. SpecA−V(a) =S
h∈aD(h);
5. D(gh) =D(g)∩D(h).
Proof These are easy consequences of the definitions, so let us just prove 2. We have to show that ifp∈SpecA thenp⊃ab ⇐⇒ p⊃a orp⊃b. The implication
⇐= is clear; conversely ifp⊃abbutp6⊃aandp6⊃b, there exist elementsa∈a−p and b∈b−p. But thenab∈ab−p, contradicting the hypothesisp⊃ab.
Observe that by 1–3 of the last lemma the sets of the form V(a), a an ideal of A, are the closed sets of a topology on SpecA. And by 4–5 of the lemma the sets D(h) form a basis of a topology on SpecA. Finally identity 4 of the lemma shows that these two topologies coincide; it will be called theZariski topologyof SpecA.
Example 5.2 (Topology of the affine line) Let us describe the Zariski topology of the affine line SpecC[x] (see example 2.3). We claim that its proper closed subsets are precisely the finite subsets of SpecC[x]− {(0)}. In fact, sinceC[x] is a PID a proper closed subset has the form V((f)) for some nonzero polynomial f ∈ C[x] (that is, V((f)) is the variety “cut out” by f). But then V((f)) is the finite subset{(x−a)|f(a) = 0}of SpecC[x]− {(0)}. Conversely, any finite subset F ⊂ SpecC[x]− {(0)} can be written as F ={(x−a) | a ∈ Z} for some finite subsetZ ⊂C. But thenF =V((f)) wheref(x) =Q
a∈Z(x−a), proving thatF is closed.
Since SpecC[x] has infinitely many points, it cannot be written as the union of two proper closed subsets and hence it is irreducible. Then any two nonempty open sets intersect non-trivially, showing that this topology is notHausdorff. Also, the maximal ideals (x−a), a ∈ C, areclosed points, i.e., {(x−a)} =V((x−a)) is a closed set. On the other hand, (0) is a dense point, i.e., {(0)} is a dense set: if (0)∈V((f)) thenf = 0 and henceV((f)) = SpecC[x], showing that the smallest closed set containing{(0)}is the whole space SpecC[x]. This is why (0) was drawn as the whole complex line C, highlighting its “diffuse” nature.
Maps between spectra are continuous with respect to the Zariski topology:
Theorem 5.3 Let φ:A→B a morphism of rings. We have that Spec (φ)−1
D(h)
=D φ(h) and hence Spec (φ): SpecB→SpecAis continuous.
Proof We have that q∈ Spec (φ)−1
D(h)
⇐⇒ φ−1(q)∈D(h) ⇐⇒ q∈D φ(h)
DRAFT ET, May 08, 2007 (21:24)
Zariski Topology 23
Two special cases of the last theorem are of particular interest.
Theorem 5.4 Let Abe a ring.
1. (An open immersion) Leth∈Aand denote byρ:A→Ah the localisation map. ThenSpec(ρ): SpecAh֒→SpecAdefines a homeomorphism between SpecAh andD(h)⊂SpecA.
2. (A closed immersion) Letabe an ideal ofAand letq:A→A/abe the quo- tient map. Then Spec(q): SpecA/a ֒→SpecA defines a homeomorphism between SpecA/aandV(a)⊂SpecA.
Proof We already know that Spec(ρ) and Spec(q) are continuous and that they are bijections, so all we need to check is that they are open (and closed) maps onto their images.
First we show that Spec(ρ) is open. Given an open set D(g/hn) of SpecAh, g∈A, by theorem 3.6 we have that a primep∈D(h) is in the image ofD(g/hn) under Spec(ρ) if and only ifph∈D(g/hn), which is equivalent to p∈D(g). Hence Spec(ρ) D(g/hn)
=D(g)∩D(h), which is open.
Now we show that Spec(q) is a closed map: an ideal ofA/acan be written as b/awherebis an ideal ofA containinga. Then the image ofV(b/a) by Spec(q) is justV(b), which is closed, and we are done.
Example 5.5 The space X = SpecC[x, y]/(x2−y2) is reducible since it can be written as the union of two proper closed sets V(x−y) and V(x+y). In C2, they correspond to the linesx=y andx=−y intersecting at the origin. Each of these “lines” V(x−y) andV(x+y) is irreducible since, for instance,V(x−y) is homeomorphic to SpecC[x, y]/(x−y), which in turn is homeomorphic to SpecC[x], an irreducible space. But X is connected, since it is the union of two connected spacesV(x−y) andV(x+y) that intersect in a point given by the maximal ideal (x, y) (the “origin” ofC2).
Here are a couple of useful identities:
Lemma 5.6 Let Abe a ring and leta andbbe ideals ofA.
1. V(a) =V(√a);
2. V(a) =V(b) ⇐⇒ √ a=√
b;
3. for p∈SpecA, the closure of {p} is given by {p}=V(p). In particular, p∈SpecAis a closed point if and only ifp is a maximal ideal.
Proof The first identity follows from corollary 4.3. For the second, it is enough to show thatV(a)⊃V(b) ⇐⇒ √
a ⊂√
b, which is also a direct consequence of corollary 4.3. Finally, 3 follows from
{p}= \
V(a)∋p
V(a) =VX
a⊂p
a
=V(p)
DRAFT ET, May 08, 2007 (21:24)
Example 5.7 IfAis a domain then (0) is a dense point in SpecA. In theorem 4.7, the prime ideal (y−x2) is not a closed point; its closure is given by
V((y−x2)) ={(y−x2)} ∪ {(x−a, y−a2)|a∈k}
which corresponds to the points (a, a2)∈k2 of the parabolay−x2 = 0, together with an extra “diffuse” point (y−x2) which is dense inV((y−x2)).
Example 5.8 Here is an amusing application of the Zariski topology. Let k be a field and let A, B ∈ Mn(k) where Mn(k) denotes the ring of n×n matrices with entries ink. We show that the characteristic polynomials of ABandBA are equal. For that letxij, 1≤i, j≤n, ben2indeterminates and letR=k[xij] be the corresponding polynomial ring. We work inMn(R). FixA∈Mn(k)⊂Mn(R) and letB be the matrix
B=
x11 . . . x1n
... ... xn1 . . . xnn
∈Mn(R)
Let I ∈ Mn(R) be the identity matrix and let f(t) = det(tI−AB) ∈ R[t] and g(t) = det(tI −BA) ∈ R[t] be the characteristic polynomials of AB and BA respectively. Since detB 6= 0, we have that B is invertible in Mn(FracR) and hence
f(t) = det(tI−AB) = det(B) det(tI−AB) det(B−1) = det(tI−BA) =g(t) We would like to “specialise” the variablesxijto valuesbij ∈kin order to obtain the result for an arbitrary matrixB0 = (bij)∈Mn(k) instead ofB. This can be done as follows. Writef(t) =tn+cn−1tn−1+· · ·+c0andg(t) =tn+dn−1tn−1+· · ·+d0
with ci, di ∈ R. Let a = (ci−di) be the ideal of R generated by the differences ci−di, 0≤i < n. We now introduce the closed subset of SpecR
V(a) ={p∈SpecR|f(t)≡g(t) (modp[t])}
where p[t] is the prime ideal ofR[t] generated by p. Observe that f(t) =g(t) ⇒ (0)∈V(a). But sinceR is a domain, (0) is a dense point and hence we must have V(a) = SpecR. In particular, the maximal ideal m = (xij −bij) of R generated by the elements xij −bij belongs to V(a). But the characteristic polynomials f0(t)∈ k[t] and g0(t)∈ k[t] ofAB0 and B0A are precisely the reductions of f(t) and g(t) modulom, hencem∈V(a) impliesf0(t)−g0(t)∈m⇒f0(t) =g0(t), as required.
Next we study the interplay between the topology of SpecA and algebraic properties ofA.
Theorem 5.9 (Topological and algebraic properties) Let AandB be rings.
1. SpecA is always quasi-compact;
2. SpecA is irreducible if and only if A has a unique minimal prime ideal.
If SpecA is irreducible then its minimal prime ideal is a dense point and will be called the generic point ofSpecA.
