The Dimension of Fibers - A Course in Commutative Algebra

Part II Dimension

7.2 The Dimension of Fibers

x^kl∈(b₁, . . . , b_n)_S+I^k⊆(ϕ(a₁), . . . , ϕ(a_m), b₁, . . . , b_n)_S.

So (7.4) is proved. By Corollary7.9, this implies dim(S)≤m+n. ut Given a homomorphismϕ:R→S of rings, we have an induced map

f: Spec(S)→Spec(R), Q7→ϕ⁻¹(Q) (7.5) (see on page47; if R andS are coordinate rings of affine varieties, f corresponds to a morphism of varieties). For P ∈ Spec(R), the set f⁻¹({P}) is called thefiberof f overP. We now wish to give an algebraic counterpart of the fiber. The answer is as follows. Consider the ideal I = (ϕ(P))_S in S generated by the image ofP, and the multiplicative subset

U :={ϕ(a) +I|a∈R\P} ⊆S/I.

Then form the ring

S_[P]:=U⁻¹(S/I).

With the canonical homomorphisms π:S →S/I and ε:S/I →S_[P], we get the following result.

Proposition 7.11. In the above situation, the map Φ: Spec S_[P]

→f⁻¹({P}), Q7→π⁻¹ ε⁻¹(Q) is an inclusion-preserving bijection.

Proof. By Lemma1.22and Theorem6.5, Φis an inclusion-preserving injec- tion Spec S[P]

→Spec(S), and its image is

im(Φ) ={Q∈Spec(S)|I⊆QandU ∩(Q/I) =∅}.

It is easy to verify that the above conditions on Q are equivalent to P =

ϕ⁻¹(Q), i.e.,Q∈f⁻¹({P}). ut

In fact, the bijectionΦfrom Proposition7.11can be shown to be a home- omorphism (this follows from Exercises 3.4 and 6.5). So S_[P] is the desired algebraic counterpart of the fiber, and thefiber dimensionis equal to the Krull dimension ofS_[P]. Motivated by Proposition7.11, we callS_[P]thefiber ring of ϕover P. Notice that S_[P] ∼= S/I ifP is a maximal ideal, since in that case allϕ(a) +Iwitha∈R\P are already invertible inS/I. The reader should be warned that the symbolS_[P] for the fiber ring is not standard notation. In Exercise7.9we will study a more abstract way of defining the fiber ring, and introduce an alternative notation that is more standard.

Theorem 7.12(Fiber dimension, lower bound). Let ϕ:R→S be a homomorphism of Noetherian rings. Moreover, letQ∈Spec(S)andP :=ϕ⁻¹(Q).

Then for the fiber ringS_[P] we have

7.2 The Dimension of Fibers 93 dim S_[P]

≥ht(Q)−ht(P). (7.6)

In fact, if Q ∈ Spec S_[P]

is the image of Q in S_[P] (i.e., Q= U⁻¹(Q/I) with the above notation), then

ht (Q)≥ht(Q)−ht(P). (7.7)

Proof. The second inequality (7.7) implies the first, so we only need to prove (7.7). We do this by reduction to the local case. We have a (well- defined) homomorphism

ψ:RP →SQ, a

b 7→ ϕ(a) ϕ(b) mappingPP intoQQ. SettingJ := (ψ(PP))_S

Q and applying Lemma7.10, we obtain

dim (S_Q/J)≥dim(S_Q)−dim(R_P) = ht(Q)−ht(P). (7.8) We claim that dim (S_Q/J) = ht(Q). To prove this, we study the spectrum of S_Q/J. With the canonical homomorphismsε: S →S_Q andπ:S_Q →S_Q/J, we get a map

Spec (SQ/J)→Spec(S), q7→ε⁻¹ π⁻¹(q) .

By Lemma1.22and Theorem6.5, this map is inclusion-preserving and injective, and its image is

M:={Q⁰ ∈Spec(S)|ϕ(P)⊆Q⁰ andQ⁰⊆Q}

Q⁰∈Spec(S)|ϕ⁻¹(Q⁰) =P andQ⁰ ⊆Q . So dim (S_Q/J) is the maximal length of a chain inM. On the other hand, by Proposition 7.11, M is in an inclusion-preserving, bijective correspondence with

Q⁰∈Spec S_[P_]

|Q⁰ ⊆Q .

So ht(Q) is also the maximal length of a chain inM, and we conclude ht(Q) =

dim (SQ/J). ut

The most important special case of Theorem7.12is the “truly geometric”

case whereϕcomes from a morphism of affine varieties. Readers may already take a look at the inequality (10.8) in Corollary10.6on page152, which gives a translation of Theorem7.12into geometric terms.

We will continue to investigate whether (or when) the inequalities in Theo- rem7.12are actually equalities (see Theorem10.5for the final result). Before we start doing that, we draw an important conclusion, which only requires the lower bound from Theorem7.12.

Corollary 7.13 (Dimension of polynomial rings). LetR6={0}be a Noethe- rian ring and R[x] the polynomial ring in one indeterminate. Then

dim (R[x]) = dim(R) + 1.

Proof. Letϕ:R→R[x] =:S be the natural embedding. For an idealI⊆R we haveS/(ϕ(I))_S ∼= (R/I) [x] and ϕ⁻¹((ϕ(I))_S) =I. LetP0$P1$· · ·$ Pn be a chain of prime ideals in Spec(R). By the above, Qi := (ϕ(Pi))_S yields a strictly ascending chain of prime ideals in S. Qn is not maximal since S/Qn ∼= (R/Pn) [x] is not a field. Therefore dim(S) ≥n+ 1, and we conclude

dim(S)≥dim(R) + 1.

For showing the reverse inequality, letQ∈Spec(S) be a prime ideal and set P :=ϕ⁻¹(Q)∈Spec(R). We are done if we can show that ht(Q)≤ht(P) + 1. S is Noetherian by Theorem 2.11, so we may apply Theorem 7.12. The inequality (7.6) yields ht(Q)≤ht(P) + dim(S_[P]), so it remains to show that dim(S_[P_])≤1. We haveS/(ϕ(P))_S ∼= (R/P) [x], and under this isomorphism the setU ={ϕ(a) + (ϕ(P))_S |a∈R\P}maps to (R/P)\ {0}, so the fiber ring is

S_[P] ∼= ((R/P)\ {0})⁻¹(R/P) [x] = Quot(R/P)[x].

So by Example5.2(6) we get dim(S_[P]) = 1. ut

By repeated application of Corollary 7.13 we obtain a new proof of the first part of Corollary5.7. In Exercise7.10, the analogous result is shown for power series rings.

When are the inequalities of Theorem 7.12 actually equalities? We first look at two examples.

Example 7.14. We assume thatK is an algebraically closed field.

(1) LetX =V_K2(x1·x2) be the “coordinate cross”, Y =K¹ and f:X →Y, (ξ1, ξ2)7→ξ1

the first projection. This is the morphism of varieties induced by ϕ:

K[x] → K[x1, x2]/(x1·x2), x 7→ x1+ (x1·x2). Every maximal ideal inK[x] corresponds to a point ξ∈K¹, and has height 1 by the geometric interpretation of height (Remark 6.11(b)). Likewise, every maximal ideal inK[x₁, x₂]/(x₁·x₂) has height 1. Forξ∈K¹\ {0}, the fiberf⁻¹(ξ) consists of one point, so both inequalities from Theorem7.12are equalities. But forξ= 0, the fiber is the entirex₂-axis, so here the inequalities are strict.

(2) The variety X in the previous example is reducible. Now consider the irreducible variety

X =V_K3(x²₁+x²₂−x²₃) =V_K3 x²₁+ (x2+x3)(x2−x3) ,

7.2 The Dimension of Fibers 95

which inR³visualizes as a circular cone, and the morphism f:X →Y :=K², (ξ₁, ξ₂, ξ₃)7→(ξ₁, ξ₂+ξ₃).

We assume thatK is not of characteristic 2. As above, we see that all maximal ideals in the coordinate ringsK[X] andK[Y] have height 2. For (α, β)∈K²withβ 6= 0, we have

f⁻¹(α, β) =

α,β²−α²

2β ,β²+α² 2β

so the fiber dimension is 0 and we have equality in Theorem7.12. But for β = 0 andα6= 0, the fiber is empty (so Theorem 7.12does not apply), and forα=β= 0 we have

f⁻¹(0,0) ={(0, η,−η)|η∈K},

which is one-dimensional. So here the inequalities are strict. / In both of the above examples, the inequalities from Theorem 7.12 are equalities on an open, dense subset ofY. But we could easily destroy equality by substitutingY with a larger affine variety of higher dimension. This shows that forY an irreducible affine variety, equality can only hold if the morphism is dominant. So Exercise4.2 tells us that a reasonable hypothesis where we can expect equality almost everywhere is that ϕ: R → S is injective. This explains why Theorem 10.5, our final result on fiber dimension, has this hypothesis.

To push the theory further, we need a few lemmas. Before stating the first, we introduce the following terminology. We will say thatgoing down holds for a homomorphismϕ:R→S of rings if for everyP ∈Spec(R) and every Q⁰ ∈Spec(S) with ϕ(P)⊆Q⁰, there exists Q∈ Spec(S) withQ ⊆Q⁰ and ϕ⁻¹(Q) =P. This is illustrated in Figure7.2.

P ϕ⁻¹(Q⁰)

Q⁰ R

Figure 7.2.Going down

The term “going down” refers to the descent from Q⁰ toQ. Exercise 8.8 contains an example where going down fails. It is easy to see that ifU ⊆Ris a multiplicative subset and going down holds for the homomorphismU⁻¹R→ ϕ(U)⁻¹S induced by ϕ, then it also holds for the original homomorphism ϕ:R→S with the additional hypothesis thatϕ(U)∩Q⁰=∅.

Lemma 7.15 (Going down and fiber dimension). In the situation of Theo- rem 7.12, let U := ϕ(R\P). If going down holds for the homomorphism RP →U⁻¹S induced byϕ, then we have equality in (7.7).

Proof. By Proposition7.11, ht(Q) is the maximal length of a chain

Q0$Q1$· · ·$Qm=Q (7.9) of prime ideals Qi ∈ Spec(S) with ϕ⁻¹(Qi) = P. We have a chain P0 $

· · · $ Pn = P in Spec(R) of length n = ht(P). Since U ∩Q0 = ∅ and ϕ(P_n−1)⊆ϕ(P)⊆Q₀, we can use the remark preceding Lemma7.15to work downwards along the chain ofP_iand findQ₋₁, . . . , Q_−n∈Spec(S) extending the chain (7.9) downwards. This yields ht(Q) ≥ n+m = ht(P) + ht(Q),

so (7.7) is an equality. ut

A ring-homomorphism R → S makes S into an R-module. Recall that modules over a ring do not always have a basis (= a linearly independent generating set). If a module does have a basis, it is called free.

Lemma 7.16 (Freeness implies going down). Let ϕ: R → S be a ring- homomorphism withS Noetherian.

(a) IfS is free as anR-module, then going down holds for ϕ.

(b) If additionally there exists a basis B of S over R with 1 ∈B, then the induced mapϕ^∗: Spec(S)→Spec(R)is surjective.

Proof. For the proof of (a), letP ∈Spec(R) andQ⁰ ∈Spec(S) withϕ(P)⊆ Q⁰. Set I := (ϕ(P))_S, and let Q ∈ Spec(S) be minimal among the prime ideals which lie betweenI andQ⁰. In particular, Qis a minimal prime ideal overI. LetQ₁, . . . , Q_n be the other minimal prime ideals over I. (There are finitely many of them by Corollary 3.14(d).) We have Tn

i=1Q_i 6⊆Q, so we may choose y ∈ Tn

i=1Qi\Q. We claim that ϕ⁻¹(Q) ⊆ P, which together withI⊆Qyieldsϕ⁻¹(Q) =P, showing (a). So takea∈ϕ⁻¹(Q). Then

ϕ(a)y∈Q∩

i=1

Qi=√ I,

where Corollary 3.14(d) was used. So there exists a positive integer k with ϕ(a)^ky^k∈I. We havey^k ∈/I, since the contrary would implyy∈Q. So the smallest j withϕ(a)^jy^k ∈I is positive. Setz:=ϕ(a)^j−1y^k. Then z /∈I but ϕ(a)z∈I, so

Exercises 97

ϕ(a)z=

i=1

x_iϕ(a_i) (7.10)

with xi ∈S andai ∈P. IfB is a basis ofS as anR-module, we can write z =P

b∈Bϕ(zb)·b and xi =P

b∈Bϕ(xi,b)·b with zb, xi,b ∈R (where only finitely many coefficients in the sums are non-zero). From (7.10) and the linear independence ofB we get

a·zb=

i=1

xi,bai∈P

for allb∈B. But there existsb∈B withzb∈/P (otherwise,z∈I), soa∈P.

This completes the proof of (a).

The hypothesis of (b) implies that there exists a homomorphismψ:S→R of R-modules withψ◦ϕ= idR. LetP ∈Spec(R). In view of (a), we only have to show that there exists Q⁰ ∈ Spec(S) with ϕ(P) ⊆Q⁰. Assume the contrary. Then (ϕ(P))_S = S, so we have 1 = Pm

i=1s_iϕ(a_i) with s_i ∈ S and a_i ∈P. Applyingψ yields 1 =Pm

i=1ψ(s_i)a_i∈P, a contradiction. This

proves (b). ut

In Chapter8, we will obtain another set of conditions under which going down holds (see Theorem8.17).

This is how far we can push the theory with the present methods. We will continue the investigation of fiber dimensions in Chapter 10. There we will prove the generic freeness lemma (Corollary 10.2), which under rather weak assumptions says that the hypotheses of Lemma 7.16(a) and (b) are satisfied after localization at all prime ideals lying in an open, dense subset of Spec(R). Putting things together, this yields exact formulas for the fiber dimension, which hold almost everywhere (see Theorem10.5).

Exercises to Chapter 7

7.1 (The Cayley-Hamilton theorem). Deduce the Cayley-Hamilton theorem (“substituting a square matrixA∈R^n×n over a ring into its own characteristic polynomial f yields f(A) = 0”) from Lemma 7.2. (Solution on page 255)

7.2 (Hypotheses of Nakayama’s lemma). Give an example which shows that the hypothesis on finite generation of M cannot be dropped from Nakayama’s Lemma7.3. (Solution on page255)

7.3 (Nakayama’s lemma and systems of generators). LetRbe a ring with Jacobson radicalJ ⊆R, and let M be a finitely generatedR-module.

Writeπ:M →M/J M for the canonical map. Observe thatM/J M =π(M)

is an R/J-module, and for an R-submodule N of M, π(N) ⊆ π(M) is an R/J-submodule.

(a) LetN ⊆M be a submodule. Show the equivalence N =M ⇐⇒ π(N) =π(M).

(b) Letx1, . . . , xn ∈M. Show the equivalence

M = (x1, . . . , xn)R ⇐⇒ π(M) = (π(x1), . . . , π(xn))_R/J. (c) Assume thatRis local with maximal idealm, and writeK:=R/m. Show

that all minimal systems of generators ofM have the same numbernof elements, namelyn= dimK(M/mM).

(d) Give an example of a ringR and a finitely generated module M, where not all minimal systems of generators have the same size.

(Solution on page255)

*7.4 (Hypotheses of the principal ideal theorem). This exercise shows that Krull’s Principal Ideal Theorem 7.4may fail for non-Noetherian rings. The example is adapted from Gilmer [20, p. 321, Exercise 21]. Let K[x, y] be a polynomial ring over a field, and consider the subalgebra R:=K[x, xy, xy², xy³, . . .]⊂K[x, y], that we have already seen as an example of a non-Noetherian domain (Example2.3).

(a) Show that there exists precisely one prime idealP ∈Spec(R) lying over the principal ideal (x)R.

(b) Show that ht(P) = 2.

(Solution on page255)

7.5 (Can the spectrum be just one chain?). This exercise originated from the question whether there exists a Noetherian ring with just three prime ideals P0 $P1$P2. When I posed this question to Viet-Trung Ngo, he immediately answered it in the negative. His answer led to the following statement, which should be proved in this exercise: If P ⊆Qare two prime ideals in a Noetherian ring R which have at most finitely many prime ideals in between, than in fact there exists no prime ideal which properly lies betweenP and Q. (Solution on page256)

7.6 (Semilocal rings). A ring R is called semilocal if it has (at most) finitely many maximal ideals. For example, semilocal rings occur as coordinate rings of affine varieties consisting of finitely many points. Here is how semilocal rings can be constructed by localization. LetP₁, . . . , P_n∈Spec(R) be finitely many prime ideals in a ring such thatP_i6⊆P_jfori6=j. Show that

Exercises 99

U :=R\

[

i=1

P_i⊆R

is a multiplicative subset. Furthermore, show that the localization U⁻¹R is semilocal with maximal idealsU⁻¹Pi. (Solution on page256)

*7.7 (An infinite-dimensional, Noetherian ring). In this exercise we study an example of a Noetherian ring which has infinite Krull dimension.

The example is due to Nagata [41, Appendix, Example E1], and the hints given for the proof are adapted from Eisenbud [17, Exercise 9.6]. LetKbe a field andR=K[x₁, x₂, . . .] a polynomial ring in countably many indetermi- natesx_i,i∈N. Consider the prime ideals

P_i:= x_i2+1, x_i2+2, . . . , x_(i+1)2

⊂R (i∈N0)

and set U := R \ ∪_i∈_N₀Pi. Show that S := U⁻¹R is Noetherian but has infinite Krull dimension.

Hint: The hard part is to show that S is Noetherian. For this, consider a non-zero ideal I ⊆ R, take f ∈ I\ {0}, and choose n ∈ N0 such that all indeterminates xj occurring inf satisfyj≤(n+ 1)². Show that there exist f1, . . . , fm∈I such that (I)R_Pi = (f1, . . . , fm)R_Pi fori≤n. Now takeg ∈I and set

J :={h∈R|h·g∈(f1, . . . , fm, f)_R}.

Use Lemma 7.7 to show that there exists h ∈ J \ ∪ⁿ_i=0Pi. Assume that J ⊆ ∪i∈N0P_i and derive a contradiction. From this, conclude that g ∈ (f₁, . . . , f_m, f)_S, and thatS is Noetherian. (Solution on page257)

7.8 (Systems of parameters). Parts (a)–(c) of this exercise give examples of affine varieties X ⊆ Kⁿ over an algebraically closed field. Consider the localizationRxof the coordinate ringR=K[X] at the pointx= (0, . . . ,0)∈ X and find a system of parameters of Rx. Does there exist a system of parameters which generates the maximal ideal?

(a) X={(ξ1, ξ2)∈K²|ξ1ξ2= 0} (see Example7.14(1)).

(b) X={(ξ1, ξ2, ξ3)∈KL³|ξ₁²+ξ₂²−ξ₃²= 0} (see Example7.14(2)).

Figure12.1on page189).

Hint:You may use Exercise7.3(c) for answering the second question.

Remark: A local ring whose maximal ideal is generated by a system of parameters is calledregular(see Definition13.2).(Solution on page258) 7.9 (The fiber ring as a tensor product). This exercise gives a more abstract description of the fiber ring. Let ϕ: R → S be a homomorphism of rings andP ∈Spec(R). Let K:= Quot(R/P) and ψ:R →K, a7→ ^a+P_1+P the canonical map.

(a) Show that the fiber ring S_[P] is the pushout of ϕ and ψ. Here is the definition of the pushout.

For two R-algebras A and B (given by homomorphisms α and β), the pushoutof αand β is defined to be a ring C together with homomor- phismsγ:A→Candδ:B →Cmaking the square in the below diagram commutative (i.e.,γ◦α=δ◦β), such that the following universal prop- erty holds: For a ringT with homomorphismsΓ:A→T and∆:B →T withΓ◦α=∆◦β, there exists a unique homomorphismΘ:C→T such that the diagram

R α

- A β

B δ

- C A

A A

A A U Γ H

HH HH

HH HHj

∆ pppppp

ppppppppppp R Θ

commutes (i.e., Θ◦γ = Γ and Θ◦δ = ∆). As usual with universal properties, this implies that the pushout (if it exists) is unique up to isomorphism. More precisely, between two pushouts there exists a unique map which is simultaneously an isomorphism of A-algebras and of B- algebras.

(b) Conclude that Spec S_[P]

is the pullback of the maps f: Spec(S) → Spec(R) andg: Spec(K)→Spec(R) induced byϕandψ. (The pullback is defined as the pushout, but with all arrows reversed, and the maps considered are morphisms of spectra of rings.)

(c) Describe the map g: Spec(K) →Spec(R) explicitly, and show that the pullback off andgis the fiberf⁻¹({P}). So Spec S_[P]∼=f⁻¹({P}) by the uniqueness of pullbacks, which re-proves Proposition7.11.

Remark: The pushout of two homomorphismsα: R →A and β: R → B is isomorphic to the tensor productA⊗RB, which is equipped with a natural structure as a ring (see Lang [33, Ch. XVI, Proposition 6.1]). So by (a) we haveS_[P_]∼=K⊗RS. In fact, a notation the fiber ring overP more commonly found in the literature isκ(P)⊗RS, whereκ(P) := Quot(R/P) =K stands for the residue class field atP. (Solution on page258)

7.10(Dimension of the formal power series ring). Let R 6={0} be a Noetherian ring andR[[x]] the formal power series ring overR. Show that

dim (R[[x]]) = dim(R) + 1.

The proof may be broken up into the following steps.

Exercises 101 (a) IfI ⊆R is an ideal, then the kernel of the epimorphism S :=R[[x]] → (R/I) [[x]], obtained from applying the canonical map R → R/I coeffi- cient-wise, is (I)_S. Conclude that dim(S)≥dim(R) = 1.

(b) Show that 1−xf is invertible inS for everyf ∈S. Letm∈Spec_max(S) be a maximal ideal. Conclude thatx ∈ m. Show that n := R∩m is a maximal ideal inR.

ht(m)≤ht(n) + 1.

This finishes the proof.

Remark:By repeatedly using this result, we obtain dim (K[[x1, . . . , xn]]) =n

for the formal power series ring in nindeterminates over a field.

(Solution on page260)

7.11(Free modules and the locus of freeness). Parts (a)–(c) of this exercise give examples of ring extensions R ⊆S. Decide if S is free as an R- module. Determine the “locus of freeness”, i.e., the setX_free⊆Spec(R) of all P ∈Spec(R) such that (R\P)⁻¹Sis free as anRP-module. Draw conclusions on the fibers of the induced morphism Spec(S)→Spec(R). K[x1, x2, . . .] is always a polynomial ring over a field.

(a) R=ZandS=Z[1/2].

(b) S =K[x1, x2]/(x²₂−x²₁(x1+ 1)) andR =K[x1], where xi ∈S denotes the residue class ofxi. See Example8.9(4) for the variety belonging toS.

(Solution on page261)

Chapter 8 Integral Extensions

The concept of an integral ring extension is a generalization of the concept of an algebraic field extension. In the first section of this chapter, we develop the algebraic theory of integral extensions, and introduce the concept of a normal ring. Section 8.2studies the morphism Spec(S)→Spec(R) induced from an integral extension R ⊆S. In Section8.3, we turn our attention to affine algebras again. We prove the Noether normalization theorem, and use it to prove, among other results, that all maximal ideals of an affine domain have equal height.

Dalam dokumen A Course in Commutative Algebra (Halaman 91-103)