singular morphisms, smoothness, and lifting lemmas

8/13/2019 Singular Morphisms, Smoothness, and Lifting Lemmas

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Singular Morphisms, Smoothness,

and Lifting Lemmas

H.J. Stein

University of California, Berkeley

Berkeley, CA 94702


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1. Introduction and Setting

Chapter 1. Introduction and Setting

Let A and B be commutative rings with unity. Let : A B be a ring homo-

morphism making B into an A-algebra. We say that is smooth if all A-algebra

homomorphisms : B C /I ofB to a quotient of the A-algebra Ccan be lifted to

C/I2. If is understood, we say that B is smooth over A. We will investigate the

degree of liftability embodied in homomorphisms which are not smooth. This work

continues along the lines of [Coleman], which extends and unifies that of [Elkik],

[Tougeron], and [Greenberg].

Let An =A[x1, . . . , xn], : An B be surjective, Ker() = (g1, . . . , gm). Let G

be the column vector (g1 . . . gm)tr, (where the superscript tr denotes the transpose),

and let (G) denote the ideal generated by the entries ofG. ForfAn, and for columnvectors and row vectorsP = (p1 . . . ps)

tr andQ = (q1 . . . q t) respectively (with entries

inAn), letf/Qdenote ( f/q1 . . . f /q t), and letP/Qdenote the matrix

whoseith row ispi/Q. We letf=f/X, andP =P/Xdenotef/Q, and

P/Q for Q = (x1 . . . xt). For a matrixM = (mij) with entries in An, let M/f

denote the matrix (mij/f). In this notation, the Jacobian matrix ofG is G =

G/X= the matrix whose rows are gi = the matrix whose columns are G/xi.

Let Matmn(A) denote the set ofm n matrices whose entries are elements ofA. If

This work was supported in part by the NSF, the Max Planck-Institut fur Mathematik, Bonn,

Germany, and the Institut des Hautes Etudes Scientifiques, France.

This thesis was processed using LAMS-TEX. Since LAMS-TEX usesAMS-TEX, I suppose that I amrequired to mention thatAMS-TEX was used.

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m and n are left out, we mean the set of matrices of all sizes. IfMis a matrix, and

i and j are integers, let Mij denote the matrix consisting of the first i rows and j

columns ofM. IfI = (I1 . . . I m) and J = (J1 . . . J n) are row vectors of integers, let

MIJdenote the matrix (MIiJj).

The following definitions are derived from those of [Coleman]. The ideal D also

appears in part II, section 2, remark 2.1 (page 89) of [Artin].

1.1 Definition. LetHbe a function taking triples(C,c,I), whereCis anA-algebra,

andc andIare ideals ofC, with c finitely generated, to triples(h1, h2, h3) of ideals

ofCcontained in Isuch that h1 h2, andh3 h2. An idealb ofB is said to have

theH-infinitesimal lifting property (orH-lifting property for short) if for any

C, c, andI with c finitely generated, and any homomorphism : B C/h1(C,c,I)

such that (b)(C/h1(C,c,I)) c/h1(C,c,I), there exists a homomorphism : B

C/h3(C,c,I) making the following diagram commute:

B

C/h3(C,c,I)

C/h1(C,c,I) C/h2(C,c,I)

If b has the H-lifting property for H(C,c,I) = (cI, I, I 2), we say that b

has the strong infinitesimal lifting property (abbreviated SILP). For the

case of H(C,c,I) = (cI,I, AnnC(c/cI2) I), we say that b has the weak in-

finitesimal lifting property, (WILP). The H-lifting property for H(C,c,I) =

(cI,I, AnnC((c + I2)/I2)) is called the very weak infinitesimal lifting property

(VWILP). When H(C,c,I) = (c2I, cI, c2I2), the H- lifting property is called the

Newtonian infinitesimal lifting property(NILP). We call the latter the Newto-

nian infinitesimal lifting property because it has the form used in the Newtons lemma

in [Coleman]. Note that ifbhas SILP then it has WILP, and if it has WILP, then it

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has VWILP. Also, ifb has SILP then it has NILP. WILP, and SILP were introduced,

and VWILP was mentioned in [Coleman].

Let B = A[X]/(G), where X = {x1, . . . , xn}, G = (g1 . . . gm)tr (the column

vector with entries gi) and gi A[X]. Let Idenote the identity matrix with mrows

and columns.

1.2 Definition.

s = s(B/A, G) = {A[X]| (I+ M) + GN0 mod (G),

for matricies N and M with MG= 0}

w = w(B/A,G) = {A[X]| (I+ M+ GN) 0 mod (G)


= (B/A, G) = {A[X]| I+ M+ GN0 mod (G),


Ds = Ds(B/A) = s(B/A, G) mod (G)

Dw

= Dw

(B/A) = w

(B/A, G) mod (G)

D = D(B/A) = (B/A,G) mod (G)

It is shown in [Coleman] that:

Ds ={b B |(b) has SILP},

and that

Dw ={b B |(b) has WILP}.

This both shows that Dw and Ds are independent of the representation ofB as an

A-algebra and determines the principal ideals having WILP and SILP.

It is also shown in [Coleman] that D(B/A) is independent of the representation

ofB as a quotient of a polynomial ring with coefficients in A.

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We continue the study of these various lifting properties as well as the study

of the mysterious ideal D. We determine necessary and sufficient conditions for an

ideal to have SILP, WILP, VWILP and NILP. We show that WILP and VWILP are

equivalent. We show that Ds is the unique maximal ideal ofB having SILP, and we

use WILP to strengthen some lifting lemmas of [Elkik]. Towards understanding D,

we show that ifA B Care rings, withC/Bsmooth, thenD(C/A) =D(B/A)C.

In particular, this means thatDlocalizes, and that the points ofB/Dare the singular

points ofB/A. We also compute several examples ofD, Ds and Dw.

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2. Determining ideals having ILPs

Chapter 2. Determining ideals having ILPs

The setting for this chapter is as follows. B/Aare rings, withB = An/(G), where

G = (g1 . . . gm)tr is a column vector of polynomials in An. Let = (1 . . . r)

tr be

a column vector of entries of B. We will determine when () has WILP, SILP, or

VWILP. Our procedure is to create a ring R which is, in a sense, universal for () to

have these properties.

2.1. Ds has SILP

We begin by determining weaker conditions for the ideal () to have SILP.

2.1.1 Lemma. The ideal () has SILP if and only if for all C, c, and I with c

finitely generated, and I2 = 0, and all : B C/cI satisfying () c/cI, there

exists: BCmaking the following diagram commute:

B

C

C/cI C/I

Proof: Clearly, if () has SILP, the conclusion holds. Conversely, suppose the

latter antecedent holds. Let C be an A-algebra, and let c and I be ideals of C

such that c is finitely generated. Let: B C/cI be an A-algebra homomorphism

satisfying () c/cI. We must show that there exists a : B C/I2 making the

following diagram commute:

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B

C/I2

C/cI C/I

Let C =C /I2, and let c and I be the ideals generated by the images ofc and

I in C, respectively (c = (c+I2)/I2, and I =I/I2). Then we have the canonical

homomorphism f: C C which reduces to f1: C/cI C/cI, f2: C/I C

/I,

and f3: C/I2 C. Furthermore, f () c/cI. Thus, by the supposition, there

exists : BC which is congruent tof modulo I. These homomorphism make

the following diagram commute:

C

B

C/I2

f3

C/cI

f1

C/I

f2C/cI C/I

Then =f13 makes the first diagram commute. Thus () has SILP.

2.1.2 Lemma. Letd= (d1 . . . dr)tr be a column vector of polynomials inA[X]which

is congruent to modulo (G). Then () has SILP if and only if for allA-algebras

Cand all ideals I ofC with I2 = 0, and all homomorphisms and making the


A[X] B

C C/(d)I

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A[X, Y]/((Y)2, Y d G). Then there exists an A-homomorphism : B R/(d)(Y),

and () (d)/(d)(Y). The ring Rsd,G appears in [Coleman] in the case r = 1. He

uses it to determine necessary conditions for principal ideals to have SILP or WILP.

2.1.3 Proposition. LetR = Rsd,G. The ideal()has SILP if and only if there exists

an A-homomorphism making the following commute:

B

R

R/(d)(Y) R/(Y)

(3)

Proof: The only if part follows immediately from the SILP property, since in

R, (Y)2 = 0. As for the if part, suppose there exists such a . Let C be an A-

algebra and let c, I C be ideals with c finitely generated. Let : B C/cI be

an A-homomorphism such that () c/cI. We must show that there exists an A-

homomorphism making (2) commute. By the previous lemma, we may assume that

I2 = 0, and that there exists a making the following commute:

A[X] B

C C/cI

such that c = (d). Then there exists an m r matrix L with entries in Isuch that

G=Ld. We define: A[X, Y]C byX=X, and Y = L. Then ((Y)2)

I2 = 0, and (Y d G) = Ld G = 0, so factors through a homomorphism

: R C. Since maps (d) into c and (Y) into I, it reduces to homomorphisms

1: R/(d)(Y) C/cI,2: R/(Y) C/I, and3: R C.

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Then1=, so we have the following commutative diagram:

R

3

B

C

C/cI C/I

R/(d)(Y)

1 R/(Y)

2

Thus = makes (2) commute, and we conclude that () has SILP.

2.1.4 Corollary. The ideal()has SILP if and only if there exist, for all0 i n,

an m mmatrixM with M G= 0, andn mmatricesNi satisfying the following:

i(I+ M) + GNi0 mod (G), (4)

whereIis them m identity matrix.

Proof: LetJ= G. We show that (4) has solution if and only if there exists a

making (3) commute.

Let the columns ofY be Y1 through Yr. Suppose that such a exists. Then

there exist matrices Ni (with entries in A[X, Y]) such that

G(X+r

i=1

NiYi) 0 mod ((Y)2, (Y d G)). (5)

Expanding G, we see that

0 G(X+

NiYi) mod (Y)2 + (Y d G) (6)

G(X) + J

NiYi mod (Y)2 + (Y d G) (7)

Y d + J

NiYi mod (Y)2 + (Y d G). (8)

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Y d= ri=1 Yidi, so the right hand side of (8) is congruent to

ri=1

(diI+ JNi)Yi mod (Y)2 + (Y d G). (9)

This implies that there exists an m 1 matrix Ewith entries in (Y)2, and an

m m matrixFwith entries in A[X, Y] such that

r

i=1(diI+ JNi)Yi = E+ F(Y d G). (10)This equation must hold in A[X, Y]. WriteF = F0 +F1 +F2, where F0 has

entries inA[X] andF1 is linear homogeneous inY, andF2 has entries in (Y)2. Write

Ni = N0i +N

1i, where N

0i has entries in A[X], and N

1i has entries in (Y). Looking

at the terms of equation (10) which are constant with respect to the entries ofY, we

see that

F0G= 0.

Considering the terms which are linear with respect to the entries ofY, we see

that

ri=1

(diI+ JN0i)Yi=F

0r

i=1

diYi+ F1G. (11)

Taking (11) modulo (G), and comparing the like coefficients of the entries ofY,

we see thatdiI+ JN

0i diF

0 mod (G).

TakingM=F0 and Ni=N0i yields the desired equations.

Conversely, suppose that there exist matrices M and Ni with MG = 0, which

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satisfy (4). We map A[X] to R by sending X to X+ NiYi. Then G maps toG(X+

NiYi) G(X) + J

NiYi mod (Y)

2 + (Y d G)

Y d + J

NiYi mod (Y)2 + (Y d G)

(diI+ JNi)Yi mod (Y)2 + (Y d G)

(diM+ E)Yi mod (Y)2 + (Y d G). (12)

for some m m matrix E with entries in (G). Since G Y d mod (Y d G), we

may replaceEby a matrixE with entries in (Y d), in which case E Yi0 mod (Y)2.

Thus the right side of (12) is congruent todiMYi mod (Y)

2 + (Y d G)

MY d mod (Y)2 + (Y d G)

MG mod (Y)2 + (Y d G)

0 mod (Y)2 + (Y d G).

The homomorphism therefore factors through (G).

The complete picture is given by the following theorem.

2.1.5 Theorem. The set Ds is an ideal and has SILP.

To prove this we first cite the following definition (from [Coleman]).

2.1.6 Definition. Let Mbe a square matrix such that MG = 0. Define M =

{An | (I M) (G)Nmod (G) for somen mmatrixN}, and defineDM

to be the image ofM inB.

It is clear that M is an ideal and is contained in s. It is shown in [Coleman]

(and also follows from the above equations) that DMhas SILP.

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For convenience we also make the following definition.

2.1.7 Definition. For a square matrixP such that P G= G, let P = { An |

P (G)Nmod (G)for somen mmatrixN}. LetD Pbe the image ofP inB .

Then M = (I+M).

The theorem follows almost immediately from the following lemma.

2.1.8 Lemma. P+ P

PP

Proof: If P, and P, then there exist matrices N and N

such that

P GNmod (G), and P GN mod (G). Then (+ )(P P) = GNP +

PGN. By differetiating differetiating the equalityP G = G we have that P G

G mod (G). Thus (+ )(P P) GNP +GN G(N P +N) mod (G), so

+ is in PP.

The first part of the theorem is given by the following corollary to the above

lemma.

2.1.9 Corollary. The set Ds(B/A) is an ideal.

Proof: If and map to elements ofDs, then there exist P and P such that

P, and P, so +

is in PP. But s contains PP, so +

is in

s. The fact thatDs is closed under multiplication follows from the fact that each

element ofDs is contained in an ideal ofB (namelyDP for some P) which in turn is

contained in Ds.

The proof of the above theorem is completed by the following corollary to the

above lemma.

2.1.10 Corollary. The idealDs(B/A) has SILP.

Proof: Suppose that C is an A-algebra, that c, I Care ideals ofC, and that

c is finitely generated. Suppose that f is an A-algebra homomorphism of B into

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C/cIsuch that (f(Ds)) c in C/cI. We must show that there exists an A-algebra

homomorphism g ofB into C/I2 such that the following diagram commutes:

B g

C/I2

f

C/cI C/I

Since (f(Ds)) c, and c is finitely generated, there must exist 1, . . . , l Ds

whose image in C/cI generate c. Then there exist Pi such that i DPi . Then

i(i) DP1Pl

, and so (f(DP1Pl)) c in C/cI. SinceDP1Pl

has SILP, we can

lift f to g.

2.2. Equations for VWILP

To determine the ideals which have VWILP, we follow the same procedure. We first

reduce to the case that I2 = 0.

2.2.1 Lemma. The ideal () has VWILP if and only if for allC, c, and I with c

finitely generated, and I2 = 0, and all : B C/cI satisfying () c/cI, there


B

C/AnnC(c)

C/cI C/I

Proof: The only difference between the proof here and the proof in the SILP case

is that we must check that C /AnnC((c + I2)/I2) =C/AnnC(c

). This is clear.

Next, we reduce to the case that c= ().

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2.2.2 Lemma. Letd= (d1 . . . dr)tr be a column vector of polynomials inA[X]which

is congruent tomodulo(G). Then () has VWILP if and only if for allA-algebras

C and all ideals I of C with I2 = 0, and all homomorphism and making the


A[X] B

C C/(d)I

there exists: BCmaking the this diagram commute:

B

C/AnnC(c)

C/(d)I C/I

Proof: Again, the only change to the proof for the SILP case is to note that

AnnC(c) AnnC(d).

As in the strong case, let d = (d1 . . . dr)

tr

be a column vector of polynomials inA[X] which is congruent to modulo (G). Let Y be an m x r matrix of indetermi-

nants. LetR = Rv = Rvd,G = A[X, Y]/((Y)2, Y d G) (note that Rs = Rv). Then

there exists an A-homomorphism : BR/(d)(Y).

We are immediately left with the

2.2.3 Proposition. The ideal () has VWILP if and only if there exists an A-

homomorphism making the following commute:

B

R/AnnR((d))

R/(d)(Y) R/(Y)

(13)

It remains to determine the equations for VWILP.

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2.2.4 Theorem. The ideal () has VWILP if and only if there exists an m m

matrixMiwithMiG= 0and ann m NiwithMiG= 0which for alli andj satisfy

the following:

j(iI+ GNi) iMj mod (G), (14)


Let J = G. We show that (14) has solution if and only if there exists a

making (13) commute.

Let the columns of Y be Y1 through Yr. Giving is the same as giving a

homomorphism from A[X] to R/AnnR((d)) such that G0 and which makes (13)

commute (after reducing modulo (G)). To make (13) commute, such a homomorphism

must map Xto something congruent to X modulo (Y). Thus, it must map X to

X+

NiYi for some n m matrices Ni. ThenG G(X+

NiYi), by which is

meant composition, not multiplication.

Thus, we see that exists

there exist n mmatricesNi (with entries in A[X, Y]) such that

G(X+r

i=1

NiYi) 0 mod AnnR(d)

there exist n m matrices Ni (with entries in A[X, Y]) such that for all

1 j r

djG(X+r

i=1

NiYi) 0 mod (Y)2 + (Y d G). (15)

Expanding G, we have that (15) holds


1 j r

dj(G + Jr

i=1

NiYi) 0 mod (Y)2 + (Y d G). (16)

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ReplacingGbyY d(since were working modulo (Y dG)), we see that (16) holds

there exist matrices n m Ni (with entries in A[X, Y]) such that for all

1 j r

dj(Y d + J

ri=1

NiYi) 0 mod (Y)2 + (Y d G) (17)

Saying that (17) holds modulo (Y)2 + (Y d G) is equivalent to saying that for

all 1 j r there exist m 1 and m m matricies Ej and Fj respectively, with

Ej 0 mod (Y)2 such that

dj(Y d + Jr

i=1

NiYi) =Ej+ Fj(Y d G) (18)

Replacing Y dbyr

i=1 diYi we see that (18) holds


1 j r there exist m 1 and m m matricies Ej and Fj respectively, with


dj

ri=1

(diI+ JNi)Yi=Ej+ Fj(Y d G) (19)

WriteFj as F0j + F

1j + F

2j, where F

0j Mat(A[X]), F

1j Mat(A[X, Y]) is linear

homogeneous in the entries ofY, andF2j Mat((Y)2). WriteNi as N

0i + N

1i, where

N0j Mat(A[X]), N1j Mat((Y)). Separating out the terms wich are constant with

respect to the entries ofY, linear with respect to the entries ofY, and the remaining

terms, we see that (19) holds

there exist matrices n m Ni (with entries in A[X, Y]) such that for all

1 j r there exist m 1 and m m matricies Ej and Fj respectively, with

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0 =F0jG (20)

dj

ri=1

(diI+ JN0i)Yi = F

0jY d + F

1jG (21)

dj

ri=1

(JN1i)Yi = Ej+ F1jY d + F

2j(Y d G) (22)

Equations (20) and (21) imply (by taking (21) mod (G)) that there exist n m

matricies N

0

i (with entries in A[X]) such that for all 1 j r there exist m mmatrices F0j Mat(A[X]) with F

0jG = 0 and

dj

ri=1

(diI+ JN0i)YiF

0j

ri=1

(diYi) mod (G) (23)

By setting like coefficients equal, this implies (14). Thus if () has VWILP, then

(14) holds.

Conversely, given (14), we take N0i equal to the given Ni, and F0j equal to the

givenMj . Then (20) holds. Since (14) holds moduloG, there must exist matrices F1j

which are linear homogeneous in the entries ofY such that (21) holds. Finally, (22)

holds by taking N1i = 0, F2j = 0, and Ej = F

1jY d. Thus if equation (14) holds,

then () has VWILP.qed

2.3. VWILP and WILP are equivalent

We now analyze WILP. Unfortunately, we cannot proceed as we did for SILP and

VWILP. Although we can (and will) take the analogous first step of reducing to a

quotient ofC (in this case we may assume cI2 = 0, instead of I2 = 0), we cannot

replace c by (d) because ifc and c are ideals, c c, it need not be the case that

AnnC( ccI2

) AnnC( c

cI2). We solve this problem by replacing the ring R with a

family of rings Rt.

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We begin with the analogous lemma.

2.3.1 Lemma. The ideal() has WILP if and only if for all C, c, and I with c

finitely generated, andcI2 = 0, and all: B C/cI satisfying () c/cI, there


B

C/AnnC(c)

C/cI C/I

Proof: The proof is the same as those in the previous sections, except that we

takeC =C/cI2 andc =c + cI2 and must note thatC/AnnC( ccI2

) =C/AnnC(c).

Now comes the messy part. Without the second lemma, we must resort to using

additional variables to represent the fact that the image of (d) containsc. This forces

us to use a set of rings Rt.

As in the previous cases, fix d = (d1 . . . dr)tr as a column vector of polynomials

in A[X] which is congruent to modulo (G). From here on, though, things will

be slightly different. Let t be a positive integer. Let Y be an mt matrix ofindeterminants. Let Z be a t r matrix of indeterminants. Let Rt = R

wt,d,G =

A[X , Y , Z]/((Zd)(Y)2, Y Z d G). Then there exist A-homomorphisms t: B

Rt/(Zd)(Y), namely the ones sendingXmod (G) toXmod (Zd)(Y)2 + (Y Zd G).

2.3.2 Proposition. The ideal()has WILP if and only if for allt Z+ there exist

A-homomorphismst making the following commute:

B t Rt/AnnRt(Zd)

t

Rt/(Zd)(Y) Rt/(Y)

(24)

Proof: The only if part is clear. As for the if part, suppose sucht exist. Let

Cbe an A-algebra, let c = (c1 . . . ct)tr be a column vector of entries ofC, and let I

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be an ideal ofC. Let : B C /(c)Ibe such that ()(c)/(c)I. By the previous

lemma, we may assume that (c)I2 = 0.

Chooseto make the following commute:

A[X] B

C C/(c)I

Then there exists an m t matrix L with entries in Isuch that G= Lc. Since

() c/cI, there also exists a t m matrix E Mat(C) such that c = Ed.

We define : A[X , Y , Z] C by X = X, Y = L, and Z = E. Then sends

(Zd)(Y)2 and (Y Zd G) to 0 and thus factors through a homomorphism: R C.

Then t mod (c)I. Since sends (Zd) to (c) and (Y) into I, it reduces to

homomorphisms1: Rt/(Zd)(Y) C/(c)I, 2: Rt/(Y) C/I, 3: Rt/AnnRt(Zd)

C/AnnC(c)I. Thus, we have the following commutative diagram:

Rt/AnnRt(Zd)

3

B

t

t

C/AnnC(c)

C/(c)I C/I

Rt/(Zd)(Y)

1 Rt/(Y)

2

Then t is the homomorphism which demonstrates that () has WILP.

Finally, the analogous corollary detailing the equations for VWILP remains to be

proved. For simplicity, we will prove it in two steps. First we will determine systems

of equations inA[X, Z] mod (G) that must hold, one system for each t. Then we will

eliminate tand Z.

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2.3.3 Corollary. The ideal () has WILP if and only if for all t Z+ there exist

m m matricesMi Mat(A[X, Z]) such that Mi = 0, and n m matricesNi

Mat(A[X, Z]) such that the following holds:

(Z)j((Z)iI+ GNi) (Z)iMj mod (G), (25)

whereI is them m identity matrix, and(Z)i is theith entry of the column vector

Z.

Proof: Let J =G. We show that (25) has solution if and only if there exists a

t making (24) commute.

Let the columns ofY be Y1 through Yt, let d = Zd, and let = Zso that in

particular di = (Zd)i. Then exists

there existn mmatricesNi (with entries in A[X , Y , Z]) such that for all

1 j t

G(X+t

i=1NiYi) 0 mod AnnRt(d

).

there exist n mmatricesNi (with entries in A[X , Y , Z]) such that

djG(X+t

i=1

NiYi) 0 mod (d)(Y)2 + (Y d G). (26)

Expanding G, we have that (26) holds


dj(G + Jt

i=1

NiYi) 0 mod (d)(Y)2 + (Y d G). (27)

Since were working modulo (Y d G), we may replace G by Y d and we see that

(27) holds

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dj(Y d + J

ti=1

NiYi) 0 mod (d)(Y)2 + (Y d G). (28)

Saying that (28) holds modulo (d)(Y)2 + (Y d G) is equivalent to saying

that there exist m1 and m m matrices Ej and Fj respectively, with Ej

0 mod (d)(Y)2 such that

dj(Y d + J

t

i=1 NiYi) =Ej+ Fj(Y d G). (29)Replacing Y d by

ti=1 d

iYi we see that (29) holds

there exist n m, m 1, and m mmatricesNi, Ej and Fj respectively

(with entries in A[X , Y , Z]) such that Ej 0 mod (d)(Y)2 and

dj

ti=1

(diI+ JNi)Yi = Ej+ Fj(Y d G). (30)

WriteFj as F0j +F

1j +F

2j, where F

kj Mat(A[X , Y , Z]) with F

0j constant with

respect to the entries ofY, F1j linear homogeneous in the entries of Y, and F2j

0 mod (Y)2. WriteNi as N0i +N

1i, where N

kj Mat(A[X , Y , Z]) with N

0j constant

with respect to the entries ofY and N1j 0 mod (Y). Then (30) holds

there exist n m, m 1, and m mmatricesNi, Ej and Fj respectively

(with entries in A[X , Y , Z]) such that Ej 0 mod (d)(Y)2 and

0 =F0jG (31)

djt

i=1

(diI+ JN0i)Yi= F

0jY d

+ F1jG (32)

dj

ti=1

(JN1i)Yi= Ej + F1jY d

+ F2j(Y d G) (33)

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Equations (31) and (32) imply (by taking (32) mod (G)) that there exist N0i and

F0j (with entries in A[X, Z]) such that F0jG = 0 and

dj

ti=1

(diI+ JN0i)YiF

0j

ti=1

(diYi) mod (G) (34)

By setting like coefficients equal, this implies (25).

Conversely, given (25), we take N0i equal to the given Ni, and F0j equal to the

givenMj . Then (31) holds. Since (25) holds moduloG, there must exist matrices F1j

which are linear homogeneous in Y such that (32) holds. Finally, (33) holds by takingN1i = 0, F

2j = 0, and Ej =F

1jY d

, which is then 0 mod (d)(Y)2. Therefore ()

has WILP.

The final equations for WILP are given by:

2.3.4 Proposition. The ideal () has WILP if and only if there exist m m and

n mmatricesMi andNi respectively with entries in A[X] such that MiG= 0 and

following holds:

j(iI+ GNi) iMj mod (G), (35)

whereIis the identity matrix.

Proof: By the previous corollary, we must show that (35) holds if and only if (25)

holds. Suppose that (35) holds. Then it holds for t = r. MappingA[X, Z] to A[X]

by sending X to Xand Zto the identity matrix yields equation (25).

Conversely, suppose that (35) holds. Fix t. Then Z= (zij), where i ranges from

1 tot, andj ranges from 1 to r. LetZi be the ith row ofZ. Multiplying (35) byzkj ,

and summing over j , we getj

zkjj(iI+ GNi) ij

zkjMj mod (G) (36)

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Letting Mk = jzkjMj , we may rewrite (36) ask(iI+ GNi) iM

k mod (G). (37)

Now multiply by zli, sum overi, and replace

i zliNi byNl and we get equation

(25).

Thus WILP and VWILP have the same equations, so

2.3.5 Theorem. The ideal() has WILP if and only if() has VWILP.

2.4. Equations for NILP

We now analyze NILP. We will proceed as we did for WILP.

2.4.1 Lemma. The ideal()has NILP if and only if for allC,c, andIwithc finitely

generated and satisfyingc2I2 = 0, and for all: B C/c2I satisfying() c/c2I,

there exists: BCmaking the following diagram commute:

B

C

C/c2I C/cI

Proof: We proceed as before, taking C = C/c2I2, c = (c + c2I2)/c2I2, and

I =I /c2I2.

Let d = (d1 . . . dr)tr be a column vector of polynomials in A[X] which is con-

gruent to modulo (G). Let t be a positive integer. Let Yi be an m t ma-

trix of indeterminants (for i = 1, . . . , t). Then ({Yi}) denotes the ideal gener-

ated by the entries of the Yis. Let Z be a t r matrix of indeterminants. Let

Rt = Rnt,d,G = A[X , Y , Z]/

(Zd)2({Yi})

2, (

YiZd(Zd)i) G

, where (Zd)i denotes

the ith entry of the column vector Zd, and the superscript n stands for NILP. Let

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t: B Rt/(Zd)2({Yi}) be the A-homomorphisms given by sending Xmod (G) to

Xmod (Zd)2({Yi}).

2.4.2 Proposition. The ideal() has NILP if and only if for allt Z+ there exist

Ahomomorphismst making the following commute:

B t Rt

t

Rt/(Zd)2({Yi}) Rt/(Zd)({Yi})

(38)

Proof: The only if part is clear. As for the if part, suppose such t exist. Let C

be an A-algebra, let c= (c1 . . . ct)tr be a column vector of entries ofC, and let I be

an ideal ofC. Let : B C/(c)2Ibe such that () (c)/(c)2I. By the previous

lemma, we may assume that (c)2I2 = 0.

Chooseto make the following commute:

A[X] B

C C/(c)2I

Then there existmtmatricesLiwith entries inIsuch thatG=

Licci. Since

() (c)/(c)2I, there also existtrand ttmatricesE, E Mat(C) respectively,

withE 0 mod (c)Isuch thatEd= c + Ec. Multiplying byI E, (whereI is

the t tidentity matrix), using the fact that c2I2 = 0, and letting E= (I E)E,

we see that there exists a t r matrixEsuch that Ed= c.

We define : A[X, {Yi}, Z] C by X = X, Yi = Li, and Z = E. Then

sends both (Zd)2({Yi})2 and ((YiZd(Zd)i) G) to 0, and thus factors through

Rt via a homomorphism : Rt C. Since (Zd) = c and ({Yi}) I, reduces

to homomorphisms1: Rt/(Zd)2({Yi}) C/(c)

2I,2: Rt/(Zd)({Yi}) C/(c)I, and

3: Rt C. Since 1 t mod (c)2I, these homomorphisms make the following

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diagram commute:

Rt

B

t

t

C

C/(c)2I C/(c)I

Rt/(Zd)

2

({Yi})

Rt/(Zd)({Yi})

Then3 t mod (c)I, so () has NILP.

Now for the equations. As in the WILP case, we first determine equations involv-

ing Z.

2.4.3 Corollary. The ideal() has NILP if and only if for allt Z+ and1 i

j t, there exist m m andn m matricesM andNijk respectively, with entries

in A[X, Z] such that MG= 0 and the following holds:

(Z)i(Z)j(I+ M) Gt

k=1

Nijk(Z)k mod (G), (39)

whereIis them m identity matrix, and(Z)i denotes theith entry of the column

vectorZ.

Proof: It suffices to prove this without the condition that i j because the left

side of (39) doesnt change when i and j are switched, so taking Nijk to be Njik for

i > j solves the remaining equations.

Let J = G. Let the columns ofYi be Yi1 through Yit. Fix t. Let d =Zd, and

let = Zso that in particular di = (Zd)i. We show that (39) has solution if and

only if there exist t making (38) commute.

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We see that t exists

there exist n mmatricesNijk (with entries in A[X, {Yi}, Z]) such that

G(X+t

i,j,k=1

NijkdkYij) 0 mod (d

)2({Yi})2 +

(

ti=1

Yiddi) G

. (40)

Expanding G, we see that (40) holds


G + J

t

i,j,k=1 NijkdkYij 0 mod (d

)2({Yi})2 +(

t

i=1 Yiddi) G . (41)

Since were working modulo ((

Yiddi) G), we may replace G by

Yid

diand

we see that (41) holds


ti=1

Yiddi+ J

ti,j,k=1

NijkdkYij 0 mod (d

)2({Yi})2 + ((

ti=1

Yiddi) G). (42)

Saying that (42) holds modulo (d)2({Yi})2 + (Yiddi G) is equivalent to

saying that there exist m 1 and m m matrices E and F respectively, with E

0 mod (d)2({Yi})2 such that

ti=1

Yiddi+ J

ti,j,k=1

NijkdkYij =E+ F

(

ti=1

Yiddi) G

. (43)

Replacing Yid byt

j=1 Yijdj in the leftmost sum of (43), so that

i Yid

di be-

comes

i,jYijd

idj we have that (43) holds

there exist n m,m 1, andm mmatricesNijk ,E, andFrespectively,(with entries in A[X, {Yi}, Z]) with E0 mod (Zd)

2({Yi})2 such that

ti,j=1

(didjI+ J

tk=1

Nijkdk)Yij =E+ F((

ti=1

Yiddi) G). (44)

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WriteF asF0 + F1 + F2, whereFk Mat(A[X, {Yi}, Z]) withF0 constant with

respect to the entries of the Yi, F1 linear homogeneous in the entries of the Yi, and

F2 0 mod ({Yi})2. Write Nijk as N

0ijk +N

1ijk , where N

kijk Mat(A[X, {Yi}, Z])

with N0ijk constant with respect to the entries of the Yi, and N1ijk 0 mod ({Yi}).

Then (44) holds

there exist n m matrices Nijk , E, and F (with entries in A[X, {Yi}, Z])

with E0 mod (Zd)2({Yi})2 such that

0 =F0G (45)

ti,j=1

(didjI+ J

tk=1

N0ijkdk)Yij =F

0t

i,j=1

Yijdidj + F

1G (46)

Jt

i,j,k=1

N1ijkdkYij =E+ F

1t

i,j=1

Yijdidj+ F

2(t

i,j=1

Yijdidj G) (47)

Equations (45) and (46) imply (by taking (46) mod (G)) that there exist matrices

N0ijk and F0 (with entries in A[X, Z]) satisfyingF0G= 0 and

ti,j=1

did

jI+ J

k

N0ijkdk

Yij =F

0t

i,j=1

Yijdidj (48)

By comparing the like coefficients of the entries of theYi, this implies (39). Thus,

if () has NILP then (39) holds.

Conversely, given (39) we takeN0ijk =Nijk , andF0 =M. Then (45) holds. Since

(39) holds modulo G, there must exist a matrix F1 which is linear homogeneous in

the entries of the Yi such that (46) holds. Finally, (47) holds by taking N1ijk = 0,

F2 = 0, and E = F1

ijYijdidj , which is then 0 mod (d

)2({Yi})2 since F1 is

linear homogeneous in the entries of the Yi. Thus, if (39) holds then () has NILP.

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The final equations for NILP are given by:

2.4.4 Theorem. The ideal()has NILP if and only if there exists an m mmatrix

MMat(A[X]), andn mmatricesN1iandN2iwith entries inA[X](for1 i r)

such that MG= 0 and for all1 i, jr the following holds:

ij(I M) G(N1ij + N2ji) mod (G), (49)


Proof: In the equations of the previous corollary, let M = M0 +M1, where

M0 is constant with respect to the entries of Z and M1 Mat(Z). Let Nijk =

N0ijk + N1ijk + N

2ijk , where N

0ijk is constant with respect to the entries ofZ, N

1ijk is

linear homogenous in the entries ofZ, and N2ijk Mat((Z)2). Let J = G. Then

there exist M and Nijk satisfying (39) if and only if there exist M0, M1, N0ijk , N

1ijk ,

N2ijk as above with M0G= M1G= 0, satisfying

0 Jt

k=1N0ijk(Z)k mod (G), (50)

(Z)i(Z)j(I M0) J

tk=1

N1ijk(Z)k mod (G), (51)

(Z)i(Z)j(M1) J

tk=1

N2ijk(Z)k mod (G). (52)

We first note that equations (50) and (52) are superfluous, since all satisfy

them by taking N2ijk =M1 =N0ijk = 0. Thus () has NILP if and only if there exist

matrices M0 Mat(A[X]) and N1ijk Mat(A[X, Z]) with M0G= 0 and N1ijk linear

homogenous in the entries ofZ, which satisfy (51) for all 1 i, j t.

Consider equation (51). Let Z= (zuv). Since N1ijk is linear homogeneous in the

entries ofZ,N1ijk =r

u,v=1 Nijkuvzuv, for some matricesNijkuvwith entries inA[X].

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Also (Z)i= ra=1 Ziaa, so we may rewrite (51) as

ab

ziazjbab(I M0) J

kuvw

Nijkuvzuvzkww mod (G), (53)

where k and urun from 1 to t, and a, b, v, and w run from 1 to r.

Thus () has NILP if and only if there exist m mand n mmatricesM0 and

Nijkuv respectively, satisfying M0G= 0 and equation (53).

We fix i and j and set coefficients equal by determining the coefficient ofzefzgh

on each side of equation (53). We may assume thate g and that ife = g thenfh.

Consider the right side of equation (53). Ife = g and f = h, the coefficient of

zefzgh isJ Nijgefh. Ife=g orf=hthe coefficient ofzefzgh isJ Nijgefh+JNijeghf.

The right side is slightly more complicated. The monomial ziazjb = zefzgh if and

only if (i,a,j,b) = (e, f , g , h) or (j, b, i, a) = (e, f , g , h). Ifi =j the latter cannot hold

(since i j and eg). So first consider the left side wheni =j. Then ife =i or

g =j the coefficient ofzefzgh is 0 and ife = i and g = j the coefficient ofzefzgh is

fh(I M0).

There are three possibilities when i = j . Ife =i or g =j the coefficient ofzefzgh

is 0. Ife = i and g=j but f=h the coefficient ofzefzgh is 2fh(I M0). Finally,

ife = i, g = j , and f=h, the coefficient ofzefzgh is fh(I M0).

Reassembling all the cases when i=j , we have that (53) is equivalent to:

0 =J Nijgefh (e=i or g=j ) (53a)

and (e= g and f=h)

0 =J(Nijgefh+ Nijeghf) (e=i or g=j ) (53b)

and (e=g or f=h)

fh(I M0) =J(Nijgefh+ Nijeghf) e= i and g = j (53c)

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When i= j , equation (53) is equivalent to:

0 =J Nijgefh (e=i or g=j) (53a)

and (e= g and f=h)

0 =J(Nijgefh+ Nijeghf) (e=i or g=j) (53b)

and (e=g or f=h)

2fh(I M0) =J(Nijgefh+ Nijeghf) e= i and g=j and f=h (53c

)

fh(I M0) =J Nijgefh e= i and g=j and f=h(53d

)

We may discard equations (53a), (53b), (53a), and (53b) since the Nijgef and

Nijeghmatrices appearing in those equations dont appear in the remaining equations,

and (53a), (53b), (53a), and (53b) are satisfied by taking those matrices equal to zero.

Equation (53c) is equivalent to the existence of a matrix M0 annihilating G such

that for each f and h, there exist matrices N1f=Nijjif and N2f=Nijijfsuch that

fh(I M0) =J(N1fh+ N2hf) (54)

Equations (53c) and (53d) are equivalent to the existence, for eachi and j ,i = j

of matrices N3fsuch that for all f , h

2fh(I M0) =J(N3fh+ N3hf) (f=h) (55a)

2f(I M0) =J(N3ff) (f=h) (55b)

Finally, we note that equations (55a) and (55b) are redundant, since given (54),

we may take O3f=O1f+ O

2f. Thus (54) implies (49), which completes the proof.

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3. Some Lifting Lemmas

Chapter 3. Some Lifting Lemmas

We extend Colemans lifting lemma to NILP ideals, and extend his version of

Elkiks lemma to non-principal WILP ideals. As in the previous chapter, B/A are

rings, B =A[X]/(G). Additionally, let Iand Jbe ideals ofA.

We begin with a definition.

3.1 Definition. Let Hbe an ideal ofA[X]. Suppose that J= (t) is principal, and

that A is complete with respect to the J-adic topology. Let I be any ideal of A.

Let L be the ideal of annihilators of powers oft. Let k be an integer such that the

intersection ofL with (tk) is0. We will say that H satifies Elkiks lemma in the

principal complete case if for any integersh andnsuch that n >max(2h, h + k),

and for any homomorphism f: A[X] A inducing a homomorphism g: B A/JnIas follows:

A[X] B

f g

A A/JnI

such thatf(H)containsJh, then there exists a lifting ofg toA making the following

commute:A[X] B A

f g

A A/JnI A/JnhI

Elkiks lemma states that all subideals of Elkiks ideal satisfy Elkiks lemma in

the principal complete case (although its not phrased this way in [Elkik]). Colemans

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version of Elkiks lemma is that D satisfies Elkiks lemma in the principal complete

case. This is an improvement since D contains Elkiks ideal and is sometimes larger.

3.2 Definition. Let Hbe any ideal ofA[X]. Let Jbe an ideal ofA. Suppose that

A is complete with respect toJ. We say that H satisfies Elkiks lemma in the

complete caseif for any integerh, there exist integersn andr such that ifm > n,

and we have homomorphisms making the following commute:

A[X] B

f g

A A/Jm

such that f(H) containsJh, then we can lift g to make the following commute:

A[X] B A

f g

A A/Jm A/Jmr

Elkik proves the following theorem, albet not stating it in this form.

3.3 Theorem. (Elkik)Suppose thatHsatisfies Elkiks lemma in the principal com-

plete case, and that ifC is anyA-algebra, the image ofM inB C satisfies Elkiks

lemma in the principal complete case forB C/Cwith respect to the image ofJ.

Then Hsatisfies Elkiks lemma in the complete case.

Since D(B/A) satisfies the conditions given, and from [Coleman], we know that

Elkiks lemma holds for D(B/A), we have

3.4 Corollary. The idealD satisfies Elkiks lemma in the complete case.

Let (A, J) be a Henselian pair. LetA be the J-adic completion ofA. LetB =B A. Let Vbe an open subscheme of Spec B which is smooth over SpecA. ThenV =V SpecA is an open subscheme of SpecB. Let Ube the open neighborhood

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of SpecA consisting of the primes ofA which do not contain J, and letU be theopen neighborhood of SpecAconsisting of the primes ofAwhich do not contain JA.Then we have the following commutative diagram:

V SpecB SpecB V s s

U i

SpecA SpecA i U3.5 Definition. We say that H satisfies Elkiks lemma in the Henselian case

ifHsatisfies Elkiks lemma in the complete case with(A, J)merely being a Henselian

pair instead of complete.

Elkik proves a quite useful theorem which states:

3.6 Theorem. (Elkik) Let f be a section ofs such that f i factors throughV.Then for any integern, there exists a section fofs such that f ifactors throughV

and such that ff mod Jn.

This immediately implies

3.7 Corollary. The idealD satisfies Elkiks lemma in the Henselian case.

Alternatively, it is possible to extend Colemans Newtons lemma to the Henselian

case more directly. This has the advantage of only requiring a homomorphism from

B toA/a2Jinstead of toA/Jn for some sufficiently largen. It has the disadvantage,

however, of only working for principal ideals with WILP (although any SILP ideal

will still work). It also requires additional hypotheses.

First, we extend Colemans lemma for SILP ideals to NILP ideals.

3.8 Lemma. LetAbe complete with respect to an idealI. Letc be any ideal ofA.

Letb be an ideal ofB having NILP. Let : BA/c2Ibe an A-homomorphism such

that ((b)) c/c2I. Then there exists an A-homomorphism : B A making the

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following commute:

B

A

A/c2I A/cI

Proof: The proof is identical to Colemans proof. We merely reproduce it here for

simplicity. By the NILP property, there exists a making the following commute:

B

A/c2I2

A/c2I A/cI

We may then proceed by induction provided that ((b)) c/c2I2. It is easy

to see that this is the case. Replaceb and c by column vectors of their generators.

Let the number of their rows be r and s respectively. Then since ((b)) (c)/(c)2I

and mod (c)I, there must exist matricesSand L with L 0 mod Isuch that

Sb= c Lc= (I L)c, where Iis ther ridentity matrix. Since I Lis invertible

inA (its inverse isI+ L + L2 + , which converges becauseL 0 mod I), we have

that (b) (c).

Thus, by induction and the completeness ofA with respect to I, there exists a

making the above diagram commute.

Our extension of Colemans Newtons lemma to the Henselian case then reads as

follows.

3.9 Proposition. Let(A, J) be a Henselian pair. Let Hbe an ideal ofB. Let abe

an ideal ofA. Let f: BA/a2Jbe anA-homomorphism such that(f(b)) a/a2J.

Suppose that HJr for somer Z, and that SpecB V(H) is smooth overA. If

1. Hhas WILP, a is principal, AnnA(a) aJ= (0), andAnnbA aJA= (0), or34


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2. Hhas NILP,

then there existsf: BA satisfyingff mod aJ.

Proof: From Colemans Newtons lemma and from our above extension to the

NILP case, we know that this holds for A complete with respect to J. We have

that aA is principal, and NILP is stable under base change [Coleman]. Then sinceAnnbA aJA= (0), we may apply Colemans Newtons lemma to B AA/(a2JA),and we get the existence off: B AAsuch thatf f1 mod aJA. ApplyingElkiks above theorem to f finishes the proof.

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4. AboutD

Chapter 4. About D

We begin with an example.

Let K be a ring, n an integer, and suppose n is invertible in K. Let B =

K[x, y]/(xn+1, xyn). ThenB is the affine line overk with the origin given multiplicity

n.

4.1 Example. D(B/K) = ((n + 1)xnyn1, (n + 1)y2n).

Proof: Let be the canonical homomorphism from K[x, y] to B, and let G =

( xn+1 xyn )tr. To computeD(B/A) we will compute the image of (B/A, G) inB .

Let J= G = (n + 1)xn 0yn nxyn1

. Then = (B/A, G) if and only ifthere exist matrices M, N Mat(K[x, y]) of the necessary sizes such that

I2+ JN+ M0 mod (G)

where MG = 0, which is to say that we need to solve this system of equations in

B. To determine M such that M G = 0, we must solve axn+1 +bxyn = 0 for a

and b. Clearly, a and b satisfy axn+1 +bxyn = 0 iffa is a multiple ofyn, and b is

the negative of the same multiple ofxn. ThusM is of the form

kyn kxn

kyn kxn

,

for k, k K[x, y]. Let N =

a b

c d

. We must solve the following system of four

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4. AboutD

equations in B:

(n + 1)xna + kyn + = 0 (1)

(n + 1)xnb kxn = 0 (2)

yna + nxyn1c + kyn = 0 (3)

ynb + nxyn1d kxn + = 0 (4)

Equation (1) holds iff

= (n + 1)xna kyn. (1a)

Factoring an xn out of (2), we see that k (n+ 1)bmod Ann(xn) = (x, yn), where

Ann(f) denotes the annihilator off in B. Thus,

k= (n + 1)b + kxx + kyyn (2a)

for somekxand kyin B, and plugging back, we see that this is necessary and sufficient.

Note that we now have the solution set for the system consisting of (1) and (2).

Factoring yn1 out of (3), we see thatya+nxc+ky0 mod Ann(yn1) = (xy).

Taking this mod (y) we see that nxcmod (y) 0. The annihilator ofxin B/(y) is

(xn), so nc mod (xn, y) 0 so

nc= cxxn + cyy (3a)

for some cx and cy in B. Plugging back into equation (3), we see that we must have

yna + ynk = 0, so k amod Ann(yn) = (x) so

k =a + kxx (3b)

for somek x. Once again, this is necessary and sufficient, and we have now solved (1),

(2), and (3) simultaneously.

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4. AboutD

Plugging all our information into (4), equation (4) becomes nxyn1d nxna

nynbkyy2n = 0. Considering this equation modulox, we find thatyn(nb+kyyn) 0

mod (x), so nb kyyn mod (x), so

b= kyyn/n + bxx (4a)

for somebx inB . Plugging back, we see that were left with nxyn1d nxna= 0, so

yn1d xn1a mod Ann(x) = (xn, yn), so yn1d 0 mod (xn1, yn), so

d= dyy+ dxxn1. (4b)

Plugging back one last time, we see that a yn1dy mod Ann(xn1) = (x2, yn), so

a= yn1dy+ axx2 + ayy

n (4c)

for some ax and ay, and lo and behold, were done.

Sorting back through the dependencies, we can eventually see that

= ((n + 1)xnyn1, (n + 1)y2n) + (G).

Coleman proves that D(B/A) is independent of the embedding ofB into affine

space over A, thus making D a canonical ideal of B associated to the structure

morphism ofB/A. Thus SpecB/Dis a canonical subscheme of SpecB corresponding

to the structure morphism of SpecB over SpecA. Coleman also proves thatD = (1)

iffB/Ais smooth. We prove here thatDlocalizes nicely. This enables us to generalize

the definition ofD to arbitrary schemes and to prove that the support of SpecB/D

is the closure of the singular points ofB over A. It also is the first step in proving

that D also pushes through smooth maps.

Let B be an A-algebra, let: AnB be surjective, with kernel generated by the

m-rowed column vector G. Let J=G.

4.2 Proposition. IffB , D(Bf/A) =BfD(B/A).

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4. AboutD

Proof: Let f An map to f B. Then: An[z] Bf, defined by sending X

to (X) and z to 1/f is surjective and has kernel (g1, . . . , gm, zf 1). Let G =

( g1 . . . gm zf 1 )tr. Let J = G.

We first show that a D(B/A) implies that a D(Bf/A). Well, ifa D(B/A)

then there exist m m and n m matrices M and N Mat(An) such that aIm

JN+ Mmod (G), and MG= 0 in An. We will show that

aIm+1 J

N 0

z2(f)N za

+

M 0

0 0

mod (G)

It is clear that this equation holds in all but the last row. As for the last row, in all

but the last column we must show that the equation

0 ( z f/X f )

N

z2f/XN

mod (G)

holds. This holds because zf 1 mod (G). In the last position of the last row,

the equation is a fza mod (G). This holds for the same reason. Thus the above

equation holds.

ClearlyM 0

0 0

G = 0 in An[z], so a D(Bf/A), so D(B/A) D(Bf/A), so

BfD(B/A) D(Bf/A).

Conversely, suppose a D(Bf/A), and let a (Bf/A, G) such that the image

of a is a. Then there exist m m and n m matrices M and N Mat(An[Z])

satisfying aIm+1 JN +Mmod (G). Consider the map from An[z] to (An)f

which sendsAn to itself and zto 1/f. Since (G) contains the kernel of this map, we

have An

[z]/(G)

=(A

n)f/(G), so we can map our relation into (A

n)f and we get

aIm+1 JN+ Mmod (G) in (An)f.

Since MG = 0 i n An[z], the product is also zero in (An)f. But in (An)f,

G =

G

0

, so ifM is the matrix consisting of all rows and columns ofM except

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4. AboutD

the last row and column, we have that MG= 0 in (An)f. LettingN be all columns

ofNexcept the last one, we get aIm J N + M mod (G) in (An)f.

There exists an s >0, a An, and N, M Mat(An), such that a

=a/fs,

N =N/fs, andM =M/f

s. Then the congruence above implies that there exists

an integer t > 0 such that fta J(f

tN) +f

tM mod (G) in An. This means

that fta = f

s+ta (B/A, G), which implies that f

s+ta D(B/A), which

implies that a BfD(B/A). Thus, D(Bf/A)/subsetBfD(B/A), which completes

the proof.

4.3 Proposition. Let f A, and suppose that the structure morphism of B/A

factors throughAf. Then D(B/Af) =D(B/A).

Proof: Let : AnB surjective,G a column vector of generators of the kernel of

, and J =G. Let m be the number of rows ofG. Then factors through (Af)n.

We have

Ancan. (Af)n

B,

with the composition being .

The kernel of must also be generated by the entries ofG since the image off

in B must be invertible.

If D(B/A) then there exists A congruent to modulo (G) such that

I GN+Mmod (G), for n m and m m matrices N and M in Mat(An)

with M G= 0. Then the image of in (Af)n also must satisfy the same equation, so

D(B/Af). Thus D(B/A) D(B/Af).

If D(B/Af), then there exists (Af)ncongruent tomodulo (G) such that

I GN+Mmod (G), fornmand mmmatricesNandMin Mat((Af)n) with

MG= 0. Multiplying by a sufficiently high power off (sayl), insures that all entries

are in the image ofAn under the canonical homomorphism. Then fl(B/A), so

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4. AboutD

the image offl is in D(B/A). Since the image off in B is invertible, we have that

D(B/A). Thus D(B/Af) D(B/A).

4.4 Corollary. Let B be an A-algebra. Let f A. Let fdenote the image off in

B. Then D(Bf/Af) = BfD(B/A).

Proof: D(Bf/Af) = Bf D(Bf/A) =Bf D(B/A).

For now, we assume that were working over the ringAso we letD(B)=D(Spec B)

= D(B/A), and we confuse ideals of rings with sheaves of ideals over the spectrum

of said rings.

4.5 Corollary. The restriction of D(B) to an open affine subscheme SpecC of

SpecB isD(C).

Proof: The open affine subscheme can be covered by open affines of the form

SpecBf, where Bf = Cf. Then D(C) SpecCf = CfD(C) = D(Cf) = D(Bf) =

BfD(B) = D(B) Spec Bf. Thus, D(C) SpecBf = D(B) Spec Bf for each open affine in

our cover, and thus D(C) = D(B) SpecC.

4.6 Corollary. The affine schemeSpecB/D(B)is precisely the closure of the set of

points ofSpecB that arent smooth overA.

Proof: Let Sing(B) be the points of SpecB which arent smooth over A. If

p SpecB is not an element of Sing(B), then there is an open affine neighborhood

SpecC ofpwhich doesnt intersect Sing(B), and therefore C is smooth over A, and

thus D(B) SpecC = D(C) = (1), because that D(B/A) = (1) iff B/A is smooth

[Coleman]. Therefore p (as an ideal of B) does not contain D(B), so p is not in

SpecB/D(B). Thus SpecB Sing(B).

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4. AboutD

Conversely, D(B) restricted to the complement of Spec B/D(B) is 1, and thus

the complement of SpecB/D(B) is smooth over A. Therefore the complement

of SpecB/D(B) is contained in the complement of Sing(B), so SpecB/D(B)

Sing(B). SpecB/D(B), being closed, then contains Sing(B).

We now understand the geometric structure of SpecB/D. Its algebraic structure

is still not well understood.

4.7 Definition. IfY /Xare schemes, letD(Y /X)be the ideal sheaf which is isomor-

phic toD(V /U) on open affinesV ofXwhich are contained in the inverse image of

open affinesU ofY. D(Y /X) is well defined by the previous results.

We will now prove that D passes through smooth maps. Let C/B/A be rings,

such that C is smooth over B. Let B =An/(G). We want to show that D(C/A) =

C D(B/A).

4.8 Lemma. IfSpecC is affine space overSpecB, then D(C/A) =C D(B/A).

Proof: LetC= B[z1,...,zl]. LetAnmap onto Bsurjectively, with kernel generatedby the entries of the m-rowed column vector G, and let J=G. ThenAn+l surjects

ontoCwith kernel also generated byG, andG/(x1,...,xn, z1,...,zl) =J = ( J 0 ).

Let (B/A), so thatImJ N+Mmod (G) for somenmand m mmatrices

N and M respectively, with MG = 0. ThenIm ( J 0 )

N

0

+ M mod (G),

so D(C/A).

Conversely, if D(C/A) then Im JN+ Mmod (G). Let L= {monomials

inz1, . . . , z l}. Then there exist ZAn, andMZand NZ in Mat(An) such that =ZL ZZ, N=

ZL NZZ, and M=

ZL MZZ. Then

ZL

ZIm JNZ MZ0 mod (G).

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4. AboutD

Since the entries ofGare in An, Js entries are also inAn. Thus we have that

ZIm JNZ MZ0 mod (G),

and MZG= 0 for each ZL. Since JN=J(Nn,m) we have that ZD(B/A) for

eachZ, so D(B/A)C.

We now achieve the same result for C etale over B.

4.9 Lemma. IfCis etale overB then D(C/A) = C D(B/A).

Proof: Let B/A[X](G), X = (x1 . . . xn), and G = (g1 . . . gm)tr. Since C is etale

over B, its locally of the form (B[y]/(h))b for some monic polynomial h such that

h is invertible once we localize at b [Milne, Thm. I.3.14]. By previous propositions

we know that D localizes, so it suffices to prove the lemma in the case of C =

(B[y]/(h))h/y =B [y, z]/(h,zh/y 1), for some monic polynomial h B [y]. Let

h denote the derivative ofh with respect to y . Let H=

h

hz 1

. Let Y = (y, z).

Then D(C/A) iff there exists a A[X , y, z ] congruent to mod (G) + (H),

and m m, m2, 2 m and 2 2 matrices M1, M2, M3, M4 respectively, with

M1G + M2H= 0, andM3G + M4H= 0, andn m, n 2, 2 mand 2 2 matrices

N1, N2, N3, N4 respectively, satisfying:

I GX 0

HX

HY

N1 N2N3 N4

+M1 M2

M3 M4

mod (G) + (H)

The element satisfies the above equation iff it satisfies the six sets of equations

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4. AboutD

following:

I G

XN1+ M1 mod (G) + (H) (1)

0 G

XN2+ M2 mod (G) + (H) (2)

0H

XN1+

H

YN3+ M3 mod (G) + (H) (3)

IH

XN2+

H

YN4+ M4 mod (G) + (H) (4)

0 =M1G + M2H (5)

0 =M3G + M4H (6)

The matrixN4 only occurs in equation (4), and HY is invertible mod (G) + (H)

because its determinant is invertible (det( HY) = h2, and zh 1 mod (G) + (H)).

Thus equation (4) is superfluous becuase we may take N4 to be any matrix whose

reduction mod (G) + (H) is (I HXN2 M4)HY

1, and then (C/A, (G, H)).

Similarly, equation (3) is also superfluous. Thus (C/A, (G, H)) iff there exist

matrices such that equations (1), (2), (5) and (6) are satisfied.Reducing equation (5) mod (G), we see thatM2H0 mod (G). Differentiating

with respect to Y yields that M2y H+M2Hy 0 mod (G), and

M2z H+M2

Hy

0 mod (G), soM2HY 0 mod (G) + (H). Since

HY is invertible mod (G) + (H), we

have that M2 0 mod (G) + (H). This implies that that equations (1) and (2) are

equivalent to

I G

XN1+ M1 mod (G) + (H) (1

)

0 G

XN2 mod (G) + (H) (2

)

Neither of these equations involves M3 or M4, so we may drop equation (6).

Equation (2) is the only one involving N2, and is satisfied by N2 = 0, so we may

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4. AboutD

drop equation (2), and were left with only equations (1) and (5) being necessary

and sufficient. Equation (5) is the only equation left involving M2, and the equation

itself is equivalent to M1G 0 mod (H), so were left with the system:

I G

XN1+ M1 mod (G) + (H) (1

)

M1G 0 mod (H) (2)

Adjoiningzand reducing modulohz 1 is the same as localizing at h, so solving

(1) and (2) over A[X , y, z ] is the same as solving the following in A[X, y]h.

I G

XN1+ M1 mod (G) + (h) (1

)

M1G 0 mod (h) (2)

Ifsatisfies (1) and (2) for some N1 andM1, then we can clear denominators,

so there exists an integer l, a A[X, y], and matrices N and M in Mat(A[X, y])

such that /hl

=, and satisfies:

I G

XN + M mod (G) + (h) (1)

MG 0 mod (h) (2)

Clearly, the image in C of the set of A[X, y] satisfying (1) and (2)

generatesD(C/A). In particular, this implies that the image ofD(B/A) inC lies in

D(C/A). We now show that this image generatesD(C/A).

Since the equations are taken modulo the monic polynomial h, and GX and G

do not involvey, we may assume that , and entries ofN, andM are polynomials

of degree less than d = deg(h). Then if =d1

i=0 iyi, N =

d1i=0 Niy

i, and

M =d1

i=0 Miyi, where the i, and the entries of theNi and theMi are elements of

A[X], we see that there exist matrices such that satisfies (1) and (2), iff there

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4. AboutD

exist matrices such that the i satisfy:

iI G

XNi+ Mi mod (G) (A)

MiG= 0 (B)

This means that the images of the i lie in the image ofD(B/A), so D(B/A)

generatesD(C/A).

4.10 Theorem. IfC is aB-algebra, andB is an A-algebra, andC is smooth over

B, then D(C/A) =D(B/A)C.

Proof: We know that this holds ifCis affine over B , or ifCis etale over B . The

result follows from the fact that if: B C is smooth then locally there exists an

integerl such that factors throughB [z1, . . . , z l], andCis etale over B [z1, . . . , z l].

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5. Examples

Chapter 5. Examples

We compute D, Ds, andDw for a variety of rings.

5.1. Computational aids

Let A be a noetherian ring, and Ibe an ideal, letA be the I-adic completion ofA,and let Sbe the set of elements ofAwhich are congruent to 1 modulo I.

5.1.1 Lemma. LetJbe an ideal ofA. TheI-adic closure ofJinAis{g A | sg Jfor somes S}.

Proof: We have the homomorphisms A S1A

A, the second of which is

faithfully flat, so all the ideals ofS1Aare closed. So ifJA is an ideal ofA, then

its closure in the I-adic topology in A, J is n(J+In) = (JA) A = (S1J) A.Thus,J ={g A | sg J(for some s S)}.

5.1.2 Proposition. IfB is a complete intersection then D(B/A) = Ds(B/A). IfB

is an integral domain then D(B/A) =Dw(B/A).

Proof: Let B be a complete intersection over A. Then there exist n and G such

thatB =An/(G), andM G= 0 implies thatM0 mod (G). LetJ= G. Then

D(B/A) iffI+MJ Nmod (G) iffIJ Nmod (G) iff(I+M) J Nmod (G).

Similarly, Dw(B/A) iff(I+ JN+ M) 0 iff (I+ JN+ M) 0 or 0

iff D.

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5. Examples

5.1.3 Proposition. Let B =A[X]/G, with G= (g1 . . . gr)tr. Let J= G. If

1. ({m1, . . . , mr |(m1 . . . mr)G= 0})B = (1),

2. there exists a vectorv = 0 with entries in B such that vJ = 0, whereJdenotes

the image ofJ in B, and

3. B is a noetherian integral domain,

then Ds(B/A) = 0.

Proof: Leta= ({m1, . . . , mr |(m1 . . . mr)G= 0}). Letdenote reduction modulo

G. Let Ds(B/A). Then there exist MandNsuch thatM G= 0 and(I M) =

JN. Item 2 implies that v(I M) = 0. Let d Z+. Multiplying on the right by

I+ M+ + Md1, we see that v(I Md) = 0, so the entries ofvare in ad for

all d. Since a= (1) and B is an integral domain, dad = 0, so v= 0. Since v = 0,

this means that = 0.

5.2. Demonstrative examples

5.2.1 Example. The idealsD andDs

need not be equal.

Proof: We show that Ds(B/K) = 0 for B = K[x, y]/(xiyj , xkyl), where i,j,l

1, k 0, i > k, j < l, Kis a field, and i, j, k, and l are nonzero in Kwhen they are

nonzero integers. Since we already have that D(B/K) = ((l + 1)ylxl1, (l + 1)x2l) for

(i,j,k,l) = (1, l, 0, l+ 1) (from the exercise beginning chapter 4), this demonstrates

the supposition.

We have that Ds iff there exist a,b,c,d,m,n B such that

00

= ixi1yj jxiyj1kxk1yl lxkyl1

a cb d

+ my lj mxikny lj nxik

= 0Let M =

mylj mxik

nylj nxik

, N =

a c

b d

, G = (xiyj , xkyl)tr, J = G,

and I be the 2 2 identity matrix. Then (I M) = JN. Multiplying on the

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5. Examples

left by I+M + + Ms1 yields (IMs) = JN for all s 1. Thus, I

JNmod (ylj , xik)s for all s, so I JN mod s (ylj , xik)s for all s. Since

s(ylj , xik)s = s(x, y)

s we must have that I JN mod s(x, y)s. The ideal

s(x, y)s inB is the image ofs(x, y)

s + (G) = (G), the (x, y)- adic closure of (G) in

K[x, y]. LetS={1+g|g (x, y)}. Then by the lemma, (G) = {g|sg (G), s S}.

Ifs Sandsg (G) thensg =pxiyj +qxkyl =xkyj(pxik +qy lj) for somepand q.

Sinces 1 mod (x, y), neitherxnory dividess, so xkyj must divideg, so there exists

ag such thatg = xkylg. Thus,sg =pxik +qy lj . The ideal (xik, ylj) is primary

inK[x, y] becuase if(xik

, ylj

), then eitherormust have no constant term.Suppose its. thent (xik, ylj) for some positive integer t. Thus, since no power

ofs can be in (xik, ylj), we must have that g (xik, ylj), and thus g (G).

Thus (G) = G, so s(x, y)s = 0 in B. Therefore I JN mod s(x, y)

s implies

that I J N. ThusDs if and only if there exist polynomialsa, b, c, dsatisfying:

= aixi1yj + bjxiyj1 (1)

0 =akxk1yl + blxkyl1 (2)

= ckxk1yl + dlxkyl1 (3)

0 =cixi1yj + djxiyj1 (4)

First assume thatk is nonzero inK. Equating equations (1) and (3), we see that

we must have thataixi1yj+bjxiyj1 =ckxk1yl+dlxkyl1, so yj1xk1(aixikyj+

bjxik+1 ckxlj+1yl dlxylj) = 0 , s o aixikyj+ bjxik+1 ckxlj+1yl+

dlxylj mod Ann(xj1yk1). The annihilator of (xj1yk1) is (xylj+1, xik+1y),

so considering the last equation modulo y, we see that bjxik+1 = 0, which means

that bj 0(y), so there exists a by such that b = byy. Plugging this into equation

(2) we see that akxk1yl = 0, so ak = 0 mod Ann(xk1yl). This annihilator is (x),

so there exists an ax such that a = axx. Plugging the equations for a and b into

equation (1) yields that = 0.

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5. Examples

Ifk= 0, our equations become:

= aixi1yj + bjxiyj1 (1)

0 =bly l1 (2)

= dlyl1 (3)

0 =cixi1yj + djxiyj1 (4)

Equation (4) implies thatciy + djx 0 mod Ann(xi1yj1). This annihilator is

(xy,ylj+1), so considering the equation modulo (y) yelds thatdjx 0(y), sod = dyy

for some dy. Plugging this into equation (3) yields that = 0.

It is clear from the equations for D and Dw that Dw D {2 | Dw}.

The following example demonstrates that we cannot replace this containment with

equality.

5.2.2 Example. The setsD andDw need not be equal, Dw need not be an ideal,

andD need not equal{2 | Dw}. Not even the set of squares ofD need equal

{2 | Dw}.

Proof: We compute D(B/A) and Dw(B/A) for B =A[x]/(x3).

Since B is defined by one equation over A, and this equation has trivial annihi-

lator, D(B/A) = (3x2).

As for Dw(B/A), Dw if and only if there exists an n B satisfying

(+ 3x2n) 0 mod (x3).

Let n = n0+ n1x + n2x2, and let =0+ 1x + 2x

2. Then2 =20+ 201x + (21+

202)x2, and 3x2n= 30n0x

2. Thus Dw if and only if there exists n0A such

that

20+ 201x + (21+ 202)x

2 + 30n0x2 0 mod (x3).

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5. Examples

This is equivalent to

20 = 0, 201= 0, and 21+ 202+ 30n0= 0.

Thus,

Dw ={0+ 1x + 2x2 |20 = 0, 201 = 0, and n0 such that

21+ 202= 30n0}.

This demonstrates that Dw need not be an ideal, since if A is taken to be

k[u, v]/(u2, v2), we see that u Dw, and vx Dw, but u+ vx Dw (if 2 = 0

ink). This also demonstrates that D need not equal D

w

, since D is always an ideal.To show that D need not be {2 |Dw} we let A= Q. Then 3x2 D, but it

doesnt have a square root in B , let alone a square root lying in Dw.

Furthermore, the squares ofD need not be the squares of elements ofDw, as is

illustrated by taking A = k[u, v]/(u2, uv), where k is any ring in which 3 is invertible

and 2 = 0. Let = u +vx +x2. Then 2 (3x2). However there exists no n0 A

satisfying 21+ 202 = 30n0, i.e. there exists no n0 A satisfying v2 + 2u= 3un0,

so Dw.

Let Dn denote the set of elements ofB generating principal ideals having NILP.

5.2.3 Example. The setsDs andDn need not be equal.

Proof: Let B =A[x]/(x4). Then Ds =D = (4x3), but x2 Dn, as is evident by

glancing at the equations for NILP.

5.2.4 Example. A nonprincipalD , and a maximal nonprincipal SILP ideal.

Proof: Take B =k[x, y]/(x2y2). Then D= Ds = (2x, 2y).

5.2.5 Example. The ideal D need not be a maximal WILP ideal, and maximal

ideals contained inDw need not have WILP.

Proof: Let B = k[x, y]/(x2, y2). Suppose that k is a field in which 2 = 0. Then

D = (xy) (from the example at the beginning of the chapter about D), but we will

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5. Examples

show thatDw = (x, y), so (x) and (y) both have WILP. We will also show that (x, y)

does not have WILP.

Writing down the equations for () having WILP, and letting N =

n1 n2

n3 n4

,

we have that () has WILP if and only if there exist n1, . . . , n4B satisfying:

2 + 2xn1= 0 (1)

2xn2= 0

2yn3= 0

2 + 2yn4= 0 (2)

We may take n2 = n3 = 0. Let = 0 + 1x+ 2y+3xy. Then 2 = 20 +

201x + 202y + (212+ 203)xy. Letn1=n0+ n

1x + n

2y + n

3xy. Then equation

(1) becomes:

20+ 201x + 202y+ (212+ 203)xy

+ 2x0n0+ 2xy0n

2+ 2xy2n

0= 0 (1

)

Since every expression in equation (1) except for the leading 20 is multiplied by

either an x or a y, we must have that 0 = 0, and then (1) has solution for all

1, 2 and 3 by letting n0=1. By symmetry, the same is true of equation (2), so

Dw = (x, y).

To see that Dw does not have WILP, we write down the equations that xand y

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5. Examples

would have to satisfy. There would have to exist matrices Nx and Ny such that:

x(xI+ 2x 0

0 2y

Nx) = 0

x(yI+

2x 0

0 2y

Ny) = 0

y(xI+

2x 0

0 2y

Nx) = 0

y(yI+

2x 0

0 2yNy) = 0

where Iis the 2 2 identity matrix.

However, the second equation above has no solutions, sincex annihilates the first

row of

2x 0

0 2y

. Thus (x, y) does not have WILP.

5.2.6 Example. A maximal nonprincipal WILP ideal.

Proof: Let B = k[x, y]/(xy(y 1)), where k is a field in which 2 = 0. We show

that Dw(B/k) = (y(y 1), x(y 1)) (y(y 1), x(2y 1)) (y(y 1), xy), and that

each of the ideals composing this union has WILP.

The element has WILP if and only if there exist n1, n2 B satisfying:

+ ( y(y 1) x(2y 1) )

n1

n2

= 0

Multiplying out, and utilizing the fact that were working modulo xy(y 1), we

see that the above equation is equivalent to the existence of polynomials n1, n2, n3

k[x, y] satisfying:

(+ n1y(y 1) + n2x(2y 1)) =n3xy(y 1) (1)

To solve this equation, we analyze it modulo various ideals, breaking it into cases.

Consider the equation modulo (x). We get

(+ n1y(y 1)) 0 mod (x)

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5. Examples

Thus either 0 mod (x) (case A), or n1y(y 1) mod (x) (case B).

Case A: 0 mod (x).

Then = xx for some x. Plugging back into equation (1) and cancelling xs,

we find that

x(xx+ n1y(y 1) + n2x(2y 1)) =n3xy(y 1). (A)

Considering equation (A) modulo (y), we see that

x(xx n2x) 0 mod (y),

so either x 0 mod (y) (case A1), or xn2 mod (y) (case A2).

Case A1: x 0 mod (y).

Then x = yxy, for some xy. Plugging back into equation (A) and cancelling

xs, we see that

xy(xyxy+ n1y(y 1) + n2x(2y 1)) =n3(y 1). (A1)

Reducing modulo (y 1), we conclude that

xy(xxy+ n2x) 0 mod (y 1).

This means that either xy 0 mod (y 1), or xy n2 mod (y 1). The

former implies that = 0 in B. The latter implies that xy =n2+xyz(y 1) for

some xyz.

Plugging back into equation (A1) and cancelling (y 1)s, we see that

(n2+ (y 1)xyz)(xn2+ xyxyz+ n1) =n3.

This equation has solution for all xyzand for all n2, so we see that in this case,

= xx = xyxy =xy(n2+ xyz(y 1)), so () = (xy).

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5. Examples

Case A2: xx n2x 0 mod (y).

In this case, we let x = n2+ yxy. Plugging into equation (A), we see that

(n2+ yxy)(xxy+ n1(y 1) + 2n2x) =n3(y 1). (A2)

Considering this equation modulo (y1), we see that eithern2+xy 0 mod (y

1) or xy+ 2n2 0 mod (y 1). Plugging back into equation (A2), it is clear that

the former case leads to (x(y 1)), and the latter case leads to (x(1 2y)).

Thus in case A (xy) (x(y 1)) (x(2y 1)).

Case B: n1y(y 1) mod (x).

In case B let =n1y(y 1) +xx for some x. Plugging into equation (1) and

cancelling xs we get that

(n1y(y 1) + xx)(x+ n2(2y 1)) =n3y(y 1). (B)

Considering equation (B) modulo (y), we see that either x 0 mod (y), or

x n2 mod (y), or equivalently, there exists a xy such that either x =yxy (case

B1) or x =n2+ yxy (case B2).

Case B1: x =yxy.

Plugging this back into equation (B), cancellingys, and reducing modulo (y 1),

we see that either (y(y 1)), or (y(y 1), xy).

Case B2: x =n2+ yxy.

Plugging into equation (B), cancelling ys and reducing modulo (y 1), we see

that either (y(y 1), x(y 1)) or (y(y 1), x(2y 1)).

Summing up all cases, we see that

Dw = (y(y 1), x(y 1)) (y(y 1), xy) (y(y 1), x(2y 1)).

We will now show that each of these ideals has WILP.

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5. Examples

The ideal (1, 2) has WILP if and only if there exist n1, n2, n3, n4B satisfying:

1(2+ n1y(y 1) + n2x(2y 1)) = 0 (1)

1(1+ n3y(y 1) + n4x(2y 1)) = 0 (2)

2(1+ n3y(y 1) + n4x(2y 1)) = 0 (3)

2(2+ n1y(y 1) + n2x(2y 1)) = 0 (4)

In each of our cases, 12 = 0, so we need only solve the equations (over B):

1(n1y(y 1) + n2x(2y 1)) = 0 (1

)

1(1+ n3y(y 1) + n4x(2y 1)) = 0 (2)

2(n3y(y 1) + n4x(2y 1)) = 0 (3)

2(2+ n1y(y 1) + n2x(2y 1)) = 0 (4)

Since each of the ideals is generated byy(y 1) and one other element, we may

take 1 =y(y 1). Plugging in, and utilizing the fact that xy(y 1) = 0, we arrive

at the equivalent equations:

n1y2(y 1)2 = 0 (1)

y2(y 1)2 + n3y2(y 1)2 = 0 (2)

2(n3y(y 1) + n4x(2y 1)) = 0 (3)

2(2+ n1y(y 1) + n2x(2y 1)) = 0 (4)

Equations (1) and (2) are satisfied if and only ifn1 =xn1 and n3 =1 +xn

3

for some n

1 and n

3 in B. Plugging this into equations (3

) and (4

) yield that(y(y 1), 2) has WILP if and only if there exist n

1, n2, n

3, and n4 satisfying:

2(y(y 1) + n4x(2y 1)) = 0 (3)

2(2+ n2x(2y 1)) = 0 (4)

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5. Examples

Sincex(y 1),xy, andx(2y 1) each annihilatey(y 1), equation (3) is satisfied

by takingn4= 0. Thus (y(y 1), 2) has WILP if and only if there exists an n2B

satisfying:

2(2+ n2x(2y 1)) = 0 ()

Then2=x(y1) satisfies () by settingn2 = 1. For2=xy, () is satisfied by

taking n2 = 1. Finally, when2 = x(2y 1), we may solve () by setting n2=1.

Thus, each of the three ideals composing Dw have WILP.

5.3. Easy examples

5.3.1 Example. The twisted cubic along the degenerate elliptic curve.

Let B = k[x,y,z]/((y z2)2 (x z3)3) (i.e. At each z value, we have the

degenerate elliptic curve centered at (z3, z2)). Let g = (y z2)2 (x z3)3. Then

Dw =D(B/k) =Ds(B/k) = (g)

= (3(x z3)2, 2(y z2), 4z(y z2) + 9z2(x z3)2)

= (3(x z3)2, 2(y z2))

.

The ideal Dw then has SILP, and therefore WILP.

5.3.2 Example. A bound on Ds for some single point schemes.

LetB =k[X]/(G) be a one point scheme, wherek is an algebraically closed field,

X = x1, . . . , xn, and G = (g1 gm)tr. Then (G) contains a power of a maximal

ideal, which we may assume is (X). Thus there exist integers ei, such that xeii

(G), and xei1i (G). Suppose that gi = xeii for i = 1, . . . , n, and that x

eii

(g1, . . . , gi1, gi+1, . . . , gm)k[X](X), that is, xeii not in the ideal generated by the

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5. Examples

remaing equations in k[X] localized at the maximal ideal (x1, . . . , xn). Then

Ds(B/k)n

i=1

((eixei1i ) + (G))

Proof: Let G = (g1 gn)tr. Let H = (gn+1 . . . gm)

tr. Then Ds iff there

exists a k[X], and matricesM= (mij) andN= (nij), such that mod (G),

MG= 0, and:

(I M)

GXHX

Nmod (G).

This implies that for i= 1, . . . , n, (1 mii) eixei1nii mod (G).

We are given that

jmijgj = 0, so by the condition that xeii (g1, . . . , gi1,

gi+1,. . . ,gm)k[X](X), we know thatmiiis not invertible in k[X](X), and hence is not

invertible in B. Thus 1 mii is invertible in B, so (eix

ei 1) + (G).

5.3.3 Example. IfB is a hypersurface over A, B = A[X]/(g), and ifg is monic,

then B/A is a complete intersection, so by the proposition at the beginning of this

chapter, D(B/A) =Ds(B/A) = ( gX).

5.3.4 Example. The ideal D(B/A) = Ds(B/A) = (ixi1yj , jxiyj1) for B =

A[x, y]/(xiyj) (by the same proposition).

5.3.5 Example. The ideal D(B/A) = (xi1yj1) for B =A[x, y]/(xi, yj), ifi andj

are invertible in A, and D(B/A) = Ds(B/A).

Proof: The column vector [xi, yj ]tr has no annihilators not equal to zero in B , so

D= Ds, and D if and only if there exists a matrix Nsuch that satisfies:

00

= ixi1 0

0 jyj1

N .

Thus D= (xi1) (yj1) = (xi1yj1).

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5. Examples

5.4. A hard example

Let a < b < c Z+. Let B be the subring of Q[z] generated by za, zb, and zc

(i.e. Q[za, zb, zc]). Assume the nonredundancy condition thatzb Q[za], and that

zc Q[za, zb]. Assume that a = 3. We will show that in this case D(B/Q) =

(z2b, zb+c, z2c) and that Ds(B/Q) = 0. Note that this nonredundancy condition

implies that Z/3Z= {0,bmod 3,cmod 3}. It also implies that 3 < b < c 2b 3

and c 2bmod (3).

5.4.1. Equations forBWe need to first embed B in affine space over A.

5.4.1.1 Lemma. The ringQ[u,v,w]/(v2 u(2bc)/3w, u(b+c)/3 vw,w2 u(2cb)/3v)

= B byuz3, vzb,wzc.

Proof: Before we begin, we should show that the above expressions are polyno-

mials, by demonstrating that (2b c)/3, (b+c)/3, and (2c b)/3 are non-negative

integers. Since b and c are the nonzero elements ofZ/3Z, we have that 2b c mod (3),

2c bmod (3) and b cmod (3), so 2b c, b+c, and 2c b are all congruentto 0 modulo (3). Thus (2b c)/3, (b+c)/3, and (2c b)/3 are integers. They are

positive because 3< b < c 2b 3.

We mapQ[u,v,w] toB byu z3, v zb,w zc. Then the above three equa-

tions map to zero. To see that they must generate the kernel, consider a polynomial

f which maps to 0. After subtracting off multiples of the first and third equations,

we can assume that f is linear in v and w, and after subtracting off multiples of the

second equation, we may assume that f contains no monomials divisible by vw, so

f=fu+ fvv + fww, wherefu, fv, fw Q[u]. Thenfu,fv, andfw all map to polyno-

mials in z3. Sincev and w map tozb and zc respectively, and by the nonredundancy

condition, b c mod (3), and neither is congruent to 0, the monomials in the images

offu, fvv, andfww are of different degrees, so the image can only be zero iffu, fvv,

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5. Examples

and fww all map to 0. This can only occur if fu = fv = fw = 0. Thus, we may

conclude that the kernel is generated by the claimed equations.

Let g1 = v2 u(2bc)/3w, g2 = u

(b+c)/3 vw, g3 = w2 u(2cb)/3v, and G =

(g1, g2, g3)tr. To compute D we still need to know the annihilator ofG.

5.4.1.2 Lemma. The annihilator ofG is generated by the vectors(w, v, u(2bc)/3)

and(u(2cb)/3,w,v).

Proof: It is easy to verify that (w , v, u(2bc)/3) and (u(2cb)/3, w , v) annihilate (v2

u(2bc)/3w, u(b+c)/3 vw,w2 u(2cb)/3v)tr. Conversely, suppose (f , g , h) annihilates

the column vector G. By subtracting off multiples of the two annihilators of the

lemma, we can reduce (f , g , h) to the point where g is a polynomial in u. But then,

g(u(b+c)/3 vw) contains monomials purely in u, where as f and h times the other

two polynomials do not. Thus, g must now be zero. Then (f, h) annihilates (g1, g3)tr.

Sinceg1and g3are relatively prime, (f, h) must be a multiple of (g3, g1). (g3, 0, g1) =

w(w , v, u(2bc)/3

)v(u(2cb)/3

, w , v), so (w , v, u(2bc)/3

) and (u(2cb)/3

, w , v) do indeedgenerate the annihilator ofG.

We can now begin computingD. We start by determining the system of equations

that we must solve.

5.4.1.3 Lemma. Let B. Then D if and only if there exist ni B for

1 i 9 andaj , bj B for1 j 3 satisfying the system of equations in figure 1.

Proof: This is the simple matter of writing down the equations for D. D if

and only if there exists a 3 3 matrixN=

n1 n4 n7n2 n5 n8n3 n6 n9

with entries inQ[u,v,w],60


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5. Examples

1

2bc

3 z2

b3 2zb

z2

bc 0 0 0 0

0 0 0 0 2bc3 z2

b3 2zb z2bc 0

0 0 0 0 0 0 0 2bc3 z2

b3

0 b+c3 z

b+c3 zc zb 0 0 0 0

1 0 0 0 b+c3 z

b+c3 zc zb 0

0 0 0 0 0 0 0 b+c3 z

b+c3

0 b2c3 z2

c3 z2cb 2zc 0 0 0 0

0 0 0 0 b2c3 z2

c3 z2cb 2zc 0

1 0 0 0 0 0 0 b2c3 z2

c3

0 0 zc 0 0 z2cb 0 0

0 0 zb 0 0 zc 0 0

2zb z2bc z2bc 0 0 zb 0 0

0 0 0 zc 0 0 z2cb 0

0 0 0 zb 0 0 zc 0

zc zb 0 z2bc 0 0 zb 0

0 0 0 0 zc 0 0 z2cb

0 0 0 0 zb 0 0 zc

z2cb 2zc 0 0 z2bc 0 0 zb

n1

n2

n3

n4

n5

n6

n7

n8

n9

a1

a2

a3

b1

b2

b3

=

0

0

00

0

0

0

0

0

0

0

0

0

0

0

0

Figure 1. The element D if and only if this system has solution.

and a 3 3 matrixMwith entries in Q[u,v,w] satisfyingM G= 0 such that:

1 0 00 1 00 0 1

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5. Examples

Computing the jacobian matrix ofG, we see that

G =

2bc3 u

2bc3 1w 2v u

2bc3

b+c3 u

b+c3 1 w v

b2c3 u

2cb3 1v u

2cb3 2w

.By the previous lemma, we know that each row ofMmust be a linear combination

of the vectors (w , v, u(2bc)/3) and (u(2cb)/3, w , v), so Mhas the form

a1a2a3

( w v u(2bc)/3 ) + b1b2b3

( u(2cb)/3 w v )for some a1, a2, a3, b1, b2, b3Q[u,v,w].

Solving these equations in Q[u,v,w]/(G) is the same as solving their image in

Q[z3, zb, zc]. Their image is:

1 0 0

0 1 0

0 0 1+

2bc3 z2b3 2zb z2bc

b+c

3

zb+c3 zc zb

b2c3 z

2c3 z2cb 2zc

n1 n4 n7

n2 n5 n8

n3 n6 n9

+

a1a2a3

( zc zb z2bc ) + b1b2

b3

( z2cb zc zb )Rewriting these equations in the matrix form of a system of linear equations for

, n1, . . . n9, a1, a2, a3, b1, b2, b3 yields the equations cited in the lemma.

Let A denote the big matrix in the above lemma. Note that

singular morphisms, smoothness, and lifting lemmas

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