singular morphisms, smoothness, and lifting lemmas
TRANSCRIPT
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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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1. Introduction and Setting
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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1. Introduction and Setting
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)
for matricies N and M with MG= 0}
= (B/A, G) = {A[X]| I+ M+ GN0 mod (G),
for matricies N and M with MG= 0}
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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1. Introduction and Setting
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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2. Determining ideals having ILPs
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
following diagram commute:
A[X] B
C C/(d)I
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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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
exists: BCmaking the following diagram commute:
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. Determining ideals having ILPs
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
following diagram commute:
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)
whereIis them m identity matrix.
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
there exist n m matrices Ni (with entries in A[X, Y]) such that for all
1 j r
dj(G + Jr
i=1
NiYi) 0 mod (Y)2 + (Y d G). (16)
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2. Determining ideals having ILPs
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
there exist n m matrices 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
Ej 0 mod (Y)2 such that
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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2. Determining ideals having ILPs
Ej 0 mod (Y)2 such that
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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2. Determining ideals having ILPs
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
exists: BCmaking the following diagram commute:
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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2. Determining ideals having ILPs
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. Determining ideals having ILPs
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
there exist n mmatricesNi (with entries in A[X , Y , Z]) such that
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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2. Determining ideals having ILPs
there exist n mmatricesNi (with entries in A[X , Y , Z]) such that
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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
there exist n mmatricesNijk (with entries in A[X, {Yi}, Z]) such that
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
there exist n mmatricesNijk (with entries in A[X, {Yi}, Z]) such that
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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)
whereIis them m identity matrix.
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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2. Determining ideals having ILPs
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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2. Determining ideals having ILPs
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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3. Some Lifting Lemmas
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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3. Some Lifting Lemmas
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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3. Some Lifting Lemmas
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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5.1.3 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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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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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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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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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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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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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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5.4. 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
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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:
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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