Table Of ContentThe Zariski-Lipman conjecture for complete
intersections
1
1 Rolf K¨allstro¨m
0
2
January 27, 2011
n
a
J
5
2 Abstract
] The tangential branch locus BXt/Y ⊂BX/Y is the subset of points in
G thebranchlocuswherethesheafofrelativevectorfieldsTX/Y failstobe
locally free. It was conjectured by Zariski and Lipman that if V/k is a
A
varietyoverafieldkofcharacteristic0andBt =∅,thenV/kissmooth
V/k
. (=regular). We prove this conjecture when V/k is a locally complete
h
t intersection. We prove also that BVt/k = ∅ implies codimXBV/k ≤ 1 in
a positive characteristic, if V/k is the fibre of a flat morphism satisfying
m
generic smoothness.
[
5 1 Introduction
v
1
Let π : X → Y be a morphism of nœtherian schemes which is locally of finite
4
2 type, ΩX/Y its sheaf of K¨ahler differentials, and TX/Y = HomOX(ΩX/Y,OX)
4 thesheafofrelativetangentvectorfields. Wehavetheinclusionofthetangential
3. branch locus in the branch locus
0
0 Bt ={x∈X|T is not free}⊂B =B ={x∈X|Ω is not free}.
π X/Y,x π X/Y X/Y,x
1
: Define as in [5, Definitions 17.1.1 and 17.3.1] a morphism π to be formally
v
i smooth at a point x in X if the induced map of local rings OY,π(x) → OX,x is
X
formallysmooth, andthatπ issmoothatxifitislocallyfinitelypresentedand
r formally smooth; say also that π is smooth if it is smooth at all points in X.
a
In the light of the Jacobian criterion, namely that B = ∅, goes a long way to
π
implying that the morphism π is smooth (Thms. 3.1 and 3.3), it is a natural to
ask, with Zariski and Lipman [12], what are the implications of Bt = ∅? The
π
example X = SpecA[x]/(x2) → Y = SpecA, i.e. the scheme of dual numbers
overacommutativeringA,showsthatifwewantπ tobesmooth,thecondition
Bt = ∅ needs at least to be supplemented with the condition that the rank of
π
T equalstherelativedimensionateachpointinX,whichcanbeimposedby
X/Y
12000 Mathematics Subject Classification: Primary: 14A10, 32C38; Secondary: 17B99
(Secondary)
1
assumingthatX/Y issmoothatgenericpointsinX. Itisaremarkablefactthat
although T cannot even directly detect torsion in Ω , it turns out that
X/Y X/Y
these conditions combined imply B =∅ (and hence imply that π is smooth) in
π
interesting cases in characteristic 0. Already the result that Bt = 0 implies
V/k
smoothness when V/k is a curve over a field of characteristic 0, due to Lipman
[loc. cit], is, I think, quite surprising and non-trivial (see Proposition 4.4).
In positive characteristic it is easy to see that smoothness at points of height
≤ 1 does not follow from Bt = ∅, so one could perhaps add the assumption
π
codim B ≥ 2; but this is still not enough. What is needed is a condition on
X π
the discriminant locus D = π(B ). Before the main results are presented we
π π
describe some terminology.
Generalities: All schemes are assumed to be nœtherian and we use the no-
tation in EGA, but see also [13, §5] and [8]. The height ht (x) of a point x
X
in X is the same as the Krull dimension of the local ring O at x, and the
X,x
dimension of X is defined as dimX = sup{ ht(x) | x ∈ X}. The dimension at
a point x in X is
dim X =sup{ ht(x ) | x ∈X and x specialises to x };
x 1 1 1
see [4, Proposition 5.1.4]. A point x in a subset T of X is maximal if for each
pointyinT thatbelongstotheclosure{x}−of{x}(inotherwords,xspecialises
to y (see [8, p. 93])), we have ht(x)≤ht(y). That is, if x(cid:48) ∈T specialises to x,
and ht(x(cid:48))≤ht(x), then x(cid:48) =x. Denote by Max(T) the set of maximal points
of T, so Max(X) consists of points of height 0. A property on X is generic if it
holds for all points in Max(X). Put
codim+ T = sup{ ht(x) | x∈Max(T)},
X
codim− T = inf{ ht(x) | x∈Max(T)},
X
socodim− T ≤ht(x)≤codim+ T whenx∈Max(T). IfT istheemptyset, put
X X
codim+ T =−1andcodim− T =∞,sinceweareinterestedinlowerandhigher
X X
bounds on codim± T, respectively. For a coherent O -module M, the stalk at
X X
a point x is denoted M and we put depth M = inf{depthM | x ∈ T}. The
x T x
fibre X over a point y in Y is the fibre product Speck × Y, where k is
y Y,y Y Y,y
the residue field at y. We define the relative dimension d of π at a point
X/Y,x
x∈X astheinfimumofthedimensionofthevectorspaceofK¨ahlerdifferentials
at all maximal points ξ that specialise to x, i.e.
d =inf{dim k ⊗ Ω | x∈{ξ}−,ξ ∈Max(X)}.
X/Y,x kX,ξ X,ξ OX,ξ X/Y,ξ
To understand this number it is useful recall that
dim k ⊗ Ω =dim Ω
kX,ξ X,ξ OX,ξ X/Y,ξ kXπ(ξ),ξ Xπ(ξ)/kY,π(ξ)
=dim k ⊗ Ω ;
kX,ξ X,ξ OX,ξ OX,ξ/OY,π(ξ)
see Proposition 2.1 for the first equality, but note that in general the numbers
d and dim X are not equal. On the other hand, if π is flat at x, then
X/Y,x x π(x)
2
dim X =dimO −dimO , and if morover π is smooth at all points
x π(x) X,x Y,π(x)
ξ ∈Max(X) that specialise to x, then d =dim X .
X/Y,x x π(x)
Recall also (this is an easy extension of [8, Chapter II, Lemma 8.9 ]):
(∗) a coherent O -module M is free at a point x if M is free of rank equal to
X ξ
dim k ⊗ M for each ξ ∈Max(X) that specialises to x.
kX,x X,x OX,x x
Theorem 1.1. Let X → S and Y → S be morphisms of nœtherian schemes
which are of locally of finite type, and π : X/S → Y/S be a flat S-morphism.
AssumethatthebranchlociB =B =∅(e.g. X/S andY/S aresmooth),
X/S Y/S
X is Cohen-Macaulay, and codim−D ≥ 1. If Bt = ∅, then for each point
Y π π
y ∈Y
codim+ B ≤1.
Xy Xy/kY,y
Remark 1.2. The condition of generic smoothness, i.e. codim−D ≥ 1, is
Y X/Y
satisfiedwhenO isregularandtheextensionofresiduefieldsk /k is
X,x X,x Y,π(x)
separable for all points x such that π(x)∈Max(Y).
Corollary 1.3. Let V/k be a variety defined by a regular sequence {f ,...,f }
1 r
in some polynomial ring k[X ,··· ,X ] and assume that T is locally free.
1 n V/k
(1) If Chark =0 then V/k is smooth.
(2) If Chark > 0, assume moreover that the ring k(f ,...,f ) ⊗
1 r k[f1,...,fr]
k[X ,...,X ] is smooth over the field k(f ,...,f ). Then
1 n 1 r
codim+B ≤1.
V V/k
Incharacteristic0,SchejaandStorch[17]provedCorollary1.3whenV/kisa
hypersurface(weincludeaproof);Moen[14]proveditwhenV isahomogeneous
complete intersection; Hochster [10] proved it when V is the spectrum of a
graded ring, and he even attempted to find a counter-example when V/k is
a locally complete intersection surface. Platte [16] found an elementary proof
in the graded case, applicable also to analytic algebras. Using the existence of
weaklysubmersiveresolutionsofsingularities(workofHironaka)combinedwith
a detailed study of mixed Hodge structures, Straten and Steenbrink [19] argue
thatcodim+ B ≤2whenT islocallyfreeandX isananalyticspacewith
X X/C X/C
at most isolated singularities; this was extended to non-isolated singularities by
Flenner [2].
I want to thank the careful referee for helping me make this paper more
accessible.
2 Base change for relative tangent vector fields
Let π : X → Y be a morphism of nœtherian schemes which is locally of finite
type and generically smooth, so Ω is free of rank d when x is a
X/Y,x X/Y,x
maximal point. The branch scheme B(i) , i = 0,..., is defined by the Fitting
X/Y
3
ideal F (Ω ), and B = B = V(F (Ω )). Similarly, the
dX/Y+i X/Y π X/Y dX/Y X/Y
tangential branch scheme is defined by Bt = V(F (T )) (see [6, Sec.
X/Y dX/Y X/Y
1.4, p. 21], [7, Ch. 20] and [11]). We will study base change diagrams
j (cid:47)(cid:47)
X X (BC)
1
π1 π
(cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47)
Y Y,
1
where X = X × Y . I was unable to find a good reference for the following
1 Y 1
well-known important fact (see however [5, Proposition 16.4.5]).
Proposition 2.1. Consider the diagram (BC).
(1) The canonical morphism
j∗(Ω )→Ω
X/Y X1/Y1
is an isomorphism.
(2) B(i) =B(i) × X .
X1/Y1 X/Y X 1
(3) Consider the canonical morphism
ψ :j∗(T )→T .
X/Y X1/Y1
If ψ is an isomorphism, then
Bt =Bt × X .
X1/Y1 X/Y X 1
Proof. For the proof we can assume that all schemes are affine, so let A → B
andA→A behomomorphismsofcommutativerings, andputB =A ⊗ B.
1 1 1 A
(1): Let d : B → Ω be a universal derivation and define the A -
B/A B/A 1
linear derivation d = id⊗d : B → A ⊗ Ω , which can be factorised
B/A 1 1 A B/A
over a universal derivation d : B → Ω by a B -homomorphism
B1/A1 1 B1/A1 1
d˜ : Ω → A ⊗ Ω . There exists a natural B -homomorphism p :
B1/A1 1 A B/A 1
A ⊗ Ω →Ω , which is the inverse of d˜.
1 A B/A B1/A1
(2): By (1), Ω = B ⊗ Ω . Let F(Ω ) denote the Fitting ideal
B1/A1 1 B B/A B/A
defining B(i) and recall that B F(M) = F(B ⊗ M) for a B-module M of
B/A 1 1 B
finite type. Then
B(i) × X =V(F(Ω ))× SpecB =Spec(B/F(Ω )⊗ B )
X/Y X 1 B/A SpecB 1 B/A B 1
B B B
=Spec 1 =Spec 1 =Spec 1 =B(i) .
B1F(ΩB/A) F(B1⊗B ΩB/A) F(ΩB1/A1) X1/Y1
(3) is proven in the same way as (2).
4
If Y → Y is flat or B = ∅, then the canonical homomorphism ψ is an
1 X/Y
isomorphism [5, Proposition 16.5.11], but in general it need be neither injective
nor surjective, contrary to the good behaviour of Ω .
X/Y
Proposition 2.2. Assume that codim− B ≥ 2 and codim− B ≥ 2,
X X/Y X1 X1/Y1
and that X and X satisfy (S ) at all points in j(X ). Then the canonical
1 2 1
morphism ψ is an isomorphism.
Of course, if π is flat, X satisfies (S ) and Y satisfies (S ) along Y , then
1 2 2 1
X satisfies (S ) along j(X ).
2 1
Lemma 2.3. (1) Let B be a ring, I an ideal, and N and M B-modules (not
necessarily of finite type). If there exists an N-regular sequence in I of
length 2, then depth Hom (M,N)≥2.
I R
In particular, if depth R ≥ 2, then depth T ≥ 2 for any subring
I I R/k
k ⊂R.
(2) Let (A,m ) → (B,m ) be a flat homomorphism of local rings, where
A B
A is regular. Let N be a B-module of finite type which is flat over A. If
depth N/m N ≥1anddepth N ≥2,thendepth m Hom (M,N)≥
mB A mB mB A B
2.
Proof. (1): Let {x ,x } be an N-regular sequence in I. Clearly, x is
1 2 1
Hom (M,N)-regular. Assume x φ (m) = x φ (m), φ ∈ Hom (M,N), m ∈
B 2 2 1 1 i B
M. Since {x ,x } is N-regular, φ (m) = x n(cid:48), n(cid:48) ∈ N. Since x is a regular
1 2 2 1 1
element this gives a well-defined homomorphism φ(cid:48) ∈ Hom (M,N), φ(cid:48)(m) =
B
n(cid:48), and φ =x φ(cid:48), hence {x ,x } is Hom (M,N)-regular. Therefore
2 1 1 2 B
Exti (k ,Hom (M,N))=0
R P R
when i≤1 and k is the residue field at a prime of height ≥2.
P
(2): Letx beN/m N-regularand{x ,x }beanN-regularsequence. Since
1 A 1 2
A is regular m =(y ,...,y ) where {y ,...,y } is an A-regular sequence, and
A 1 r 1 r
since N is flat it is also an N-regular sequence. Then {y ,...,y ,x } is an
1 r 1
N-regular sequence. Assume x φ = x φ , where φ ,φ ∈ m Hom (M,N).
2 2 1 2 1 2 A B
As {x ,x } is N-regular, φ (m) ∈ x N, and since x is N-regular φ = x φ(cid:48),
1 2 2 1 1 2 1 2
whereφ(cid:48) ∈Hom (M,N). Thereforeφ ∈x Hom (M,N)∩m Hom (M,N).
2 B 2 1 B A B
(cid:80)
Assume that y f = x f, f,f ∈ Hom (M,N). Since {y ,...,y ,x } is
i i 1 i B 1 r 1
N-regular we have x N ∩ m N = x m N, hence f (m) ∈ x N, and since
1 A 1 A i 1
x is N-regular, f = x f(cid:48) where f(cid:48) ∈ Hom (M,N). This implies φ ∈
1 i 1 i i B 2
x Hom (M,N)∩m Hom (M,N) = x m Hom (M,N), and thus {x ,x }
1 B A B 1 A B 1 2
is m Hom (M,N)-regular. (cid:3)
A B
Proof of Proposition 2.2. Let i : X0 = X \B → X be the inclusion
X/Y
morphism, j : X0 → X0 the base-change of j over i, and let i : X0 → X be
0 1 1 1 1
the canonical morphism, so that i◦j =j◦i . We have:
0 1
(i) Ω is locally free so j∗(T )=T .
X0/Y0 0 X0/Y X10/Y1
5
(ii) j is quasi-compact and separated, so cohomology commutes with base-
change over the flat morphism i, in particular j∗ ◦i = (i ) ◦j∗ as a
∗ 1 ∗ 0
functor on quasi-coherent sheaves on X0.
(iii) codim− B ≥ 2 and X satisfies (S ), so T = Hom (Ω ,O )
X X/Y 2 X/Y OX X/Y X
satisfies(S )(Lem.2.3), implyingT =i i∗(T ); similarly, sinceX
2 X/Y ∗ X/Y 1
satisfies(S )andcodim− B ≥2wehaveT =(i ) i∗(T )=
2 X1 X1/Y1 X1/Y1 1 ∗ 1 X1/Y1
(i ) (T ).
1 ∗ X10/Y1
(i-iii) imply
j∗(T )=j∗i i∗(T )=(i ) j∗i∗(T )=(i ) j∗(T )=(i ) (T )
X/Y ∗ X/Y 1 ∗ 0 X/Y 1 ∗ 0 X0/Y 1 ∗ X10/Y1
=T .
X1/Y1
(cid:3)
Define the following subsets of X :
1
A=suppKer(ψ), B =suppCoker(ψ).
Proposition 2.4. Let π :X →Y be a finitely presented morphism of schemes.
(1) If π is flat, Y is regular, and X contains no embedded associated compo-
1
nents, then codim+ A≤1.
X1
(2) IfIm(ψ)satisfies(S )andX containsnoembeddedassociatedcomponent
2 1
of codimension ≥2 (e.g. X is normal), then codim+ B ≤1.
1 X1
(3) Assume that ψ is injective, X satisfies (S ), and codim− B ≥2. If
1 2 X1 X1/Y1
T is locally free, then T is locally free.
X/Y X1/Y1
Proof. (1): Let ψ¯ : j−1(T ) → T be the natural morphism. We
have m j−1(T ) ⊂XK/Yer(ψ¯) foXr1/eYa1ch point x. Assume on the con-
Y,π(x) X/Y,x x
trary that x ∈ Max(A) ⊂ X is a point of height ≥ 2, so m T =
1 Y,π(x(cid:48)) X/Y,x(cid:48)
Ker(ψ) when ht(x(cid:48)) ≤ 1 and x is a specialisation of x(cid:48). By Lemma 2.3
x(cid:48)
depthm j−1(T ) ≥ 2, hence m j−1(T ) = Ker(ψ¯) , implying
Y,π(x) X/Y,x Y,π(x) X/Y,x x
ψ is injective, and contradicting the assumption x∈A.
x
(2): Assume on the contrary that there exists a point x ∈ Max(B) ⊂ X of
1
height ≥ 2. Since Im(ψ) has depth ≥ 2 the exact sequence 0 → Im(ψ) →
x x
T → Coker(ψ) → 0 is split, so we get an injective homomorphism
X1/Y1,x x
Coker(ψ) → T , and since x ∈ Max(B), it follows that Coker(ψ)
x X1/Yx,x x
can be identified with a submodule of codimension ≥ 2 in T . By as-
X1/Y1,x
sumption O contains no associated prime of height ≥ 2, hence T =
X1,x X1/Y1,x
Hom (Ω ,O ) also has no associated prime of height ≥ 2, which
OX1,x X1/Y1,x X1,x
gives a contradiction.
(3): Since j∗(T ) is locally free and X satisfies (S ), j∗(T ) satis-
X/Y 1 2 X/Y
fies (S ), and the conditions in (2) are satisfied, so codim B ≤ 1. Since
2 X1
codim B ≥2 it follows that B =∅, hence ψ is an isomorphism.
X1 X1/Y1
6
Proposition 2.5. Let π : X → Y be a flat locally of finite type morphism
of nœtherian schemes and consider the base change diagram (BC). Assume
that X /Y and X/Y are generically smooth, X and Y satisfies (S ), and
1 1 1 2
codim− B ≥2.
X1 X1/Y1
(1) In a neighbourhood of j(X ), codim− B ≥2 and X satisfies (S ).
1 X X/Y 2
(2)
Bt =Bt × X .
X1/Y1 X/Y X 1
Inparticular,themoduleT islocallyfreeifandonlyifT islocally
X1/Y1 X/Y
free in a neighbourhood of j(X ).
1
Proof. (1): It is well-known that by flatness, X satisfies (S ) at points in
2
j(X ) ⊂ X. When Y → Y is flat the assertion is obvious, so by Stein fac-
1 1
torisationwecanassumethatY →Y isaclosedimmersion, hencej :X →X
1 1
is a closed immersion. The assertion is also obvious when D ∩ i(Y ) = ∅,
π 1
so assume that there exists a point x in B that specialises to a point
X/Y
x ∈ j(X ). We can assume that x ∈ Max(B ) and we can also find
0 1 X/Y
x ∈Max(B ), such that j(x ) specialises to x and x specialises to j(x ),
1 X1/Y1 1 0 1
and j(x ) ∈ Maxj(B ) = Max(j(X )∩B ). In other words, x is a
1 X1/Y1 1 X/Y 1
maximal point in the set of points in j(X )⊂X that are specialisations of the
1
point x, therefore by the going-down theorem for flat morphisms
ht (x )≤ht (x).
X1 1 X
If on the contrary ht (x) ≤ 1, then x (cid:54)∈ B so Ω is free of rank
X 1 X1/Y1 X1/Y1,x1
d . SinceX /Y isgenericallysmoothandlocallyoffinitetype,d
X1/Y1,x1 1 1 X1/Y1,x1
equals the Krull dimension of a generic fibre of X /Y , and since X/Y is flat
1 1
the Krull dimension of the generic fibres of X/Y that specialise to the same
generic fibre of X /Y also equals d , and as X/Y is generically smooth,
1 1 X1/Y1,x1
we conclude that d = d . Now since Ω = j∗(Ω )
X/Y,j(x1) X1/Y1,x X1/Y1,x1 X/Y x1
(Prop. 2.1, (1)), it follows that the O -module Ω is generated by
X,j(x) X/Y,j(x1)
d elements, implying that Ω is free, and hence j(x ) (cid:54)∈ B
X/Y,j(x1) X/Y,j(x1) 1 X/Y
(Prop. 2.1, (2)). Since x specialises to j(x ) it follows that x(cid:54)∈B , resulting
1 X/Y
in a contradiction. Therefore ht (x)≥2.
X
(2): By (1) and Proposition 2.2 the canonical morphism ψ : j∗(T ) →
X/Y
T is an isomorphism so the assertion is implied by Proposition 2.1.
X1/Y1
3 Differential criterion of smoothness
The relation between the branch locus and the locus of non-smooth points of
a morphism is of course much discussed in the literature, but there still seems
to remain room for clarification. In [18, §2] one can find a nice summary of
characterisations of smoothness in terms of the vanishing of Andr´e-Quillen ho-
mology and the Jacobian condition B =∅. We are however more interested in
π
the “Jacobian characterisation” [5, Proposition 17.15.15]. A proof is included
7
since I find the argument in EGA difficult to disentangle and I could not find
any other satisfactory treatment of this important result in the literature. The
proof relies on the fundamental theorem that smoothness implies flatness, and
conversely, if a morphism is flat at a point and smooth along the fibre at the
point, then the morphism is smooth at that point [3, Th´eor`eme 19.7.1].
Theorem 3.1. Let π : X → Y be a morphism of schemes which is locally
finitely presented. Let x be a point in X and put y = π(x). The following are
equivalent:
(1) π is smooth at the point x.
(2) π is flat at x, x (cid:54)∈ B , and is smooth at all points ξ ∈ Max(X) that
π
specialises to x.
(3) π is flat at x, x(cid:54)∈B , and rkΩ =dim X .
π X/Y,x x y
(4) π is flat at x, x (cid:54)∈ B , and X /k is smooth at all points η ∈ Max(X )
π y Y,y y
that specialises to x.
Remark 3.2. The proof in [5] that the rank of Ω is as asserted in (3)
X/Y,x
seems to contain a gap. It relies on [5, Proposition 17.10.2], and presupposes
that π be smooth not only at the point x, but also at all specialisations of x.
Proof. (1) ⇒ (2) and (3): Put A = O and B = O , so A → B is formally
Y,y X,x
smooth,andsincemoreoverthishomomorphismisfinitelypresesented,itisalso
flat[3,Th17.5.1]. Letξ beapointthatspecialisestox,andputB(cid:48) =O and
X,ξ
A(cid:48) = O . It follows directly from the definition of formal smoothness that
Y,π(ξ)
the composition A → B → B(cid:48) and the base change A(cid:48) → B(cid:48) are also formally
smooth (see also [13, Theorem 28.2]); hence π is smooth at ξ. Since A → B
is formally smooth it follows that Ω is projective [13, Theorem 28.5], and
B/A
since B is a local ring, Ω is free; hence x∈/ B . We now determine the rank
B/A π
r =rankΩ . Letkbeanalgebraicclosureofk ,putB¯ =k⊗ k ⊗ B,
andletkBBb/eAtheresiduefieldofB¯. PropositionY2.,y1impliesthatΩkYB¯,y/kYi,syfreAeof
rank r, hence by the second fundamental exact sequence in [13, Theorem 25.2],
noting that k /k is formally smooth since k is algebraically closed,
B
m
r =dim B¯ +tr.degk /k.
k m2 B
B¯
Formal smoothness is preserved under base change, hence the map k → B¯ is
formally smooth; hence B¯ is a regular local ring [13, Lemma 1], so dimk mm2B¯ =
dimB¯. Letnowx beapointinX suchthatdim X =ht (x )=dimO B¯ .
Put R = k⊗ 1O and let ky denote its rexsidyue fieldX.yTh1en B¯ = RXyf,xo1r
kY,y Xy,x1 R P
some prime ideal P in R, and dim X = dimR, since R is a flat base change
x y
of k →O . Since, moreover, π is finitely presented and k is algebraically
Y,y Xy,x1
closed, it follows that k = k , and by Hilbert’s Nullstellensatz [13, Theorem
R
5.6] dimR/P =tr.degk /k. Since R is a catenary ring, we then get
B
dimR=ht (x)+dimR/P =dimB¯+tr.degk /k =r.
Xy B
8
(2) ⇒ (1): Since π is flat at x, it suffices by [3, Th 19.7.1] to prove that
k → k ⊗ O is formally smooth to conclude that O → O is
Y,y Y,y OY,y Xy,x Y,y X,x
formally smooth, and by [13, Theorem 30.3] (see also [5, Proposition 15.15.5])
this follows if Ω is free of rank dim X .
OXy,x/kY,y x y
Since x ∈/ B , by Proposition 2.1 the O -module Ω is free and
π Xy,x Xy/kY,y,x
rkΩ = rkΩ . Since π is smooth at points ξ ∈ Max(X) that spe-
Xy/kY,y,x X/Y,x
cialise to x, and Ω = O ⊗ Ω , it follows as in the proof of
X/Y,ξ X,ξ OX,x X/Y,x
(1) ⇒ (3) that this rank equals dim X (or see [5, Prop. 17.15.5]). Since π
ξ π(ξ)
is flat, dim X =dim (X ), hence rkΩ =dim X .
x y ξ π(ξ) Xy/kY,y,x x y
(3)⇒(4): Since Ω is free of rank dim X it follows that Ω is
X/Y,x x y Xy/kY,y,x
free of rank dim X (Prop. 2.1). In the same way as in the proof of (2)⇒(1)
x y
it follows that k →O is formally smooth; hence any localisation k →
Y,y Xy,x Y,y
O is also formally smooth.
Xy,η
(4)⇒(1): Since k →O is formally smooth it follows that Ω
Y,y Xy,η Xy/kY,y,η
isfreeofrankdim X =dim X . NowΩ isfreeandbyProposition2.1its
η y x y X/Y,x
rankisdim X . Itfollowsasintheproofof(2)⇒(1)thatthehomomorphism
x y
O →O is formally smooth.
Y,y X,x
Oftenthecondition(2)or(3)inTheorem3.1serveasadefinitionofsmooth-
ness (see e.g. [8]). Alternatively, π is smooth if it is flat and all its fibres are
smooth [5, Th´eor`eme 17.5.1]. In either case, the condition that π be flat can
be a nuisance. Put Γ = Ker(π∗(Ω ) → Ω ). Assuming X/S is
X/Y/S Y/S X/S
smooth, (3) in the following theorem shows that the non-smoothness locus of
π is exactly suppΓ ∪B . Therefore, if Γ = 0 it follows that the
X/Y/S π X/Y/S
Jacobian criterion B =∅ implies smoothness, i.e. flatness is automatic. More-
π
over,Γ =0wheneitherX/Y isgenericallysmoothandalocallycomplete
X/Y/S
intersection,orX/S isalocallycompleteintersection(see[11,Proposition2.11]
for a discussion of this assertion).
Theorem 3.3. Assume that X/S is smooth at the point x. The following are
equivalent:
(1) π :X →Y is smooth at x.
(2) π∗(Ω ) −→p Ω has a left inverse.
Y/S x X/S,x
(3) x∈/ B and Γ =0.
π X/Y/S,x
Proof. (1)⇔(2): See[5,Th. 17.11.1]. (3)⇒(2): Thiscanbeseendirectlyfrom
the fundamentalexact sequence ofdifferentials [8, Proposition 8.11]. (1)⇒(3):
(1) implies by Theorem 3.1 that x ∈/ B , and since (1) implies (2) we also get
π
Γ =0.
X/Y/S,x
4 Proof of main results
Although our main result does not rely on the following preliminary result, it
does provide insight into how the assumption ‘locally complete intersection’ is
applied.
9
Lemma 4.1. (Lichtenbaum-Schlessinger [12, Prop. 5.2]) Let X/k be a l.c.i.
scheme locally of finite type over a field. Assume that X/k is generically smooth
(e.g. kisperfectandX isreduced)andT islocallyfree. Thencodim+ B ≤
X/k X X/k
2.
Proof. We follow the proof in [loc. cit.]. The problem being local at a point
x, we can assume that there exists a regular immersion i : X → X over
r
k, where X /k is smooth. Since I/I2 is locally free and X/k is generically
r
smooth we have the presentation 0 → I/I2 → i∗(Ω ) → Ω → 0,
Xr/k X/k
so p.d.Ω ≤ 1. So Ω is free if and only if Ext1 (Ω ,O ) =
X/k,x X/,k OX,x X/k,x x
Ext1 (Ω ,O ) = 0, and therefore B = suppExt1 (Ω ,O ). As-
sumOeXnowXt/hkatxX∈xMax(B ),soxisanaXss/okciatedpointoOfXExtX1/k(Ω X ,O )
and therefore depthExt1 X(/Ωk ,O ) = 0. Dualising the presOeXntatXio/nk givXes
theexactsequence0→TOX →X/ik∗(ΩX x )∗ →(I/I2)∗ →Ext1 (Ω ,O )→
X/k Xr/k OX X/k X
0, since Ω is locally free; hence
Xr/k
p.d.Ext1 (Ω ,O ) ≤2.
OX X/k X x
BytheAuslander-Buchsbaumformula,depthO =depthExt1 (Ω ,O ) +
p.d.Ext1 (Ω ,O ) ≤2, and since X is CoXh,xen-Macaulay, hOtX (xX)/≤k 2.X x
OX X/k X x X
There is also a relative version for morphisms between smooth schemes,
which is used in the proof of the main theorem.
Proposition 4.2. Let X be a Cohen-Macaulay scheme and X/S and Y/S be
smooth S-schemes. If Bt =∅, then
X/Y
codim+ B ≤2.
X π
Lemma 4.3. Assume that X/S and Y/S are smooth morphisms and that X/Y
is generically smooth. Then B =suppC .
π X/Y
Proof. This follows from [11, Prop. 1.3 ], but we give a more concrete indepen-
dent proof. Let φ : G → G be a homorphism of free modules over a local
1 2
commutative ring A of rank g and g , respectively. Put M = Coker(φ) and
1 2
D(M)=Coker(φ∗) (the transpose of M). We assert:
M (cid:54)=0⇔Coker(D(M)) cannot be generated by g −g elements. (∗∗)
1 2
Letting G¯ be the image of the homomorphism φ we get the exact sequence
1
0→G¯∗ →G∗ →D(M)→0.
1 1
For an A-module N we denote by β (N)=dim Tori (k,N) the ith Betti num-
i k A
ber. The exact sequence results in a long exact sequence in homology, from
which follows that
g −β (D(M))−β (G¯∗)+β (D(M))=0,
1 0 0 1 1
10
