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which is composed of the inclusion
r’1 r’1 r+1 r+1
s’1 t’1 — m
(Rn )— , —
— ) ’’ R— yI — zJ ’’ yI∪{s,i} — zJ∪{t,j} ,
R (R
the homomorphism •x,r , and the residue class map with respect to the corresponding
submodule of M . Obviously

•(yI — zJ ) = residue class of d([I, s, i|J, t, j]),
I ∈ S(r’1, s’1), J ∈ S(r’1, t’1).
By expansion one obtains
σ(i, I\i)[i|j] d([I\i|J]) = 0,
i∈I
(1)
σ(l, L\l)[k|l] d([K|L\l]) = 0
l∈L
for all I ∈ S(r+2, m), j ∈ S(1, n), J ∈ S(r+1, n), k ∈ S(1, m), K ∈ S(r+1, m), L ∈
S(r+2, n). We therefore get

σ(i, I\i)[i|j] yI\i — zJ ∈ Ker •,
i∈I

σ(l, L\l)[k|l] yK — zL\l ∈ Ker •
l∈L
for all I ∈ S(r, s’1), j ∈ S(1, n), J ∈ S(r’1, t’1), k ∈ S(1, m), K ∈ S(r’1, s’1),
L ∈ S(r, t’1). By elementary determinantal calculations it follows that
— —
[K|L] yI — zJ , [K|L] yI — zJ ∈ Ker •
for all K ∈ S(r, s’1), L ∈ S(r, n), K ∈ S(r, m), L ∈ S(r, t’1) and all I ∈ S(r’1, s’1),
J ∈ S(r’1, t’1). This proves the inclusion “⊃” of (a).
Put R = R(X; δst ). Then • induces an R-homomorphism
r’1 r’1
s’1
(Rt’1 )— ’’ M (s, t)

R
m n
whose kernel contains the kernel of the map • in (14.2). So •(s, i; t, j)
i=s+1 j=t+1
induces a surjective map
m n
I(x; δs+1,t , δs,t+1 ) ’’ M (s, t).
s+1 t+1

The proof of the proposition is complete once we have shown that M(s, t) contains a free
R-module of rank (m ’ s)(n ’ t). This clearly holds if the residue classes of the elements
d([1, . . . , r ’ 1, s, i|1, . . . , r ’ 1, t, j]) are linearly independent over R. Assume that
±
0 if s = t = r,

m n
aij d([1, . . . , r ’ 1, s, i|1, . . . , r ’ 1, t, j]) ∈ M (s, t’1) if t > r,


i=s+1 j=t+1 M (s’1, t’1) if s > r, t = r
with elements aij ∈ R. The left side of this relation has ±aij δst as its component
belonging to dXij while the corresponding component of the right side lies in the ideal
I(X; δst )/Ir+1 (X). Since δst is not a zero-divisor of R(X; δst ) we get
aij ∈ I(X; δst )/Ir+1 (X). ”
180 14. The Module of K¨hler Di¬erentials
a

(14.6) Corollary. For every prime ideal P in B[X] containing I(X; δst ) one has

depth M (s, t)P = depth RP ’ (s ’ r) ’ (t ’ r).


Proof: Assume ¬rst that B is a ¬eld or B = Z. From (14.1) we get

depth I(x; δs+1,t , δs,t+1 )P = depth R(X; δst )P ,

so
depth M (s, t)P = depth R(X; δst )P

by (14.5),(b). Using the dimension formula (5.12),(a) we obtain

depth R(X; δst )P = depth RP ’ (s ’ r) ’ (t ’ r).

It follows in particular that M(s, t) is a torsionfree R(X; δst )-module if B = Z, so it is
B-¬‚at in case B is an arbitrary noetherian ring, what we will assume from now on. Put
Q = P © B and consider the ¬‚at extension BQ ’ RP . Since M(s, t)P and R(X; δst )P
are also ¬‚at over BQ , the depth formula we have used already in the previous sections
(cf. the proof of (3.14) for example) yields

depth M(s, t)P = depth BQ + depth(M(s, t)P — (BQ /QBQ )),
depth R(X; δst )P = depth BQ + depth(R(X; δst )P — (BQ /QBQ )),
depth RP = depth BQ + depth(RP — (BQ /QBQ )).

The claim now follows from what we have derived in the ¬eld case. ”
(14.7) Proposition. For every prime ideal P in B[X] containing Ir (X)

depth(„¦1 )P ≥ depth RP ’ grade(Ir (X), R) + 2.
R/B


Consequently
grade(I, „¦1 ) ≥ grade(I, R) ’ grade(Ir (X), R) + 2
R/B

for all ideals I in B[X], I ⊃ Ir (X).
Proof: Once more we consider the ¬rst syzygy M of „¦1
R/B and its ¬ltration
M(s, t) : r ¤ s < m, r ¤ t < n . From (14.6) and the depth analogue to Lemma
(13.1) it follows that

depth MP ≥ depth RP ’ (m ’ 1 ’ r) ’ (n ’ 1 ’ r)
= depth RP ’ (m + n ’ 2r + 1) + 3
= depth RP ’ grade(Ir (X), R) + 3.

This proves the proposition. ”
181
C. The Syzygetic Behaviour of the Di¬erential Module

(14.8) Remarks. (a) The module of relations of M is generated by the linear
relations (1) in the proof of (14.5), in other words: Let F = R m , G = Rn . Then the
sequence
r+2 r+1 r+1 r+2 r+1 r+1
•1 •0
— — —
G— ’’ F — G— ,
F —G — G )•( F —F — G ) ’’ F—
(

where •0 = •x,r , •1 = (•x,1 — 1) • (1 — •x,1 ) (cf. 13.B) is exact. To demonstrate this,
one has only to look into the proof of (14.5). Another way to obtain exactness is as
follows: The sequence is easily seen to be a complex which is (split) exact in depth 0.
Furthermore one may treat Coker •1 in the same manner as the module M (cf. (14.5),
(14.6) and the proof of (14.7)), to get that

depth(Coker •1 )P ≥ min(3, depth RP )

for all prime ideals P ∈ Spec R. So Coker •1 is torsionfree and thus Im •1 = Ker •0 . ”
(b) The module M (= Ir+1 (X)/Ir+1 (X)(2) ) is a direct B-summand of the symbolic
graded ring
()
Ir+1 (X)(i) /Ir+1 (X)(i+1) ,
GrIr+1 (X) B[X] =

and inherits a standard basis from this ASL in a natural way (cf. (10.6) where this has
been de¬ned for arbitrary B). The ¬ltration considered above is compatible with the
standard basis: each submodule M(s, t) is generated as a B-module by the elements of
the standard basis it contains. The rank argument in the proof of (14.5) can be replaced
by a comparison of standard bases.
Of course each of the quotients Ir+1 (X)(i) /Ir+1 (X)(i+1) inherits a standard basis
from the symbolic graded ring, and it should be possible to construct similar ¬ltrations for
them. These ¬ltrations may yield lower bounds for the depth of Ir+1 (X)(i) /Ir+1 (X)(i+1)
as indicated below (10.8). ”

C. The Syzygetic Behaviour of the Di¬erential Module

The inequalities of (14.7) actually are equalities. This, of course, determines the
syzygetic behaviour of „¦1 . On the other hand we do not use the full truth about
R/B
depth(„¦R/B )P , P a prime ideal in B[X], to describe the syzygetic behaviour of „¦1 .
1
R/B
Besides (14.7) we only need:
(14.9) Proposition. For every minimal prime ideal P of Ir (X)

depth(„¦1 )P = 2.
R/B


Proof: We use induction on r. Let r = 1. A simple localization argument shows
that we may assume P to be the only prime ideal in B[X] containing I1 (X). From (14.7)
we get
depth(„¦1 )P ≥ 2,
R/B

and by (14.4) („¦1 )Q is free for all prime ideals Q in B[X] which are di¬erent from P .
R/B
1
If depth(„¦R/B )P ≥ 3 then every RP -sequence consisting of three elements would be an
182 14. The Module of K¨hler Di¬erentials
a

(„¦1 )P -sequence. According to (5.11), X11 , X12 + X21 , Xmn form an RP -sequence. To
R/B
show that it is not an („¦1 )P -sequence we put ω = X12 dX11 . Then
R/B


Xmn ω =X11 (X21 dXmn ’ X2n dXm1 + Xm2 dX1n )
+ (X12 + X21 )(’Xm1 dX1n + Xmn dX11 )
’ X21 d([1 m|1 n]),

but ω ∈ X11 (Rm — (Rn )— ) + (X12 + X21 )(Rm — (Rn )— ) + M . Thus depth(„¦1 )P = 2.
/ R/B
Assume now that r > 1. Let xmn denote the residue class of Xmn in R. By (2.4)
we have an isomorphism

R[x’1 ] ∼ Rr (Y )[Xm1 , . . . , Xmn , X1n , . . . , Xm’1,n ][Xmn ],
’1
mn =


Y being an (m ’ 1) — (n ’ 1) matrix of indeterminates over B, which maps the extension
of Ir (X) to the extension of Ir’1 (Y ). Put S = Rr (Y ), Q = P R[x’1 ] © S. Since
mn

’1 ∼ ∼
„¦1 ’1
R/B —R R[xmn ] = „¦R[x’1 ]/B = („¦S/B —S R[xmn ]) • F,
mn



with a free R[x’1 ]-module F , and S ’ R[x’1 ] is a ¬‚at extension, we obtain
mn mn


depth(„¦1 )P = depth(„¦1 )Q + depth RP ’ depth SQ .
R/B S/B


Clearly Q is a minimal prime ideal of Ir’1 (Y ), so

depth SQ = grade Ir’1 (Y ) = grade Ir (X) = depth RP .

Using the inductive hypothesis we get the required result. ”
Now it is easy to prove
(14.10) Theorem. „¦1
R/B is a second syzygy but not a third one.

Proof: If P is a prime ideal in B[X], P ⊃ Ir (X), then depth RP ≥ grade(Ir (X), R),
so depth(„¦1 )P ≥ 2 in view of (14.7). In all other cases („¦1 )P is RP -free. Conse-
R/B R/B
quently „¦R/B is a second syzygy and, if it were a third one, then depth(„¦1 )P ≥ 3 for
1
R/B
all prime ideals P ⊃ Ir (X), which obviously contradicts (14.9). ”
(14.11) Remarks. (a) To prove the inequality “≥” in (14.9), we only need the lower
bound for grade(I1 (X), „¦1 ) coming from Proposition (10.8) (cf. the introduction).
R/B
This lower bound also su¬ces (combined with the usual localization argument) in showing
that „¦1R/B is a second syzygy (cf. the proof of (14.10)). ”

(b) In the next section we shall give an explicit presentation of „¦1
R/B as a second
syzygy. ”
(14.9) will also be used to prove the main result of this section:
183
D. Comments and References

(14.12) Theorem. Let P be a prime ideal in B[X] which contains Ir (X). Then

depth(„¦1 )P = depth RP ’ grade(Ir (X), R) + 2.
R/B


Consequently
grade(I, „¦1 ) = grade(I, R) ’ grade(Ir (X), R) + 2
R/B

for all ideals I in B[X], I ⊃ Ir (X).
Proof: We only need to prove the ¬rst equality. Since „¦1 R/B is B-¬‚at, the usual
techniques (used for instance in the proof of (14.7)) allow us to restrict ourselves to the
case in which B is a ¬eld. The inequality “≥” has been established in (14.7). Assume
that
depth(„¦1 )P ≥ depth RP ’ grade(Ir (X), R) + 3
R/B

and take a (grade(Ir (X), R) ’ 3)-th syzygy N of „¦1 . Then depth NP = depth RP and
R/B
consequently

depth(„¦1 )Q ≥ depth RP ’ grade(Ir (X), R) + 3 ≥ 3
R/B

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