Proceedings of the 44th Symposium on Ring Theory and ...

Therefore, thanks to <strong>the</strong> uniform bounds [8, Theorem 2.3] <strong>of</strong> reg G(Q) for parameter
ideals Q in a generalized Cohen-Macaulay ring A, we readily get <strong>the</strong> finiteness in <strong>the</strong> set
Λ i (A) for all 1 ≤ i ≤ d.
We are now in a position to finish <strong>the</strong> pro<strong>of</strong> <strong>of</strong> Theorem 1.
Pro<strong>of</strong> <strong>of</strong> Theorem 1. We may assume that A is complete. Also we may assume A is not
unmixed, because Λ 1 (A) is a finite set (cf. [2, Proposition 4.2]). Let U denote <strong>the</strong> unmixed
component <strong>of</strong> <strong>the</strong> ideal (0) in A. We put B = A/U and t = dim A U (≤ d − 1). We must
show that B is a generalized Cohen-Macaulay ring and t = 0.
Let Q be a parameter ideal in A. We <strong>the</strong>n have
l A (A/Q n+1 ) = l A (B/Q n+1 B) + l A (U/Q n+1 ∩ U)
for all integers n ≥ 0. Therefore, <strong>the</strong> function l A (U/Q n+1 ∩ U) is a polynomial in n ≫ 0
with degree t and <strong>the</strong>re exist integers {s i Q (U)} 0≤i≤t with s 0 Q (U) = e0 Q (U) such that
t∑
( ) n + t − i
l A (U/Q n+1 ∩ U) = (−1) i s i Q(U)
t − i
for all n ≫ 0, whence
Consequently
l A (A/Q n+1 ) =
(−1) d−i e d−i
Q
i=0
d∑
( ) n + d − i
(−1) i e i Q(B)
+
d − i
i=0
t∑
( ) n + t − i
(−1) i s i Q(U)
.
t − i
{ (−1) d−i
(A) = e d−i
Q
(B) + (−1)t−i s t−i
Q
(U) if 0 ≤ i ≤ t,
(−1) d−i e d−i (B) if t + 1 ≤ i ≤ d.
Q
Therefore, if t < d − 1, we have e 1 Q (A) = e1 Q (B), so that Λ 1(B) = Λ 1 (A) is a finite
set. If t = d − 1, we get −e 1 Q (A) = −e1 Q (B) + s0 Q (U). Since e1 Q (A), e1 Q (B) ≤ 0 and
s 0 Q (U) = e0 Q (U) ≥ 1, Λ 1(B) is a finite set also in this case. Thus <strong>the</strong> set Λ 1 (B) is finite in
any case, so that <strong>the</strong> ring B a generalized Cohen-Macaulay ring.
We now assume that t ≥ 1 and choose a system a 1 , a 2 , · · · , a d <strong>of</strong> parameters in A so
that (a t+1 , a t+2 , · · · , a d )U = (0). Let l ≥ 1 be an integer such that m l is standard for <strong>the</strong>
ring B and choose integers n ≥ l. We look at parameter ideals Q = (a n 1, a n 2, · · · , a n d ) <strong>of</strong> A.
Then
t∑
( ) t − 1
(−1) d−t e d−t
Q (B) = h j (B)
j − 1
by [10, Korollar 3.2], which is independent <strong>of</strong> <strong>the</strong> integers n ≥ l. Therefore, since
we see
j=1
i=0
s 0 Q(U) = e 0 (a n 1 ,an 2 ,··· ,an t ) (U) = n t·e 0 (a 1 ,a 2 ,··· ,a t )(U) ≥ n t ,
(−1) d−t e d−t
Q (A) = (−1)d−t e d−t
Q
(B) + s0 Q(U)
t∑
( ) t − 1
=
h j (B) + n t·e 0 (a
j − 1
1 ,a 2 ,··· ,a t )(U) ≥ n t ,
j=1
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