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

Let B = A/U(a d−1 ). We <strong>the</strong>n have
∑d−3
( ) d − 4
e 2 QB(B) =
h j (B)
j − 1
for <strong>the</strong> same reason as for <strong>the</strong> equality e 2 QC (C) = ∑ d−3
e 2 QC (C) = ∑ d−3
j=1
j=1
j=1
( d−4
j−1)
h j (C) (in fact, to show
( d−4
j−1)
h j (C), we only need that a 1 is superficial with respect to Q). Therefore,
by <strong>the</strong> hypo<strong>the</strong>sis <strong>of</strong> induction on d, we may choose elements β 2 , β 3 , · · · , β d ∈ B so
that QB = (β 2 , β 3 , · · · , β d )B and <strong>the</strong> sequence β 2 , β 3 , · · · , β d forms a d-sequence in B.
We put b 1 = a d−1 and write β j = b j with b j ∈ Q for 2 ≤ j ≤ d, where b j denotes <strong>the</strong>
image <strong>of</strong> b j in B. We now put q ′ = (b 1 , b 2 , · · · , b d ). Then q ′ is a parameter ideal in A and
because U(b 1 ) ∩ Q = (b 1 ), we get
Q ⊆ [q ′ + U(b 1 )] ∩ Q = q ′ + [U(b 1 ) ∩ Q] ⊆ q ′ + (b 1 ) = q ′ ;
hence Q = q ′ . Thus <strong>the</strong> sequence b 2 , b 3 , · · · , b d forms a d-sequence in A/(b 1 ), so that
b 1 , b 2 , · · · , b d forms a d-sequence in A, because b 1 is A-regular. This complete <strong>the</strong> pro<strong>of</strong>
<strong>of</strong> Theorem 15 and that <strong>of</strong> Theorem 2 as well.
□
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Meiji University
1-1-1 Higashi-mita, Tama-ku, Kawasaki 214-8571, Japan
Email: kozeki@math.meiji.ac.jp
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