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

Pro<strong>of</strong> <strong>of</strong> <strong>the</strong> Theorem 15. We proceed by induction on d. Choose a 1 , a 2 , · · · , a d ∈ A so
that Q = (a 1 , a 2 , · · · , a d ) and for each 1 ≤ i ≤ d−2, <strong>the</strong> i+2 elements a 1 , a 2 , · · · , a i , a d−1 , a d
form a superficial sequence with respect to Q. We will show that <strong>the</strong>re exist b 2 , b 3 , · · · , b d ∈
A such that b 1 = a d−1 , b 2 , b 3 , · · · , b d forms a d-sequence in A and Q = (b 1 , b 2 , · · · , b d ). We
put A = A/(a 1 ), Q = QA, and C = A/H 0 m(A) (= A/U(a 1 )).
Suppose that d = 3. Then
e 2 QC(C) = e 2 Q (A) − h0 (A) = e 2 Q(A) − h 0 (A) = h 1 (A) − h 0 (A) = 0,
because h 1 (A) = h 0 (A) (recall that QH 1 m(A) = (0) by Proposition 13). Hence, thanks
to Proposition 8, a 2 , a 3 forms a d-sequence in C, because a 2 is superficial for <strong>the</strong> ideal
QC = (a 2 , a 3 )C. Therefore, since a 1 H 1 m(A) = (0), we have
U(a 2 ) ∩ [Q + U(a 1 )] = (a 2 ),
by Proposition 16. Let Q = (a 2 , a 3 , b 3 ) and B = A/U(a 2 ). Then since e 2 QB (B) = 0, by
Proposition 8 <strong>the</strong> sequence b 2 = a 3 , b 3 forms a d-sequence in B, because b 2 is superficial
for QB. Therefore, since U(a 2 ) ∩ Q ⊆ U(a 2 ) ∩ [Q + U(a 1 )] = (a 2 ), <strong>the</strong> sequence b 2 , b 3
forms a d-sequence in A/(a 2 ), so that b 1 = a 2 , b 2 , b 3 forms a d-sequence in A, because b 1
is A-regular.
Assume that d ≥ 4 and that our assertion holds true for d−1. Then, thanks to Theorem
13 and its pro<strong>of</strong>, we have
e 2 Q(A) = e 2 (A) = ∑d−3
( ) d − 4
Q e2 QC(C) ≤
h j (C)
j − 1
=
=
j=1
∑d−3
( ) d − 4
h j (A)
j − 1
because Q·H j m(A) = (0) for 1 ≤ j ≤ d − 3. Hence
∑d−3
( ) d − 4
e 2 QC(C) =
h j (C).
j − 1
j=1
j=1
∑d−2
( ) d − 3
h j (A) = e 2
j − 1
Q(A),
Therefore, because QC = (a 2 , a 3 , · · · , a d )C and <strong>the</strong> sequence a 2 , a 3 , · · · , a i , a d−1 , a d is superficial
in <strong>the</strong> ideal QC for all 1 ≤ i ≤ d − 2 where a j denotes <strong>the</strong> image <strong>of</strong> a j in C, <strong>the</strong>
hypo<strong>the</strong>sis <strong>of</strong> induction on d yields that <strong>the</strong>re exist γ 2 , γ 3 , · · · , γ d−1 ∈ C such that <strong>the</strong> sequence
γ 1 = a d−1 , γ 2 , γ 3 , · · · , γ d−1 forms a d-sequence in C and QC = (γ 1 , γ 2 , · · · , γ d−1 )C.
Let us write γ j = c j for each 2 ≤ j ≤ d − 1 with c j ∈ Q, where c j denote <strong>the</strong> image
<strong>of</strong> c j in C. We put q = (a 1 , a d−1 , c 2 , c 3 , · · · , c d−1 ). Then q is a parameter ideal in A,
a 1 H 1 m(A) = (0), and a d−1 , c 2 , c 3 , · · · , c d−1 forms a d-sequence in C. Therefore
j=1
U(a d−1 ) ∩ [Q + U(a 1 )] = U(a d−1 ) ∩ [q + U(a 1 )] = (a d−1 )
by Proposition 16, whence U(a d−1 ) ∩ Q = (a d−1 ).
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