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

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Pro<str<strong>on</strong>g>of</str<strong>on</strong>g> <str<strong>on</strong>g>of</str<strong>on</strong>g> Claim 7. Take y ∈ [(a) : b]∩I n . Then, since by ∈ (a), we see bt·yt n = byt n+1 = 0<br />

in G(I). Hence yt n ∈ [H 0 M (G(I))] n, because bt is a homogeneous parameter for <str<strong>on</strong>g>the</str<strong>on</strong>g> graded<br />

ring G(I). Recall now that n ≥ 2, whence [H 0 M (G(I))] n = (0), so that yt n = 0. Thus<br />

y ∈ (a) + I n+1 , whence y ∈ (a) + [((a) : b) ∩ I n+1 ], as claimed.<br />

□<br />

Since x ∈ (a) + [((a) : b) ∩ I 2 ], thanks to Claim 7, we get x ∈ (a) + I n+1 for all n ≥ 1,<br />

whence x ∈ (a), so that <str<strong>on</strong>g>the</str<strong>on</strong>g> map ρ is injective. Thus<br />

as A-modules.<br />

[H 1 M(G)] 0<br />

∼ =<br />

[(a) : b] ∩ I<br />

(a)<br />

Theorem 8. Suppose that d = 2 <strong>and</strong> depth A > 0. Let Q = (x, y) be a parameter ideal<br />

in A <strong>and</strong> assume that x is superficial with respect to Q. Then<br />

−h 1 (A) ≤ e 2 Q(A) ≤ 0<br />

<strong>and</strong> <str<strong>on</strong>g>the</str<strong>on</strong>g> following three c<strong>on</strong>diti<strong>on</strong>s are equivalent.<br />

(1) e 2 Q (A) = 0.<br />

(2) x, y forms a d-sequence in A.<br />

(3) x l , y l forms a d-sequence in A for all integers l ≥ 1.<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. By Lemma 5 we have<br />

e 2 Q(A) = −l A<br />

( [(x l ) : y l ] ∩ (x, y) l<br />

(x l )<br />

)<br />

≤ 0<br />

for all integers l ≫ 0. To show that −h 1 (A) ≤ e 2 Q (A), we may assume that H1 m(A) is<br />

finitely generated. Take <str<strong>on</strong>g>the</str<strong>on</strong>g> integer l ≫ 0 so that <str<strong>on</strong>g>the</str<strong>on</strong>g> system a = x l , b = y l <str<strong>on</strong>g>of</str<strong>on</strong>g> parameters<br />

<str<strong>on</strong>g>of</str<strong>on</strong>g> A is st<strong>and</strong>ard. Then since<br />

[(a) : b] ∩ Q l<br />

(a)<br />

⊆ (a) : b<br />

(a)<br />

we get −h 1 (A) ≤ e 2 Q (A).<br />

Let us c<strong>on</strong>sider <str<strong>on</strong>g>the</str<strong>on</strong>g> sec<strong>on</strong>d asserti<strong>on</strong>.<br />

(1) ⇒ (3). Take an integer N ≥ 1 so that<br />

for all l ≥ N (cf. Lemma 5); hence<br />

∼ = H<br />

0<br />

m (A/(a)) ∼ = H 1 m(A),<br />

e 2 Q(A) = −l A<br />

( [(x l ) : y l ] ∩ (x, y) l<br />

(x l )<br />

[(x l ) : y l ] ∩ (x, y) l = (x l ).<br />

Claim 9. [(x l ) : y l ] ∩ (x, y) l = (x l ) for all l ≥ 1.<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g> <str<strong>on</strong>g>of</str<strong>on</strong>g> Claim 9. We may assume that 1 ≤ l < N. Take τ ∈ [(x l ) : y l ] ∩ (x, y) l . Then,<br />

since y N (x N−l τ) = y N−l x N−l (y l τ) ∈ (x N ), we have x N−l τ ∈ [(x N ) : y N ] ∩ (x, y) N = (x N ).<br />

Thus τ ∈ (x l ), because x is A-regular (recall that depth A > 0 <strong>and</strong> x is superficial with<br />

respect to Q).<br />

□<br />

–160–<br />

)<br />

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