A basis for the non-crossing partition lattice top homology by Zoque E.

By Zoque E.

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10. 10 In chapter 2, exercise 6, we deduced that M [x] M ⊗A A[x]. Since M and A[x] are Noetherian Amodules (by assumption and Hilbert’s Basis Theorem respectively), M ⊕ A[x] will be a Noetherian Amodule. 1. This completes the proof. 11 ∞ It’s not necessary that A is Noetherian. Consider, for example, A = i=1 Z/2Z, a direct product of infinitely many copies of the field F = Z/2Z = {0, 1}. The strictly ascending chain of ideals 0 ⊂ F × 0 ⊂ F × F × 0... shows that A is not Noetherian. It’s also evident that every element of A is idempotent (A is Boolean).

Since B is a finitely generated algebra over A, we may write B = A[b1 , b2 , . . , bm ], and this implies that S −1 B = K[b1 , b2 , . . , bm ] is a finitely generated algebra over K. By the Noether Normalization Lemma, there exist y1 /s1 , y2 /s2 , . . , yn /sn ∈ S −1 B that are algebraically independent over K and are such that S −1 B is integral over K[y1 /s1 , y2 /s2 , . . , yn /sn ]. It’s now easy to see that y1 , y2 , . . , yn are algebraically independent over K and S −1 B is integral over K[y1 , y2 , .

But then, if yz ∈ Ann(x), and y ∈ / Ann(x), we would have Ann(xy) = Ann(x). The equation 0 = (yz)x = z(xy) implies z ∈ Ann(xy) = Ann(x). Hence, p = Ann(x) is prime and this completes the proof. 15 Assume that the primary decomposition of a is a = q 1 ∩ q 2 · · · ∩ qn , where r(qi ) = pi and p1 , p2 , . . , pm constitute the isolated part of a (these are minimal). 9. The second equality Sf (a) = (a : f n ) follows directly from the selection of f . 9, any ideal of the form S −1 a has primary decomposition m S −1 a = S −1 qi i=1 if n a= qi i=1 and p1 , p2 , .

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