By Reiner I. (ed.)

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0 n with n = proj dim M . Since depth R = 0, the maximal ideal is in Ass R . e. 4, 'n maps Fn isomorphically onto a free direct summand of Fn;1, in contradiction to proj dim M = n. Therefore n = 0, and furthermore depth M = depth R = 0 since M is a free R -module. Let now depth R > 0. Suppose rst that depth M = 0. 9 yields depth M1 = 1 for a rst syzygy M1 of M . Since proj dim M1 = proj dim M ; 1, it is enough to prove the desired formula for M1 . Thus we may assume depth M > 0. Then 2= Ass R and 2= Ass M .

M Proof. For the implication (a) ) (b) it is enough to write every homogeneous element y 2 R as a polynomial in x1 . . xn with coe cients in R0 , and this is very easy by induction on deg y. The rest is evident. 4 holds for graded rings in general. 5. Let R be a graded ring. Then the following are equivalent: (a) every graded ideal of R is nitely generated (b) R is a Noetherian ring (c) R0 is Noetherian, and R is a nitely L generated R0-algebra L (d) R0 is Noetherian, and both S1 = 1i=0 Ri and S2 = 1i=0 R;i are nitely generated R0-algebras.

Let #a denote the multiplication by a on K (f ), and x the left multiplication by x on K (f ). Then #a = df x + x df as is easily veri ed. Thus multiplication by a is null-homotopic on K (f ). Of course #a M and HomR (#a M ) are the multiplications by a on K (f M ) and K (f M ), and the rest of (a) follows immediately. Part (b) is a general fact: if ' is a null-homotopic complex homomorphism, then the map induced by ' on homology is zero. For (c) we choose a = 1, and apply (a) and (b). Let L1 and L2 be R -modules, and f1 : L1 !