By Paul-Jean Cahen, Marco Fontana, Evan Houston, Salah-Eddine Kabbaj

Lawsuits of the second one foreign convention on Cumulative Ring concept held in June 1992 at Fes, Morocco. Paper.

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**Example text**

It follows that (0,), = f-l*(f3) for some f3 E Ext~Jn-1(Lc;:, k). But then (2')' is represented by (((0')') = f-l* o (*(jj) = 0, where jj E Ext~J2n-1(Lc;:,nn(k)) corresponds to f3. 0 7 Examples and diagrams Before proceeding further we pause here to consider some examples. Except in a few special cases, cohomology rings are very difficult to calculate. Fortunately, one of the special cases is that of an elementary abelian p-group. These groups play a major role in the cohomology theory, and so we want to look at their cohomology rings here.

Here is the result. 3 Let G = (g I gpn = 1), n > o. Then ifpn ifpn with deg 'Yi = i. Proof. The chain map for 'Y2 is given by Hence 'Y02 = 'Yi+2 for i ;? o. If pn degree l. = 2, > 2. = 2, then we get 'Y01 = 'Y2 : X i+ 2 ---; Xi, ao f---+ ao for i ;? O. 'Yi+1 for i ;? 0, by a similar chain map in It remains to be seen that 'Yf = 0 if pn > 2. For this we construct the chain map which covers 'Y1. In diagrammatic form it looks like this: Xo k • + + • X2 Xl • Xo k • + • 1 + • + • • 45 Section 7. Examples and diagrams Therefore 'h Xl ------+ ao SO 1'I f------+ apn_2 if pn > 2.

D Modules and group algebras 38 For M = M' = N = N' = k we get the next corollary. 9 H*(G, k) = Extkdk, k) is a graded commutative ring. Remark The ring ExtkC(M, M) is in general not graded commutative. In fact, ExtgdM, M) = Homkc(M, M) is in general not commutative. 10 Le, is well-defined up to isomorphism. Proof. 5. In fact, the module Le, is the third object in the triangle for the morphism (. Hence the isomorphism class of Le, in kCstmo() is determined by (. Since nn(k) has no nonzero projective summands, Le, is in fact defined up to isomorphism in kCmo() by (.