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Extra info for Handbook of Categorical Algebra 2: Categories and Structures

Sample text

Conditions (2) and (3) just state the completeness of Sunder composition. 8 implies the next G. 10 (Diamond Lemma). Let S ~ F(X) \ {OJ. Assume (X) to be a semigroup partially ordered by ~, meeting the minimality condition, and such that for any s E S there exists a leading word s in s, s having the coefficient 1. Then the following are equivalent: (a") The algebm R = F(X; s = 0 (s E S)) is a free F-module and the image (under the injective map) of the set M consisting of the words of < X > without any occurrences of subwords s, s E S, forms the basis of this F-module.

35. For n > 3 or m > 2, C = CF(n, m, 0) is not weakly noethe- nan. Proof. Consider the following cases. Case 1. Suppose m > 2, then a < c < b. Put Xt = b(ac)tba n- 1 and verify that the ideal generated by Xk'S, 1 ~ k < t, does not contain Xt. This will prove the theorem. By way of contradiction, suppose that t-l Xt = L L /iUiXkVi, (4) k=1 iEh where Ii E F, Ui, and Vi are words and h's, 1 ~ k < t, are finite sets. Since the defining relations of C are homogeneous, Ui and Vi can be considered as words in a and c.

B - a51ds~b E Iv (v = a51ds2b). b. It follows that uS2b, aSlw < v, and so the difference is contained in Iv. , v = a51ds 2b, al = a, bl = ds 2b, a2 = a51d and b2 = b. The following lemma is clear. 6. Let v E G and a, b E G#. Then aIvb ~ I avb . 7. This lemma shows that if condition (1) is met by some Si, ai, and bi (i = 1,2), then it is also satisfied by Si, aai, and bib (i = 1,2 and a,b E G#). s2. In this case, (1) expresses the completeness of 5 under composition. 4 is exactly our Composition Lemma for the semigroup algebra FG.