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**Example text**

Furthermore the functor is simple. For, if F is a nontrivial subfunctor of 5 M , then F(M) ^ 0 and F(N) = 0 if N is indecomposable and is not isomorphic to M. 2), we see that F(M) = SM(M). Thus SM is a simple functor. We can see that the converse is also true. 12) LEMMA. An object in mod(£) is simple as an object in Mod(£) if and only if it is isomorphic to SM for some indecomposable CM module M. PROOF: Let 5 ^ 0 € mod(C) be a simple object in Mod(C). 2), so that it has SM(M) as a quotient module.

We can see that fi < n. In fact, if// > n, then yM C xnM = y2M, hence M = 0. Take an element a in M with ya € x^M — x**+lM, and write ya = x^ft for some ft G M. Note that neither a nor /? is in xM. Actually, if ft € xM, then ya would be in x*i+l1 a contradiction. If a G xMj then x^ft € xyM C x^^M, hence 0 € xM, which is also a contradiction by the above. Notice also that xn~fiM C yM C x^M. In fact, from yM C x^M, we have xnM = y2M C yx^M, hence xn~fiM C yM. From the inclusion xn~fiM C x^M, we see that n — fi > fi, hence the following inequality: (0 0 < fi < (n - l)/2.

That is, each vertex in T has only a finite number of arrows ending in it or starting from it. PROOF: Let [M] be any vertex in F where M is an indecomposable CM module. 2) there is an AR sequence ending in M. It thus follows from the AR quivers 39 above remark that there are only a finite number of arrows ending in [M]. Similarly if M is not isomorphic to the canonical module KR, then there are only finite arrows from [M]. The problem is to show that the number of arrows into [R] or from [KR] is finite.