By Yves Diers

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**Additional info for Categories of Boolean Sheaves of Simple Algebras**

**Example text**

Suppose 8* is not onto 4' 62, and let Lo be in Q\8*,b. We choose h in C(l2) such that h is 1 in a neighbourhood of A0 and h(8*4) _ {0 }, and a non-zero g in C(1) such that hg = g. Since h(6*+) = {0 } it follows that (8h)"(f) = {0 1 so that Oh is in the radical of B. From (1 - 8(h))8(g) = 0 we obtain 8(g) = 0. This contradicts 6 being a monomorphism. Hence 0*1 = 62. If f is in C(62), then IlfII = suplf(62) I = suplf(9*l) I = supl(Of)(l) I = spectral radius of Of < II Of II. This completes the proof.

Therefore hZ = Z for all non-zero h in D because D = D-1P. The that Z is D divisible. If f torsion free property also follows from the relation D = D-1P, and the proof is complete. We apply this theorem to obtain [28, Theorem 2]. There are many continuous linear operators T with the properties required in the example. Any quasinilpotent operator T on a Banach space X such that (TX) = X and T is one-to-one is a suitable candidate (see §3, [68], [1], or [28]). 8. 8. Example. Let 0 be the algebra of germs of analytic functions on the closure O of the open unit disc 0 = I IA EC : I p I < 1 and let 6) be the subalgebra of 0 of polynomials in one variable.

V n 1 (b) R = R ®... ® Rn where R. (C(ft))) _ , (c) RJRk = {0 } if j # k, (d) the restriction of P. (J({Aj}))= {0}. Proof. (i) This follows from Theorem 9. 3(i) by regarding C(Q) as a Banach C(62)-module and B as a C(62)-module by f. b = 6(f). b for all f in C(62) and b in B. (ii) Choose fl, ... , fn in C(62) so that fjfk = 0 if j * k and is equal to 1 in a neighbourhood of Xj for each j. Let D = Cf1 ® ... ® Cfn G J(F). Then D is a subalgebra of C(62) because = fj modulo J(F). Let µ0 be the restriction of 6 to D.