By Allen Nussbaum

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It is elementary to show that any right Goldie domain has dimension 1. 14 THEOREM The following conditions on a ring R are equivalent: (1) R is semiprime right Goldie; (2) R is semiprime, Z(R) = 0 and dim R < oo; (3) R has a semisimple right quotient ring S. 14 until Chapter 4. There, the orders in simple Artin rings will be discussed at great length. The following concept plays a fundamental role both in the 28 Simple Noetherian rings study of orders in semisimple rings and the representation theory for simple rings.

B. 25(3) can be generated by 2 elements. Thus emerges an intimate connection between these rings and the classical Dedekind domains. 28 THEOREM (Webber [70]) Let A be a simple, right hereditary, right Noetherian ring. Let I, J and K be any three nonzero right ideals of A, J essential in A. Then, there exists a right ideal L suCh that I E8 K :::::; J E8 L. Proof If I :::::; J, we are finished. Otherwise, choose j 0 E J which is regular and set J 0 = j 0 I. Since I :::::; J 0 , if we set L 0 = K, I E8 K :::::; J 0 E8 L 0 • Simple Noetherian rings 36 Moreover, J 0 c J.

Next, for all r E R, tr - rt E I and has degree less than n. ). Moreover, if t = an + rn-lan-l + ... + ro, tr = rt Vr E R (I) implies that by equating the coefficients of both sides of (I). Thus, if n =F 0, ais inner, contradicting (b). ). , a(r) = ar - ra, Vr E R. ) . ). 1 1E I} Thus, (a) ~ (b). 2a THEOREM Suppose R is a commutative domain with the characteristic of R, char R, = p 0. )R is simple iff R is a field and dimcR is infinite, where C = constants a = {r E R I a(r) = 0}. Proof ~: First, a(rP) = 0 for all (nonunits) r E R.