By Andrew O Lindstrum

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**Example text**

2: This proves so would E 5. T We define Now f is a left R-mapping, show that s = f is the desired element° Fix b ~ B, and choose any r E I b • Ib~ ~ ~. rb~ E R, so rb e ~-I(R) • R = T. Thus rb~ = rbs, We have shown that b~0 - bs annihilates annihilator Then rb s R, and so r(b~0 - bs) = 0. I b ~ Ib~ ~ ~, which has zero in R~ by part 3) of the lemma. Thus b~ = bs, and we are done. i The next definition Definition: [51]. The e x t e n d e d c e n t r o i d C of R is the center of R~. It is not difficult elements is due to M a r t i n d a l e of R~ arising to see that C consists p r e c i s e l y from (R,R)-bimodule m a p p i n g s of those of some I c ~ to 41 R.

If rx = 0, for r s R, then r = 0. We claim that For rx = 0 gives rxR = (0) = rRx = rR(JxX), w h e r e Jx ~ ~ is an ideal such that 0 # Jx x % R. prime, it follows that r = 0. Now I c ~ since R is prime, define f: I ÷ R by rxf = r, all rx ~ I. S i m i l a r l y yx = i. Gin n is clearly a n o r m a l so we may f is w e l l - d e f i n e d rx = 0 implies r = 0; thus y = T E R~. so xy = i. Since R is M o r e o v e r R(xy-l) since = 0, and Thus x is a unit in R~. subgroup, as it is p r e c i s e l y the set of elements of G w h i c h b e c o m e inner on R~.

We show that R~ is not simple. 5, R~ and R2*H = S w o u l d be Morita However, Morita oH is a domain, ~'2 equivalent and Stafford to a domain. ly in this situation next Alternatively, that there is no element however, theorem is due to Osterburg [65], proved earlier by l~iiyashita i) t(R) of trace I. more can be said. although part I The 3) actually was [56]. is a simple ring, ideals equivalent. one could show direct- Let R be simple w i t h 1 and G outer. 9: we see Then and is the intersection of all of R G 2) R G is p r i m i t i v e 3) the fixed ring of the center of R is the center of the fixed ring; that is, Z(R) G = Z(R G) 26 proof: Let B be a non-zero G-invariant right A = {u e R*G since fB ~ A.

### Abstract Algebra (Holden-Day Series in Mathematics) by Andrew O Lindstrum

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