By Toma Albu, Robert Wisbauer (auth.), S. K. Jain, S. Tariq Rizvi (eds.)

"[The booklet] exhibits advancements in lots of topics of this very lively box of earrings and modules and it includes a wealth of recent principles, innovations and effects introduced by way of essentially the most very important researchers within the field..."

**--Mathematica**

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**Extra resources for Advances in Ring Theory**

**Example text**

Preliminaries We start with the following definition. 1. We say R is a right (resp. left) FI (faithful idempotents) ring if e E R such that e = eZ i= 1 and r(eR) = 0 (resp. f(Re) = 0) implies HomR(eR, (1 - e)R) i= 0 (resp. HomR(Re, R(1 - e)) i= 0). If R is both a left and a right FI ring, then we say R is an FI ring. 2. The following conditions are equivalent: (i) R is a right PI ring. (ii) If e = eZ E R such that r(eR) = 0 and HomR(eR, (1 - e)R) = 0, then e = 1. (iii) If 0 i= e = eZ E R such that (1 - e)R contains no nonzero ideals, then either e = 1 or HomR(eR, (1 - e)R) i= O.

Thus, in this case, 4> = 4>i and consequently n 4>(2)ak ekk + bke~k+1)) = 4>i( ae ii + beTi+1) k=l n aeill)akekk + he~k+1))' k=l Assume, now, that I i+1 -# o. Define 4>: : Ji ---+ Q as follows: if b E J i and 4>i (aeii + beTH 1) = peii + qeTH 1 for some a E Ii, then set 4>: (b) = q. The map 4>: is well defined, for if, aeii E (I, J)i and 4>( aeii) = peii + qeTi+l then, as aeiiei+1i+l = 0, we have q = o. 4>: is clearly a K-homomorphism. Since Q is injective as K -module, there exist a E Q such that 4>: (b) = ab for all b E 1;.

WIEGANDT where Ua; is a neighborhood of Ya; for i = 1, ... ,n. Let us choose an index 0:0 such that 0:0 :::: O:i for i = 1, ... ,n, and let Uao be a neighborhood of Yao such that 1raoa;(Uao) ~ Ua;, i = 1, ... ,no Now (n n Y n 1r;;ol(Uao ) ~ Y n 1r;;;l(UaJ) i=l is a neighborhood of (Ya)' Further, we have 1rao f(X) = fao(X) and fao(X) is dense in X ao . Hence 1rao f(X) n Uao i 0, and so also Thus every neighborhood of (Ya) intersects f(X), which means that f(X) is dense in Y, that is, f is an epimorphism in C.