# A First Course in Rings and Ideals by David M. Burton

By David M. Burton

This quantity is designed to function an advent to the fundamental rules and methods of ring concept. it's meant to be an expository textbook, instead of a treatise at the topic. The mathematical historical past required for a formal knowing of the contents isn't wide. We suppose that the typical reader has had a few past touch with summary algebra yet remains to be particularly green during this appreciate. as a result, approximately every little thing herein might be learn through anyone accustomed to uncomplicated group-theoretic ideas and having a nodding acquaintance with linear algebra. the extent of fabric may still turn out appropriate for complex undergraduates and starting graduate scholars.

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**Example text**

If the ele~ent a happens to have a multiplicative i~verse, t~en t~e 1 regularity condition is satisfied by setting a' = a- ; in Vlew of thlS, a' IS . ~ ' I r Definition 2-7. Let R and R' be two rings. '. for every pair of elements a, b E R. A homomorphísm which is also one-to-one as a map on the underlying sets is called an isomorphism. We emphasize that the + arid . occurring on the left-hand sides of the equations in Definition 2-7 are those of R, whereas the + and . occurring on the right-hand sídes are those of R'.

A straightforward caIculation, wruch we omit, shows that R x {O} is isomorphic to the given ring R under the mapping f: R -¡. R x {O} defined by fea) = (a, O). identity. g(a) 33 IDEALS AND THEIR OPERATIONS + b)u) f(au + bu) + f(bu) = g(a) + g(b). g(ab) f(abu) = f(abu Z) = f(au)(bu») = f(au)f(bu) = g(a)g(b). The crucial third equality is justified by the fact that u E cent R, hence, commutes with b. As regards the uniqueness assertion, let us assume that there is another homomorphic extension offto the set R'i9all it h.

D), 34 PROBLEMS FIRST COURSE IN RINGS AND IDEALS therefore regard R as being an ideal of the ring R'. 1 J fór a suitable ideal J ~f R'. It is thus possible to choose an element (e, n) in J so that (O, -1) = (r, O) + (e, n), for sorne r E R. The last-written equation tells us that e = - r and n = -1; what is important is the resulting conc1usion that (e, -1) E J. For arbitrary rE R, the product (r, O){e, -1) = (re - r, O) will consequently be in both R and J (each being an ideal of R'). The fact that R n J = {O} forces (re - r, O) = (O, O); hence, re = r.