David M. Burton's A First Course in Rings and Ideals PDF

By David M. Burton

ISBN-10: 0201007312

ISBN-13: 9780201007312

This quantity is designed to function an advent to the elemental principles and strategies of ring concept. it truly is 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 really huge. We think that the common reader has had a few earlier touch with summary algebra yet continues to be really green during this recognize. as a result, approximately every thing herein could be learn via anyone acquainted with simple 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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Suppose that f is a homomorphism from the ring R onto the ring R'. We have already observed that each ideal l of the ring R determines an ideal f(I) of the ring R'. It goes without saying that ring theory would be considerably simplified ifthe ideals of R were in a one-to-one correspondence with those of R' in this manner. Unfortunately, this need not be the case. The difficulty is refiected in the fact that if l and J are two ideals of R with l ~ J ~ l + kerJ, thenf(I) = f(J). The quickest way to see this is to notice tbat f(l) ~ f(J) ~ f(l + kerf) = f(I) + f(kerf) 31 IDEALS AND THEIR OPERA TIONS FIRST COURSE IN RINGS AND IDEALS = f(l), from which we conclude that all the inclusions are actually equalities.

Conslder the sets Although Definition 2-6 appears to have a somewhat artificial air, we might remark that the set of alllinear transfonrtations on a finite dimensional vector space over a field forms a regular ring (Problem 20, Chapter 9). This in itself would amply justify the study of such rings. 1 = (filfE R;f(O) = O}, J = {ji2 + ni21fE R;f(O) = O; n E Z}, We now turn our attention to functions between rings arid, more specifically, to functions which preserve both the ring operations. l: r ~,J¡ = fa E Rla(~ J;)~ 1} = {q:E RlaJ¡' ~ Ifor all i} n (1 :r 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. In a like fashion, we obtain er = r, proving that R admits the element e as an identity. (that is, the set consisting of aH n-tuples with zeroes in all places but the ith) forms an ideal of R naturally isomorphic to R¡ under the mapping which sends (O, ... , O, a¡, O, ... ,O) to the element a¡. Since (al' a 2 , ••• , a n ) = (al' O, O, ...

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

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