By David M Bressoud; S Wagon
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Extra resources for A course in computational number theory
The following statements are equivalent. (1) R is a field. (2) R contains no proper, nontrivial ideals. 52 1. PRELIMINARIES-GROUPS AND RINGS Proof. Let R be a field and I a nonzero ideal of R. Then I contains a nonzero element x. Since R is a field, x has an inverse x-I in R. But then, since I is an ideal of R, the element 1 = x X-I is in I. Now, for all r E R, r = r 1 E I so that I = R. Now suppose that R contains no proper, nontrivial ideals. Let x be a nonzero element of R. Then the principal ideal generated by x, Rx, is nonzero and hence must equal R.
R .... N .... t .... then, for each i, r == ri (mod nil as claimed. D. 6]. If nl, n2, . ,nu are pairwise relatively prime natural numbers, and n = nl n2 ... n u , then Proof. Define () : Z - Znl X Zn2 X ••• X Znu by (}(m) = (kl (m), k2(m), . 3] that () is a homomorphism of groups with kernel the subgroup nZ of Z. We must show that () is surjective. Let (kl(rI), k 2(r2), ... • X Znu with ri E Z for i = 1, ... / By the Chinese Remainder Theorem, there is an element r E Z such that, for all i, r == ri (mod ~).
Prove that 0 is a surjective homomorphism with kernel (l). 3J 1, ... 6], define 0 : Z -+ Z2 X Z5 X Z9 by Oem) = (m2, m5, m9) where mk denotes the equivalence class of m in the group Zk. Find an m E Z such that Oem) = (12, 25, (9). 6J Recall that a simple group is a group with no proper, nontrivial normal subgroups. Prove that, if G is a simple, nonabelian group, then G is not a solvable group. 1J Suppose that H and K are groups and that G == H x K. Prove that G is solvable if, and only if, both Hand K are solvable.
A course in computational number theory by David M Bressoud; S Wagon