Math H113 Spring, 2003
Thu May 1 18:27:53 2003 : can you post solutions to 7.6.2, 8.1.6, and 8.2.4? thanks
Problem 2 of §7.6: I’ll take it as given that Ris a commutative ring (Exercise
15 of §1) and that we know about idempotents (Exercise 1 of §7.6). We can try
to do this exercise by induction on the number of elements of R. It might be
informative to know going in that the order of Ris a power of 2: This follows
from the observation that 2a= 0 for all a∈R, which we can prove by writing
4a= 4a2= (a+a)2=a+a. Since 2a= 0 for all a∈R,Ris naturally a
vector space over the field Z/2Z, so it’s isomorphic to (Z/2Z)n(for some n)
as an additive group. In the induction that we contemplate, we can start with
the case where Rhas 2 elements, in which case it’s clear that R≈Z/2Zas a
ring. Now suppose that Rhas more than 2 elements, and pick e∈Rdifferent
from 0, 1. Then eis an idempotent, so Ris the product of Re and R(1 −e)
by Exercise #1. Both factors are non-zero; indeed, they contain the non-zero
elements eand 1 −e, respectively. Hence each factor has fewer elements than R.
By induction, each factor is isomorphic as a ring to a product of copies of Z/2Z,
so the same statement is true for R.
Problem 6 of §8.1: We are given relatively positive integers aand band wish to
study the set of integers of the form an +bm with nand mnon-negative. We’re
supposed to be able to get all integers greater than ab −a−band not ab −a−b.
(No information is requested on integers smaller than ab −a−b.) Equivalently,
we can study the set of integers an +bm with nand mpositive; these are gotten
by adding a+bto the integers an +bm with nand mnon-negative. This second
way of doing things seems promising because translating ab −a−bup by a+b
turns it into the simpler-looking ab. We have to show that ab is not of the form
an +bm (with nand mpositive) but that integers bigger than ab are of this
form.
If an +bm =ab, then bdivides an, so it divides nbecause it’s prime to a. Thus
nis a multiple of b. Similarly mis a multiple of a. Since we are requiring n
and mto be positive, an is at least as big as ab, and so is bm. Hence an +bm
is at least 2ab and can’t be ab.
Assume now that dis bigger than ab. Because aand bare relatively prime, there
are integers xand yso that ax +by = 1. Clearly one of ax, by is positive and
