Math330 Solutions HW 11 Fall 2008 13.6. There are many examples. One is the variable x in the polynomial ring Z[x]. Another is the integer 2 in the ring Z. 13.20. We want to find all units, zero-divisors, idempotents, and nilpotent elements in L Z3 Z6. 1. The units of Z3 are 1 and 2, since 1 × 1 = 1 and 2 × 2 = 1. Similarly, the units for L Z6 are the elements 1 and 5. So the units of Z3 Z6 are :(1, 1), (1, 5), (2, 1), (2, 5). 2. There are no zero divisors of Z3 but Z6 has three, the elements 2, 3, and 4. This means L that, for example, the pair (a, 2) is a zero divisor of Z3 Z6 where a is any element of Z3 (we can multiply by (0, 3)). The zero divisors are {(a, b)|a ∈ Z3, b ∈ {2, 3, 4}}. 3. Recall that an element of a ring is called idempotent if a2 = a. The idempotents of Z3 are the elements 0, 1 and the idempotents of Z6 are the elements 1, 3, 4. So the L idempotents of Z3 Z6 are {(a, b)|a = 0, 1; b = 1, 3, 4}. 4. An element of a ring is called nilpotent if an = 0 for some integer n. The only L nilpotent element in either Z3 or Z6 is 0, so the only nilpotent element of Z3 Z6 is (0, 0). √ √ a−b k 13.36 In the ring Zp[k] the inverse of (a + b k) is a2−b2k which is defined if and only if a2 − b2k 6= 0. 13.44 We need an example of an infinite integral domain with characteristic 3. We don’t know that many examples of infinite integral domains, so a good guess to start would be with the polynomial ring Z[x]. But this has characteristic zero. So we can consider the polynomial ring Z3[x]. This is an infinite integral domain (see page 241) and has characteristic 3. 14.4 The subring S = {(a, a)|a ∈ Z} is not an ideal of Z L Z . We use the ideal test on page 253. First (0, 0) is in S so S is nonempty. We check that (a, a) − (b, b) = (a − b, a − b) is in S whenever (a, a) and (b, b) are in S. This is true since a − b is an integer whenever a, b are integers. So S is a subring. Let’s show that S is not an ideal. Let r = (0, 1) then r(a, a) = (0, a) which is not in S whenever a 6= 0. Thus S fails the ideal test. 0 0 0 0 14.12 We use the ideal test of page 253. Let a1b1 + ... + anbn and a1b1 + ... + ambm be 0 0 0 0 two elements of AB. Then (a1b1 + ... + anbn) − (a1b1 + ... + ambm) = a1b1 + ... + anbn + 0 0 0 0 (−a1)b1 + ... + (−am)bm which is an element of AB. Also let r be any element in R. Then r(a1b1 + ... + anbn) = (ra1)b1 + ... + (ran)bn by distributive laws, and ra ∈ A since A is an ideal. Thus r(a1b1 +...+anbn) is in AB. Similarly, (a1b1 +...+anbn)r = a1(b1r)+...+an(bnr) is an element in AB. So by the ideal test, AB is an ideal. 1.