MAT 3343, APPLIED ALGEBRA, FALL 2003

Answers to Problem Set 1 (due Sept. 16)

Problem 1, Handout 1

(a) (2) Suppose a ∈ I ∩ J and k ∈ ℤ. We want to show that ka ∈ I ∩ J. By assumption, we know that a ∈ I and a ∈ J. Therefore, by property (2) of I and J, ka ∈ I and ka ∈ J. Therefore ka ∈ I ∩ J.

(3) Since I is an ideal, we know 0 ∈ I by property (3) of I. Similarly 0 ∈ J by property (3) of J. It follows that 0 ∈ I ∩ J.

(b) (1) Suppose a, b ∈ I + J. We want to show that a + b ∈ I + J. We know that a = a_{1} + a_{2}, for some a_{1} ∈ I and a_{2} ∈ J. Similarly, b = b_{1} + b_{2} for some b_{1} ∈ I and b_{2} ∈ J. It follows that a + b = (a_{1} + a_{2}) + (b_{1} + b_{2}) = (a_{1} + b_{1}) + (a_{2} + b_{2}) ∈ I + J by property (1) of I and J.

(2) Suppose a ∈ I + J and k ∈ ℤ. We want to show that ka ∈ I + J. But a = a_{1} + a_{2} for some a_{1} ∈ I and a_{2} ∈ J, therefore ka_{1} ∈ I and ka_{2} ∈ J by property (2) of I and J. Thus ka = ka_{1} + ka_{2} ∈ I + J, as desired.

(3) We want to show that 0 ∈ I + J. But 0 ∈ I and 0 ∈ J, thus 0 = 0 + 0 ∈ I + J.

Problem 2, Handout 1 Suppose n|m and m|p. By definition of divisibility, there are integers k and l such that nk = m and ml = p. It follows that n(kl) = (nk)l = ml = p, and thus n|p.

Problem 3, Handout 1 “⇒”: Suppose aℤ ⊆ bℤ. Because a = a · 1, we have a ∈ aℤ, hence a ∈ bℤ. This means that a = bk for some k ∈ Z, hence b|a.

“⇐”: Suppose b|a. By definition of divisibility, we have a = bl for some l ∈ ℤ. We want to show that aℤ ⊆ bℤ. So take any element x ∈ aℤ. Then x = ak for some k ∈ ℤ. It follows that x = ak = blk, hence x ∈ bℤ.

Problem 4, Handout 1 (a) If we drop the word “non-empty”, the principle is not valid. Counterexample: the empty set ∅ has no least element. (b) If we drop the word “positive”, the principle is not valid. Counterexample: the set ℤ of all integers has no least element. (c) If we replace “integer” by “rational number”, the principle is not valid. Counterexample: the set A = {x ∈ ℚ | x > 0} has no least element. Proof: take any x ∈ A. Then x/2 ∈ A as well. Since x/2 < x, x is not a least element. (d) If we replace the word “positive” by “non-negative”, the principle is valid. Proof: Let A be any non-empty set of non-negative integers. Then either 0 ∈ A, in which case 0 is the least element of A. Otherwise, 0 ∉ A, in which case A is a non-empty set of positive integers and the original wellfoundedness principle applies.

Problem 1.2 # 7 (a) Suppose d > 0, d|(11k+4), and d|(10k+3) for some integer k. Then d|10(11k + 4) + (−11)(10k + 3) by Theorem 2(4), p.37. Therefore d|110k + 40 − 110k − 33, hence d|7. Since 7 is prime, we have d = 1 or d = 7.

(b) Suppose d > 0, d|(35k + 26), and d|(7k + 3) for some integer k. Then d|1(35k + 26) + (−5)(7k + 3) by Theorem 2(4), p.37. Therefore d|35k + 26 − 35k − 15, hence d|11. Since 11 is prime, it follows that d = 1 or d = 11.

Problem 1.2 #16 Suppose n|k(n + 1). Then there exists a ∈ ℤ such that na = k(n + 1), hence na = kn + k, hence n(a − k) = k, hence n|k.