Homework 11 Solutions
نویسنده
چکیده
38. Following the hint, suppose I is a non-zero ideal (since the zero ideal, (0), is clearly principal). Then there exists some a ∈ I with a 6= 0. Then, either a is positive, or −a is positive, and also in I since I is a subring of Z. Thus, I contains positive elements. Now, let S = {x ∈ I |x > 0}. By the Well-Ordering Property of Z, S contains a smallest element, call it c. We claim that, in fact, (c) = I. To prove this, we will show that each set (c) and I contains the other. First, since c ∈ I, every multiple of c is also in I, so (c) ⊂ I. On the other hand, let a be any element of I. Then applying the Division Algorithm to divide a by c, we find that there exist q and r in Z with a = cq + r and 0 ≤ r < c. So, r = a − cq. Suppose r 6= 0. Then c ∈ I implies that cq ∈ I, and a ∈ I, so r = a − cq ∈ I. But r is a positive integer, smaller than c, in I, contradicting the choice of c as the minimal positive integer in I. Therefore, r = 0, so a = cq, which means that a is a multiple of c, i.e. a ∈ (c). Thus, I ⊂ (c). And since (c) ⊂ I and I ⊂ (c), I = (c), and is a principal ideal.
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تاریخ انتشار 2010