Prove that the ring of integers is a principal ideal ring
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A ring R is a principal ideal domain(PID) if it is an integral domain(25.5) such that every ideal of R is a principal ideal. 27.5 Proposition. The ring of integers Z is a PID.
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We have shown that I is generated by a, and hence, Z is a principal ideal ring.
To prove that the ring of integers is a principal ideal ring, we need to show that every ideal in the ring is a principal ideal.
Let I be an ideal of the ring of integers, denoted by Z. We consider two cases:
Case 1: I = {0}
In this case, I is the trivial ideal of Z, and it is a principal ideal since it is generated by 0.
Case 2: I ≠ {0}
In this case, I contains a nonzero element. We choose a nonzero element a ∈ I with the smallest absolute value, i.e., if I contains a nonzero element b with |b| < |a|, then we replace a by b. We claim that I is generated by a, i.e., I = (a).
To prove this claim, we first show that (a) ⊆ I. Let b ∈ (a) be an arbitrary element.
Then there exists an integer k such that b = ka. Since a ∈ I and I is an ideal, we have ka ∈ I for all integers k.
Therefore, b ∈ I.
Next, we show that I ⊆ (a). Let c ∈ I be an arbitrary element.
We divide c by a using the division algorithm: c = qa + r, where q, r are integers such that 0 ≤ |r| < |a|.
Since a ∈ I and I is an ideal, we have qa ∈ I.
Therefore, r = c − qa ∈ I. However, by the choice of a, we have |r| < |a|. This contradicts the minimality of |a| unless r = 0. Hence, c = qa ∈ (a).
Therefore, we have shown that I is generated by a, and hence, Z is a principal ideal ring.
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