Introduction to Abstract Algebra: with Applications

Let \(D\) be a Euclidean domain and let \(\nu\) be a Euclidean valuation on \(D\text{.}\) Suppose \(I\) is a nontrivial ideal in \(D\) and choose a nonzero element \(b \in I\) such that \(\nu(b)\) is minimal for all \(a \in I\text{.}\) Since \(D\) is a Euclidean domain, there exist elements \(q\) and \(r\) in \(D\) such that \(a = bq +r\) and either \(r=0\) or \(\nu(r) \lt \nu(b)\text{.}\) But \(r = a - bq\) is in \(I\) since \(I\) is an ideal; therefore, \(r = 0\) by the minimality of \(b\text{.}\) It follows that \(a = bq\) and \(I = \langle b \rangle\text{.}\)

(1) Suppose that \(a \mid b\text{.}\) Then \(b = ax\) for some \(x \in D\text{.}\) Hence, for every \(r\) in \(D\text{,}\) \(br =(ax)r = a(xr)\) and \(\langle b \rangle \subset \langle a \rangle\text{.}\) Conversely, suppose that \(\langle b \rangle \subset \langle a \rangle\text{.}\) Then \(b \in \langle a \rangle\text{.}\) Consequently, \(b =a x\) for some \(x \in D\text{.}\) Thus, \(a \mid b\text{.}\)

(2) Since \(a\) and \(b\) are associates, there exists a unit \(u\) such that \(a = u b\text{.}\) Therefore, \(b \mid a\) and \(\langle a \rangle \subset \langle b \rangle\text{.}\) Similarly, \(\langle b \rangle \subset \langle a \rangle\text{.}\) It follows that \(\langle a \rangle = \langle b \rangle\text{.}\) Conversely, suppose that \(\langle a \rangle = \langle b \rangle\text{.}\) By part (1), \(a \mid b\) and \(b \mid a\text{.}\) Then \(a = bx\) and \(b = ay\) for some \(x, y \in D\text{.}\) Therefore, \(a = bx = ayx\text{.}\) Since \(D\) is an integral domain, \(x y = 1\text{;}\) that is, \(x\) and \(y\) are units and \(a\) and \(b\) are associates.

(3) An element \(a \in D\) is a unit if and only if \(a\) is an associate of \(1\text{.}\) However, \(a\) is an associate of \(1\) if and only if \(\langle a \rangle = \langle 1 \rangle = D\text{.}\)

Suppose that \(\langle p \rangle\) is a maximal ideal. If some element \(a\) in \(D\) divides \(p\text{,}\) then \(\langle p \rangle \subset \langle a \rangle\text{.}\) Since \(\langle p \rangle\) is maximal, either \(D = \langle a \rangle\) or \(\langle p \rangle = \langle a \rangle\text{.}\) Consequently, either \(a\) and \(p\) are associates or \(a\) is a unit. Therefore, \(p\) is irreducible.

Conversely, let \(p\) be irreducible. If \(\langle a \rangle\) is an ideal in \(D\) such that \(\langle p \rangle \subset \langle a \rangle \subset D\text{,}\) then \(a \mid p\text{.}\) Since \(p\) is irreducible, either \(a\) must be a unit or \(a\) and \(p\) are associates. Therefore, either \(D = \langle a \rangle\) or \(\langle p \rangle = \langle a \rangle\text{.}\) Thus, \(\langle p \rangle\) is a maximal ideal.

Let \(p\) be irreducible and suppose that \(p \mid ab\text{.}\) Then \(\langle ab \rangle \subset \langle p \rangle\text{.}\) By Corollary 7.18, since \(\langle p \rangle\) is a maximal ideal, \(\langle p \rangle\) must also be a prime ideal. Thus, either \(a \in \langle p \rangle\) or \(b \in \langle p \rangle\text{.}\) Hence, either \(p \mid a \) or \(p \mid b\text{.}\)

We claim that \(I= \bigcup_{i = 1}^\infty I_i\) is an ideal of \(D\text{.}\) Certainly \(I\) is not empty, since \(I_1 \subset I\) and \(0 \in I\text{.}\) If \(a, b \in I\text{,}\) then \(a \in I_i\) and \(b \in I_j\) for some \(i\) and \(j\) in \({\mathbb N}\text{.}\) Without loss of generality we can assume that \(i \leq j\text{.}\) Hence, \(a\) and \(b\) are both in \(I_j\) and so \(a - b\) is also in \(I_j\text{.}\) Now let \(r \in D\) and \(a \in I\text{.}\) Again, we note that \(a \in I_i\) for some positive integer \(i\text{.}\) Since \(I_i\) is an ideal, \(ra \in I_i\) and hence must be in \(I\text{.}\) Therefore, we have shown that \(I\) is an ideal in \(D\text{.}\)

Since \(D\) is a principal ideal domain, there exists an element \(\overline{a} \in D\) that generates \(I\text{.}\) Since \(\overline{a}\) is in \(I_N\) for some \(N \in {\mathbb N}\text{,}\) we know that \(I_N = I = \langle \overline{a} \rangle\text{.}\) Consequently, \(I_n = I_N\) for \(n \geq N\text{.}\)

Let \(D\) be a PID. We want to use the criterion from Theorem 6.12. By Corollary 7.22, every irreducible in \(D\) is prime, and so it suffices to show the existence of factorizations into irreducibles in \(D\text{.}\)

Let \(a\) be a nonzero element in \(D\) that is not a unit. If \(a\) is irreducible, then we are done. If not, then there exists a factorization \(a = a_1 b_1\text{,}\) where neither \(a_1\) nor \(b_1\) is a unit. Hence, \(\langle a \rangle \subset \langle a_1 \rangle\text{.}\) By Lemma 7.20, we know that \(\langle a \rangle \neq \langle a_1 \rangle\text{;}\) otherwise, \(a\) and \(a_1\) would be associates and \(b_1\) would be a unit, which would contradict our assumption. Now suppose that \(a_1 = a_2 b_2\text{,}\) where neither \(a_2\) nor \(b_2\) is a unit. By the same argument as before, \(\langle a_1 \rangle \subset \langle a_2 \rangle\text{.}\) We can continue with this construction to obtain an ascending chain of ideals

By Lemma 7.23, there exists a positive integer \(N\) such that \(\langle a_n \rangle = \langle a_N \rangle\) for all \(n \geq N\text{.}\) Consequently, \(a_N\) must be irreducible. We have now shown that \(a\) is the product of two elements, one of which must be irreducible.

Now suppose that \(a = c_1 p_1\text{,}\) where \(p_1\) is irreducible. If \(c_1\) is not a unit, we can repeat the preceding argument to conclude that \(\langle a \rangle \subset \langle c_1 \rangle\text{.}\) Either \(c_1\) is irreducible or \(c_1 = c_2 p_2\text{,}\) where \(p_2\) is irreducible and \(c_2\) is not a unit. Continuing in this manner, we obtain another chain of ideals

This chain must satisfy the ascending chain condition; therefore,

for irreducible elements \(p_1, \ldots, p_r\text{.}\)