Theorem 7.19.
Every Euclidean domain is a principal ideal domain.
Section 7.3 Principal Ideal Domains
Let \(R\) be a commutative ring with identity. Recall that a principal ideal generated by \(a \in R\) is an ideal of the form \(\langle a \rangle = \{ ra : r \in R \}\text{.}\) An integral domain in which every ideal is principal is called a principal ideal domain, or PID.
Proof.
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 \(I = \langle b \rangle\text{.}\)
Lemma 7.20.
Let \(D\) be an integral domain and let \(a, b \in D\text{.}\) Then
- \(a \mid b\) if and only if \(\langle b \rangle \subset \langle a \rangle\text{.}\)
- \(a\) and \(b\) are associates if and only if \(\langle b \rangle = \langle a \rangle\text{.}\)
- \(a\) is a unit in \(D\) if and only if \(\langle a \rangle = D\text{.}\)
Proof.
(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 \(\langle a \rangle = \langle 1 \rangle = D\text{.}\)
Theorem 7.21.
Let \(D\) be a PID and \(\langle p \rangle\) be a nonzero ideal in \(D\text{.}\) Then \(\langle p \rangle\) is a maximal ideal if and only if \(p\) is irreducible.
Proof.
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.
Corollary 7.22.
Let \(D\) be a PID. If \(p\) is irreducible, then \(p\) is prime.
Proof.
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 \(p \mid b\text{.}\)
Lemma 7.23.
Let \(D\) be a PID. Let \(I_1, I_2, \ldots\) be a set of ideals such that \(I_1 \subset I_2 \subset \cdots\text{.}\) Then there exists an integer \(N\) such that \(I_n = I_N\) for all \(n \geq N\text{.}\)
Proof.
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 \(n \geq N\text{.}\)
Any commutative ring satisfying the condition in Lemma 7.23 is said to satisfy the ascending chain condition, or ACC. Such rings are called Noetherian rings, after Emmy Noether.
Theorem 7.24.
Every PID is a UFD.
Proof.
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
\begin{equation*}
\langle a \rangle \subset \langle a_1 \rangle \subset \langle a_2 \rangle \subset \cdots.
\end{equation*}
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
\begin{equation*}
\langle a \rangle \subset \langle c_1 \rangle \subset \langle c_2 \rangle \subset \cdots.
\end{equation*}
This chain must satisfy the ascending chain condition; therefore,
\begin{equation*}
a = p_1 p_2 \cdots p_r
\end{equation*}
for irreducible elements \(p_1, \ldots, p_r\text{.}\)
Remark 7.25.
It is important to notice that every Euclidean domain is a PID and every PID is a UFD. However, the converse of each of these statements fails, as demonstrated by the following example for the second statement.
Example 7.26.
Not every UFD is a PID. In Corollary 6.33, we will prove that \({\mathbb Z}[x]\) is a UFD. However, \({\mathbb Z}[x]\) is not a PID. Let \(I = \{ 5 f(x) + x g(x) : f(x), g(x) \in {\mathbb Z}[x] \}\text{.}\) We can easily show that \(I\) is an ideal of \({\mathbb Z}[x]\text{.}\) Suppose that \(I = \langle p(x) \rangle\text{.}\) Since \(5 \in I\text{,}\) \(5 = f(x) p(x)\text{.}\) In this case \(p(x) = p\) must be a constant. Since \(x \in I\text{,}\) \(x = p g(x)\text{;}\) consequently, \(p = \pm 1\text{.}\) However, it follows from this fact that \(\langle p(x) \rangle = {\mathbb Z}[x]\text{.}\) But this would mean that 3 is in \(I\text{.}\) Therefore, we can write \(3 = 5 f(x) + x g(x)\) for some \(f(x)\) and \(g(x)\) in \({\mathbb Z}[x]\text{.}\) Examining the constant term of this polynomial, we see that \(3 = 5 f(x)\text{,}\) which is impossible.
Subsection Historical Note
Amalie Emmy Noether, one of the outstanding mathematicians of the twentieth century, was born in Erlangen, Germany in 1882. She was the daughter of Max Noether (1844–1921), a distinguished mathematician at the University of Erlangen. Together with Paul Gordon (1837–1912), Emmy Noether’s father strongly influenced her early education. She entered the University of Erlangen at the age of 18. Although women had been admitted to universities in England, France, and Italy for decades, there was great resistance to their presence at universities in Germany. Noether was one of only two women among the university’s 986 students. After completing her doctorate under Gordon in 1907, she continued to do research at Erlangen, occasionally lecturing when her father was ill.
Noether went to Göttingen to study in 1916. David Hilbert and Felix Klein tried unsuccessfully to secure her an appointment at Göttingen. Some of the faculty objected to women lecturers, saying, “What will our soldiers think when they return to the university and are expected to learn at the feet of a woman?” Hilbert, annoyed at the question, responded, “Meine Herren, I do not see that the sex of a candidate is an argument against her admission as a Privatdozent. After all, the Senate is not a bathhouse.” At the end of World War I, attitudes changed and conditions greatly improved for women. After Noether passed her habilitation examination in 1919, she was given a title and was paid a small sum for her lectures.
In 1922, Noether became a Privatdozent at Göttingen. Over the next 11 years she used axiomatic methods to develop an abstract theory of rings and ideals. Though she was not good at lecturing, Noether was an inspiring teacher. One of her many students was B. L. van der Waerden, author of the first text treating abstract algebra from a modern point of view. Some of the other mathematicians Noether influenced or closely worked with were Alexandroff, Artin, Brauer, Courant, Hasse, Hopf, Pontryagin, von Neumann, and Weyl. One of the high points of her career was an invitation to address the International Congress of Mathematicians in Zurich in 1932. In spite of all the recognition she received from her colleagues, Noether’s abilities were never recognized as they should have been during her lifetime. She was never promoted to full professor by the Prussian academic bureaucracy.
In 1933, Noether, a Jew, was banned from participation in all academic activities in Germany. She emigrated to the United States, took a position at Bryn Mawr College, and became a member of the Institute for Advanced Study at Princeton. Noether died suddenly on April 14, 1935. After her death she was eulogized by such notable scientists as Albert Einstein.
