3.2: Factorization in Euclidean Domains
In this section, we'll seek to answer the questions:
- What is a unique factorization domain? What examples of UFDs do we possess?
- What is the ascending chain condition on ideals? What are Noetherian rings?
- What does the ascending chain condition have to do with unique factorization?
In this section, our explorations of the structural arithmetic properties that guarantee unique factorization culminate in Theorem 3.2.7 . Specifically, we'll see that all Euclidean domains possess the unique factorization property. To prove this theorem, we will rely in part on an interesting property of chains of ideals in Euclidean domains.
3.2.1 Unique Factorization Domains
We begin by describing exactly what we mean by unique factorization. The reader may find it helpful to compare Definition: Unique Factorization Domain to The Fundamental Theorem of Arithmetic.
An integral domain \(R\) is called a unique factorization domain (or UFD ) if the following conditions hold.
- Every nonzero nonunit element of \(R\) is either irreducible or can be written as a finite product of irreducibles in \(R\text{.}\)
-
Factorization into irreducibles is unique up to associates. That is, if \(s\in R\) can be written as
\begin{equation*} s = p_1 p_2 \cdots p_k \text{ and } s = q_1 q_2 \cdots q_m \end{equation*}
for some irreducibles \(p_i, q_j\in R\text{,}\) then \(k=m\) and, after reordering, \(p_i\) is an associate of \(q_i\text{.}\)
Using \(\mathbb{Z}\) as an example, illustrate the definition of UFD by factoring 20 into two sets of different irreducibles which nonetheless can be paired up as associates.
We are already familiar with several examples.
The integers \(\mathbb{Z}\) form a UFD.
Every field is a UFD.
3.2.2 The Ascending Chain Condition and Noetherian Rings
We now set our sights on a proof of Theorem 3.2.7 . In order to prove it, we will make use of an important property of ideals in Euclidean domains. First, a definition.
A commutative ring \(R\) is called Noetherian if it satisfies the ascending chain condition on ideals.
That is, \(R\) is Noetherian if whenever
\begin{equation*} I_1 \subseteq I_2\subseteq I_3\subseteq \cdots \end{equation*}
is an ascending chain of ideals in \(R\text{,}\) then there exists some \(n\) for which \(I_n = I_{n+1} = I_{n+2} = \cdots\text{.}\)
\(\subseteq\): These rings are named in honor of Emmy Noether , one of the preeminent mathematicians of the 20th century. In addition to making substantial contributions to physics, she formalized the axiomatic definition of ring that is still in use today.
Consider the ideals \(I_1 = \langle 30 \rangle\) in \(\mathbb{Z}\) and \(J_1 = \langle 32 \rangle\text{.}\) Find the longest ascending chains of ideals starting first with \(I_1\) and then with \(J_1\) that you can. When does each chain stabilize?
We next show that every PID is Noetherian.
Every principal ideal domain is Noetherian.
- Hint
-
Let \(I_1\subseteq I_2\subseteq I_3 \subseteq \cdots\) and set \(I = \cup I_j\text{.}\) Show that \(I\) is an ideal, and use your assumptions!
Every Euclidean domain is Noetherian.
3.2.3 Euclidean Domains are UFDs
We now begin collecting results to prove that every Euclidean domain is a UFD. The first condition in the UFD definition is that every nonzero nonunit factors as a product of irreducibles. We first show that every nonzero nonunit is divisible by at least one irreducible (Lemma 3.2.1 ), which we apply to show that every nonzero nonunit can be written as a finite product of irreducibles (Theorem 3.2.4 ).
Let \(R\) be a principal ideal domain, and \(r\in R\) a nonzero nonunit. Then \(r\) is divisible by an irreducible.
- Hint
-
Let \(r\in R\) be reducible and write \(r = r_1 r_2\text{.}\) Continue to factor reducibles and build an ascending chain of ideals.
Let \(R\) be a PID. Then every nonzero nonunit element of \(R\) is either irreducible or can be written as a finite product of irreducibles in \(R\text{.}\)
The second condition that must be satisfied for a domain to be a UFD is that the product of irreducibles must be unique (up to associates). In order to prove that, we will make use of Theorem 3.2.5 , which states that in PIDs, primes and irreducibles are identical concepts.
Let \(R\) be a PID and let \(p\in R\) be irreducible. Let \(a\in R\) be such that \(p\nmid a\text{.}\) Then \(1\in I = \{ax+py : x,y\in R\}\) and thus there exist \(s,t\in R\) such that \(1 = as+pt\text{.}\)
Let \(R\) be a PID and let \(p\in R\text{.}\) Then \(p\) is prime if and only if \(p\) is irreducible.
Observe that Theorem 3.2.5 implies that if \(R\) is a PID and \(p\in R\) is irreducible with \(p|ab\text{,}\) then \(p|a\) or \(p|b\text{.}\)
Our crucial final step on the road to Theorem 3.2.7 is the following.
Every PID is a UFD.
- Hint
-
For part 2 of the definition, use induction on the number of irreducible divisors of an arbitrary nonzero nonunit. Mimic the proof of Theorem 1.3.4 .
Every Euclidean domain is a unique factorization domain.
Let \(F\) be a field. Then \(F[x]\) is a UFD.
That is, if \(f(x) \in F[x]\) with \(\deg(f(x)) \ge 1\text{,}\) then \(f(x)\) is either irreducible or a product of irreducibles in \(F[x]\text{.}\) What is more, if
\begin{equation*} f(x) = p_1(x) p_2(x) \cdots p_k(x) \text{ and } f(x) = q_1(x) q_2(x) \cdots q_m(x) \end{equation*}
are two factorizations of \(f\) into irreducibles \(p_i, q_j\text{,}\) then \(m=k\) and after reordering, \(p_j\) and \(q_j\) are associates.
- Hint
-
Handle existence and uniqueness separately. For each, (strong) induction on \(\deg(f(x))\) will work. Or do something entirely different.
Thus, we see that the existence of a well-behaved division algorithm and (a lack of zero divisors) is sufficient to guarantee unique factorization. However, it is not necessary. The following theorem is included for reference, but is not intended to be proved.
If \(R\) is a UFD, then \(R[x]\) is a UFD.
Thus, \(\mathbb{Z}[x]\) is a UFD. That is, every nonconstant polynomial in \(\mathbb{Z}[x]\) is either irreducible or can be factored uniquely into a product of irreducibles. However, as we will see later, \(\mathbb{Z}[x]\) is not a PID.