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Abstract algebra A Kursobjekt for Math 331

Aaron Greicius

Section 2.12 PIDs and UFDs

Subsection 2.12.1 UFDs

Proof.

Proof.

The proof of this result is identical to that of Theorem 2.8.10, which treats the special case of polynomial rings over a field. If you look there, you’ll notice that the only thing used is the fact that \(F[x]\) is a PID when \(F\) is a field.
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For convenience, however, we reprise some of the proof here in cursory form. Implication (1)\(\implies\)(2) is true in any commutative ring, as is (3)\(\implies\)(1). For the last link, (2)\(\implies\)(3), to show that \((p)\) is maximal if \(p\) is irreducible, we use the fact that \(R\) is a PID to assert that \((p)\subseteq I\) implies \((p)\subseteq (a)\) for some \(a\in R\text{.}\) But in that case \(a\mid p\text{,}\) in which case either \(a\) is a unit (and \((a)=R\)), or \(a\) is an associate of \(p\text{,}\) in which case \((a)=(p)\text{.}\)
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Example 2.12.3. \(\Z[\sqrt{-5}]\) is not a PID.

Let \(R=\Z[\sqrt{-5}]\text{.}\) Show that the elements \(2\) and \(3\) are irreducible, but not prime. Conclude that \(R\) is not a PID (nor a Euclidean domain).
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Solution.
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Definition 2.12.4. Unique factorization domain.

A unique factorization domain (UFD) is an integral domain \(R\) that satisfies the following properties.
  1. Factorization into irreducibles.
    If \(a\in R\) is nonzero and not a unit, then there are irreducible elements \(p_1,p_2,\dots, p_n\) in \(R\) satisfying
    \begin{align} a \amp = p_1p_2\cdots p_n\text{.}\tag{2.12.1} \end{align}
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  2. Uniqueness of factorization.
    The factorization (2.12.1) is unique in the following sense: if we have
    \begin{align*} a \amp =p_1p_2\cdots p_n=q_1q_2\cdots q_m\text{,} \end{align*}
    where the \(p_i\) and \(q_j\) are irreducible, then \(n=m\text{,}\) and after a reordering there are units \(u_i\in R^*\) such that \(q_i=u_ip_i\) for all \(1\leq i\leq n\text{.}\) In other words, in any factorization of \(a\) into irreducibles, the number of irreducible factors is constant, and the irreducible elements appearing in the factorization are unique up to multiplication by a unit.
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Proof.

Since (1)\(\implies\)(2) is true in any commutative ring, it suffices to show (2)\(\implies\)(1).
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Assume \(p\) is irreducible, and that \(p\mid ab\) for some elements \(a,b\in R\text{.}\) This means we have
\begin{align*} ab \amp =cp \end{align*}
for some \(c\in R\text{.}\) Let \(c=\prod_{i\in I}p_i\) be an irreducible factorization of \(c\text{,}\) so that
\begin{align*} ab \amp =p\prod_{i\in I}p_i \end{align*}
is an irreducible factorization of \(ab\text{.}\) This shows that \(p\) is (up to multiplication by a unit) one of the irreducible elements appearing in the factorization of \(ab\text{.}\) Since we can also obtain an irreducible factorization of \(ab\) by taking the product of the two irreducible factorizations of \(a\) and \(b\text{,}\) it follows that \(p\) must be associate to one of the irreducible factors of \(a\) or of \(b\text{,}\) and thus that \(p\mid a\) or \(p\mid b\text{.}\)
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Remark 2.12.6. Irreducible is prime in UFD.

According to Theorem 2.12.5, in a UFD, the notions of irreducible and prime elements are equivalent. Accordingly, we are tempted to treat the two terms as synonymous when working in a UFD. Since however, the two properties are not generally equivalent for integral domains, we will do our best to resist this temptation.
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Proof.

The proof is left as an exercise.
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Definition 2.12.8. Valuation at irreducible.

Let \(R\) be a unique factorization domain, and let \(p\) be an irreducible element of \(R\text{.}\) Given a nonzero element \(a\in R\text{,}\) the valuation of \(a\) at \(p\), denoted \(v_p(a)\text{,}\) is the largest nonnegative integer \(n\) satisfying \(p^n\mid a\text{.}\)
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Proof.

  1. Write \(r\sim s\) to denote that \(r\) and \(s\) are associates in \(R\text{.}\) This relation is easily seen to be an equivalence relation. Given an irreducible factorization
    \begin{align*} a \amp =\prod_{j=1}^s q_i\text{,} \end{align*}
    let \(p_1, p_2,\dots, p_r\) be any representatives of the distinct associate classes of the \(q_i\text{.}\) Sorting the \(q_i\) by their associate classes, we have
    \begin{align} a \amp =\prod_{j=1}^s q_i\tag{2.12.5}\\ \amp = \prod_{i=1}^r \prod_{q_j\sim p_i} q_i\tag{2.12.6}\\ \amp = \prod_{i=1}^r\prod_{q_j\sim p_i}u_{ij}p_i \tag{2.12.7}\\ \amp = u\prod_{i=1}^rp_i^{n_i}\text{,}\tag{2.12.8} \end{align}
    where \(u=\prod u_{ij}\) and \(n_i\) is the number of factors \(q_j\) that are associates of \(p_i\text{.}\)
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    Fix \(i\in \{1,2,\dots, r\}\) and write
    \begin{align*} a \amp =p_i^{n_i}\cdot \prod_{j\ne i}p_j^{n_j}\text{.} \end{align*}
    Since \(p_i\) is not associate to \(p_j\) for \(j\ne i\text{,}\) it follows that \(p_i\nmid \prod_{j\ne i}p_j^{n_j}\text{,}\) and hence that \(n_i=v_{p_i}(a)\text{.}\)
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  2. Given factorizations of \(a\) and \(b\) as specified, first observe that \(v_p(a)=v_p(b)=0\) for all \(p\) not associate to one of the \(p_i\text{.}\) It follows that \(d'\) is a common divisor of \(a\) and \(b\) if and only if \(v_{p_i}(d')\leq \min\{n_i, m_i}\) for all \(1\leq i\leq r\text{,}\) and \(v_p(d)=0\) for all \(p\) not associate to one of the \(p_i\text{,}\) if and only if \(d'\mid d\text{,}\) where
    \begin{align*} d \amp = \prod_{i=1}^rp_i^{n_i} \text{.} \end{align*}
    This shows \(d\) is a GCD of \(a\) and \(b\text{.}\) A similar argument shows that the given \(m\) is a LCM of \(a\) and \(b\text{.}\)
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Proof.

Proof of uniqueness.
The existence of irreducible factorizations is the trickier argument. Let’s look at uniqueness, assuming existence. Let \(a=\prod_{i\in I}p_i\) be an irreducible factorization of the nonzero, non-unit element \(a\text{.}\) We prove the uniqueness property of this factorization by induction on \(\abs{I}\text{.}\) If \(\abs{I}=1\text{,}\) then \(a=p\) is irreducible. Suppose we have another factorization
\begin{align*} p \amp =p_1p_2\cdots p_n\text{.} \end{align*}
First observe that we must have \(n=1\text{.}\) Otherwise, we would have \(p=p_1c\text{,}\) where \(p_1\) and \(c=\prod_{i\ne 1}p_i\) are both non-units, contradicting the fact that \(p\) is irreducible.
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Assume now that the uniqueness property holds for all irreducible factorizations into \(n\) irreducible factors, and assume \(a=\prod_{i=1}^{n+1}p_i\text{,}\) with \(n\geq 1\text{.}\) If we have another factorization
\begin{align*} a= \amp \prod_{i=1}^m q_i\text{,} \end{align*}
then by induction we must have \(m\geq n+1\text{.}\) Since \(p_1\) divides the product of the \(q_i\text{,}\) and since \(p_1\) is prime (irreducible implies prime in PIDs), \(p_1\) must divide one of the \(q_i\text{.}\) Without loss of generality, we may assume that \(p_1\mid q_1\text{.}\) Since \(q_1\) is irreducible and \(p_1\) is not a unit, we conclude that \(q_1=up_1\) for some unit. But then we can cancel \(p_1\) from both sides to obtain
\begin{align*} a' \amp = \prod_{i=2}^np_i=u\prod_{i=2}^{m+1}q_j\\ \amp =(uq_2)q_3\cdots q_m\text{.} \end{align*}
Since \(a'\) is a product of \(n\) irreducible elements, we may apply induction to conclude that \(n=m-1\text{,}\) and after a reordering, \(p_i\) is associate to \(q_i\) for all \(2\leq i\leq n\text{.}\) (Note that the presence of \(uq_2\) instead of \(q_2\) here is not an issue here, since being associates is an equivalence relation.) But then we have \(n+1=m\text{,}\) and \(p_i\) is associate to \(q_i\) for all \(1\leq i\leq n\text{,}\) as desired.
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Proof of existence.
Suppose, by contradiction that there is a nonzero, non-unit element \(a\) that does not admit an irreducible factorization. Since, in particular, \(a\) cannot be irreducible, we have \(a=bc\) for two non-unit elements \(b,c\in R\text{,}\) and at least one of \(b\) and \(c\) fails to admit an irreducible factorization (otherwise \(a\) would have an irreducible factorization). Call this element \(a_1\text{,}\) and proceed in the same way to produce a sequence \(a=a_0, a_1, a_2,\dots\) satisfying \(a_n\mid a_{n-1}\) and \(a_{n-1}\nmid a_n\) for all \(n\geq 1\text{,}\) and hence a chain of strictly proper inclusions
\begin{align*} (a) \amp \subsetneq (a_1)\subsetneq (a_2)\subsetneq \dots\text{.} \end{align*}
It is easy to see that \(I=\bigcup_{i=1}^\infty (a_i)\) is an ideal of \(R\text{.}\) Since \(R\) is a PID, we have \(I=(b)\) for some element \(b\in R\text{.}\) But then \(b\in (a_i)\) for some \(i\text{,}\) from whence it follows that \(I=(b)=(a_i)\text{,}\) and \((a_{i+1})=(a_i)\text{.}\) A contradiction! We conclude that every nonzero, non-unit element of \(R\) does admit an irreducible factorization.
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Note: it is possible to give a proof of existence that uses Zorn’s lemma. Can you produce it? The one we provide has the advantage of not using Zorn’s lemma, which you will recall is equivalent to the axiom of choice. If you look closely, however, you will see that our proof does involve some fishy choices to construct the sequence \((a_n)\text{.}\) Indeed, it is making use of what is called the axiom of dependent choice (DC), which is logically weaker than the full axiom of choice.
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Looking at the proof of the uniqueness of irreducible factorizations in PIDs, we see that the only property we used, assuming that factorizations exist, is that irreducible elements are prime. This is a useful shortcut for proving that a given integral domain is a UFD, and so we extract it as its own result.
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Subsection 2.12.2 PIDs and UFDs among quadratic rings of integers

In the past three sections we have established the chain of implications
\begin{align} \text{Euclidean domain} \amp \implies \text{PID} \implies \text{UFD}\implies \text{integral domain}\text{.}\tag{2.12.9} \end{align}
As always in mathematics, when confronted with a stated implication, we should ask whether the converse is also true. No implication in the chain above is an equivalence. Let’s assemble some counterexamples:
  • We have seen that the ring \(\Z[\sqrt{-5}]\) is integral, but not a UFD, since not all irreducible elements are prime.
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  • Soon we will see that the ring \(\Z[x]\) is a UFD, but it is not a PID: it is easy to see that \((2,x)\) is not a principal ideal.
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  • As mentioned earlier, the ring \(\Z[\omega]\text{,}\) where \(\omega=\tfrac{1}{2}+\tfrac{\sqrt{-19}}{2}\) is a PID, but not a Euclidean domain (see the Dummit and Foote text).
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Remark 2.12.12. Quadratic ring of integers.

Let \(D\) be a square-free integer. We have seen that \(F=\Q[\sqrt{D}]\) is a field. The ring of integers \(\mathcal{O}_{\sqrt{D}}\) of \(F\) is defined as \(\Z[\omega]\text{,}\) where
\begin{align*} \omega \amp = \begin{cases} \sqrt{D} \amp \text{if } D\not\equiv 1\pmod 4 \\ \tfrac{1}{2}+\tfrac{\sqrt{D}}{2} \amp \text{if } D\equiv 1\pmod 4 \text{.}\end{cases} \end{align*}
This family of rings provides a nice testing ground for exploring Euclidean domains, PIDs, and UFDs, in that the rings are fairly well behaved, and yield examples that distinguish between these various types of integral domains. Below you find a summary of what is currently known about the rings \(\mathcal{O}_{\sqrt{D}}\text{.}\)
  • For all square-free integers \(D\text{,}\) \(\mathcal{O}_{\sqrt{D}}\) is a PID if and only if it is a UFD. Thus we need not cast our net among this family of rings to find examples of UFDs that are not PIDs.
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  • If \(D\) is negative, then \(\mathcal{O}_\sqrt{D}\) is a PID if and only if \(D\in \{-1,-2,-3,-7,-11,-19,-43,-67,-163\}\text{,}\) an \(\mathcal{O}_D\) is a Euclidean domain if and only if \(D\in \{-1,-2,-3,-7,-11\}\text{.}\)
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  • Turning to the positive case, it is conjectured, but not known, that there are infinitely many positive square-free integers \(D\) such that \(\mathcal{O}_{\sqrt{D}}\) is a PID. Sequence A003172 on the Online Encylopedia of Infinite Sequences (OEIS) is an updated list of the known examples of positive \(D\) for which \(\mathcal{O}_{\sqrt{D}}\) is a PID.
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  • Continuing with positive case. It is known that with the possible exception of two values of positive, square-free integers \(D\text{,}\) \(\mathcal{O}_{\sqrt{D}}\) is a PID if and only if it is a Euclidean domain. Additionally, the generalized Riemann hypothesis implies that for positive square-free \(D\text{,}\) \(\mathcal{O}_{\sqrt{D}}\) is a PID if and only if it is a Euclidean domain.
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