Section 3.4 Euclidean domains, PIDs, UFDs and all that jazz
You may recall from
Theorem 3.2.7, that a Noetherian integral domain in which every irreducible is prime enjoys unique factorization, but that determining those properties seems a formidable task. So we consider rings with stronger properties which imply those conditions, and also look for ways in which to build new UFDs from old ones.
The definition of a Euclidean domain varies a bit from source to source, but the main takeaway is that it is an integral domain that admits a division algorithm. The division algorithm can be leveraged to produce a Euclidean algorithm, and hence the notion of a greatest common divisor. We shall take the definition:
Definition 3.4.1.
An integral domain \(R\) is a Euclidean domain if it is equipped with a function (norm) \(d:R\setminus\{0\} \to \Z_{\ge
0}\) so that given two elements \(a,b \in R\) with \(b
\ne0\text{,}\) there exist \(q,r \in R\) with \(a=bq+r\text{,}\) and either \(r=0\) or \(d(r) \lt d(b).\)
Example 3.4.2.
The integers \(\Z\) with the absolute value function \(d(a) = |a|\) is a Euclidean domain.
Example 3.4.3.
If \(k\) is a field, then the polynomial ring \(k[x]\) with the degree function as the function \(d\) is a Euclidean domain.
Definition 3.4.4.
An integral domain
\(R\) is a
PID (principal ideal domain) if every ideal in
\(R\) is principal. An integral domain
\(R\) is a
UFD (unique factorization domain) if it satisfies the conclusion of
Theorem 3.2.7.
The following theorem presents the familiar relationship among these notions.
Theorem 3.4.5.
Every Euclidean domain is a PID, and every PID is a UFD.
The proof of this theorem is standard, but we want to pull out a couple of propositions which are of independent value.
Proposition 3.4.6.
Proof.
We first recall that prime and irreducible elements are nonzero and non-units. Let \(R\) be an integral domain, and \(\pi \in R\) a prime element. To show \(\pi\) is irreducible, suppose that \(\pi = ab\text{.}\) Then of course \(\pi \mid ab \) and since \(\pi\) is prime, it divides \(a\) or \(b,\) say \(a = \pi a_0.\) Then
\begin{equation*}
\pi = ab = \pi a_0b\text{,}
\end{equation*}
so
\begin{equation*}
\pi(a_0b -1) =0.
\end{equation*}
Since \(R\) is an integral domain and \(\pi \ne 0,\) we have \(a_0b = 1\) which implies \(b \in R^\times.\)
Let \(R\) be a UFD, and and \(\pi \in R\) a irreducible element. To show that \(\pi\) is prime, suppose that \(\pi\mid ab\text{.}\) If \(ab=0,\) then at least one of \(a\) or \(b\) equals zero, and since \(\pi\mid 0\) we are done in this case, so suppose that \(ab \ne 0.\) Since \(\pi \mid ab,\) we know that \(ab\) is not a unit so \(ab\) can be factored into irreducibles:
\begin{equation*}
ab = \pi_1\cdots\pi_r \cdot \pi'_1\cdots \pi'_s
\end{equation*}
with \(\pi_i\mid a,\) \(\pi'_j\mid b\) and with the possibility that \(r\) or \(s\) could be zero (though not both). Since \(\pi \mid ab\text{,}\) unique factorization says that \(\pi\) is an associate of some \(\pi_i\) or \(\pi'_j\) implying \(\pi \mid a\) or \(\pi \mid b.\)
Proposition 3.4.7.
In a PID, irreducibles generate maximal ideals.
Proof.
Let \(\pi\) be an irreducible in the PID \(R\text{,}\) and suppose that
\begin{equation*}
(\pi) \subseteq I \subseteq R.
\end{equation*}
Since \(R\) is a PID, \(I = (r)\) for some \(r\in
R,\) and the inclusion \((\pi) \subseteq (r)\) implies that \(\pi = rs\) for some \(s\in R.\) Since \(\pi\) is irreducible, either \(r\) or \(s\) is a unit. If \(r\) is the unit, then \(I = R.\) If \(s\) is the unit, then \(r\) and \(\pi\) are associates, so \(I = (r) = (\pi)\text{.}\) Thus \((\pi)\) is maximal ideal.
Proposition 3.4.8.
Proof.
Suppose that
\(\pi\) is a prime element, so in particular,
\(\pi\) is nonzero and a non-unit. Thus
\(P=(\pi)\) is a proper ideal. Suppose that
\(ab \in P=(\pi).\) Then
\(ab
=\pi c\) for some
\(c\in R.\) By
Proposition 3.4.6,
\(\pi\) is irreducible, so occurs in the factorization of
\(a\) or
\(b\text{.}\) But that means that
\(a\in P\) or
\(b \in P.\)
Conversely, suppose that \(P=(\pi)\) is a prime ideal. To show that \(\pi\) is a prime element, suppose that \(\pi
\mid ab.\) Then \(ab \in P=(\pi).\) Since \(P\) is a prime ideal, either \(a\) or \(b\) is in \(P=(\pi)\text{,}\) meaning \(\pi \mid a\) or \(\pi \mid b.\)
Proof of Theorem 3.4.5.
Suppose that \(R\) is a Euclidean domain, and let \(I\) be an ideal of \(R.\) Without loss, be may assume that \(I \ne (0).\) Using the norm, \(d\text{,}\) with which the Euclidean domain is equipped, choose a nonzero element \(b \in I\) so that \(d(b)\) is minimal. We claim that \(I = (b).\) Of course, \((b) \subseteq I\text{,}\) so choose a nonzero element \(a \in I\text{.}\) Then we may write \(a = bq +r\) with either \(r = 0\) or \(d(r) \lt
d(b).\) As \(b\) was chosen to have \(d(b)\) minimal, we must have \(r=0\) which implies that \(a \in (b).\) Thus \(I = (b)\text{,}\) and \(R\) is a PID.
It is certainly easiest to recognize a ring as a PID by noting that it is Euclidean, and that it has unique factorization since it is a PID. But there are counterexamples which show that these classes of rings are distinct.
The harder of the two is to produce PIDs which are not Euclidean. The following are examples from algebraic number theory:
\begin{equation*}
R = \Z[\frac{1+\sqrt{-d}}{2}] \text{ where } d=19, 43, 67,
163
\end{equation*}
and the proofs are all similar in nature (and in standard references), but philosophically, the challenge in showing that something is not Euclidean is showing that there is no choice of norm \(d\) which satisfies the required division algorithm property, and not simply that some natural function fails.
Adding to the challenge of finding counter examples is that no polynomial ring \(A[x]\) will work for we have the following proposition.
Proposition 3.4.9.
Let \(A\) be a commutative ring with identity, and assume that the polynomial ring \(A[x]\) is a PID. Then \(A\) is a field, so that \(A[x]\) is a Euclidean domain.
Proof.
The following theorem is of great utility in producing new UFDs from old ones.
Theorem 3.4.10.
Let \(R\) be a UFD. Then the polynomial ring \(R[x]\) is a UFD
(Main idea).
The proof of this result is not intrinsically difficult, but it does take some time and care to develop. The key is that associated to any integral domain (e.g., \(\Z\)) is its field of fractions (e.g., \(\Q\)). So associated to the UFD \(A\) is a field \(K\) which contains an isomorphic copy of \(A\text{.}\) Since \(K[x]\) is a Euclidean domain, it is also a UFD and it is natural to compare factorizations of a polynomial with coefficients in \(A\) in the rings \(A[x]\) versus \(K[x].\)
The crucial result is Gauss’s lemma which discusses how irreducibility changes as one views factorizations in \(A[x]\) versus \(K[x].\) For example, the polynomial \(p(x) =
2x\) is irreducible in \(\Q[x],\) but reducible in \(\Z[x].\) Fortunately this example is as bad as things get and it is straightforward to address.
Exercise 3.4.1.
(a)
Show that the ideal \((2,x)\) is not principal in \(\Z[x].\)
(b)
Show that the ideal \((x,y)\) is not principal in \(\Q[x,y].\)
Remark 3.4.12. Something to ponder.
Theorem 3.4.10 tells us that since
\(\Z\) and
\(\Q\) are UFDs, so are
\(\Z[x], \Q[x],
\Z[x,y], \Q[x,y]\text{,}\) as well as many others.
Suppose that \(p(x) \in \Z[x]\text{.}\) The question to consider is how does irreducibility of \(p\) change as we view in in these larger domains. For example,
If \(p\) is irreducible in \(\Z[x]\text{,}\) is it irreducible in \(\Q[x]?\)
If \(p\) is irreducible in \(\Q[x]\text{,}\) is it irreducible in \(\Z[x]?\)
If \(p\) is irreducible in \(\Z[x]\text{,}\) is it irreducible in \(\Z[x,y]?\)
Consider why the last question is important. We know that \(\Z[x]\) and \(\Z[x,y]\) are both UFDs and we can view \(\Z[x] \subset \Z[x,y]\text{.}\) Are irreducibles in \(\Z[x]\) still irreducible in \(\Z[x,y]\text{,}\) or are we forced to start from scratch in finding irreducibles in this larger domain?
