Principal Ideal Domains (PIDs)

Section 10.2 Principal Ideal Domains (PIDs)

“One chance is all you need.”
―Jesse Owens

Subsection Principal Ideals

Definition 10.7. Principal Ideal.

Let \(I\) be an ideal of a ring \(R\text{.}\) The ideal \(I\) is principal if \(I = \igen a\) for some \(a \in R\text{,}\) that is, \(I\) is generated by a set with a single element.

Exercise 10.8. Principal Ideals in \(\Z\).

Let \(\igen m,\igen n\) be ideals in \(\Z\text{.}\) Then

\(\displaystyle \igen m+\igen n=\igen{\gcd(m,n)}\)
\(\displaystyle \igen m\cap\igen n=\igen{\lcm(m,n)}\)
\(\igen m\igen n=\igen {mn}\) Conclude that \(\igen m\igen n=\igen m\cap\igen n\) if and only if \(m\) and \(n\) are relatively prime.

Example 10.9. Examples of Principal Ideals.

Every ideal of \(\Z\) is principal with \(I=\igen n\) for some \(n\in \Z\)
¹
This means \(\Z\) is a PID, something that will be defined in Section 10.2
For any field \(F\text{,}\) every ideal of \(F[x]\) is principal.
For any field \(F\text{,}\) every ideal in \(F[x_1,\ldots,x_n]\) is finitely generated, but not necessarily principal. This is a consequence of a deep theorem called the Hilbert Basis Theorem, which you will see in Math 905.

Subsection Principal Ideal Domains

Definition 10.10. PID.

A ring \(R\) is called a principal ideal domain (PID) if it is a domain with the property that every ideal is principal, i.e., for each ideal \(I\text{,}\) we have \(I = \igen a\) for some \(a \in R\text{.}\)

Exercise 10.11. Polynomial Rings PID iff Field.

A polynomial ring \(R[x]\) is a PID if and only if \(R\) is a field.

Qual Watch.

Proving Exercise 10.11 was [cross-reference to target(s) "june-2016-8" missing or not unique] on the [cross-reference to target(s) "june-2016" missing or not unique] qualifying exam.

Theorem 10.12. Euclidean Domains are PIDs.

If \(R\) is a Euclidean domain, then \(R\) is a PID.

Proof.

Let \(N\) be the norm function making \(R\) into a Euclidean domain. Pick an ideal \(I\text{.}\) If \(I\) is the zero ideal, \(I = \igen0\text{.}\) Otherwise pick a non-zero element \(b\) of \(I\) with \(N(b)\) as small as possible. (Such a \(b\) exists by the well-ordering of \(\Z_{\geq 0}\text{.}\)) I claim \(I = \igen b\text{.}\) It is clear that \(\igen b \subseteq I\text{.}\) Pick \(a \in I\text{.}\) Then

\begin{equation*} a = bq + r \end{equation*}

and either \(r = 0\) or \(N(r) < N(b)\text{.}\) But note that \(r = a - bq \in I\text{,}\) and we cannot have both \(r \ne 0\) and \(N(r) < N(b)\) by our choice of \(b\text{.}\) So it must be that \(r = 0\text{,}\) and hence \(a\in \igen b\text{.}\)

The converse is not true in general.

Example 10.13. PIDs Need not be Euclidean Domains.

The ring \(\Z\left[ \frac{1 + \sqrt{-19}}{2}\right]=\left\{a+ b\frac{1 + \sqrt{-19}}{2} \mid a,b\in \Z\right\}\) is a PID, but not a Euclidean domain. It is the simplest example of such a ring, but the proofs of these claims are not easy. I will not cover a proof of this fact.

Proposition 10.14. Nonzero Primes are Maximal in PIDs.

If \(R\) is a PID, then every nonzero prime ideal is maximal.

Qual Watch.

Proving Proposition 10.14 was [cross-reference to target(s) "jan-2023-4" missing or not unique] on the [cross-reference to target(s) "jan-2023" missing or not unique] qualifying exam.

Definition 10.15. Associate.

Let \(R\) be a domain. Two elements \(r,s \in R\) are associates if there is a unit \(u\) of \(R\) such that \(s = ur\text{.}\)

Lemma 10.16. Associates and Principal Ideals.

Two elements \(x,y\) of a domain \(R\) are associates if and only if \(\igen x = \igen y\text{.}\)

Proof.

If \(\igen x = \igen y\) then \(x \in \igen y\) and so \(x = yu\) for some \(u\text{.}\) Similarly \(y = xs\) and hence \(y = yus\text{.}\) Since \(R\) is a domain, either \(y = 0\) or \(su = 1\text{.}\) If \(y= 0\text{,}\) then \(x = 0 = 1 y\) and otherwise \(u\) is a unit.

If \(x = uy\) for a unit \(u\text{,}\) then \(y = u^{-1}x\) and so \(x \in \igen y\) and \(y \in \igen x\text{,}\) from which is follows that \(\igen x \subseteq \igen y\) and \(\igen y \subseteq \igen x\text{.}\)

Proposition 10.17. GCDs and Units in PIDs.

If \(R\) is a PID and \(a,b\in R\text{,}\) then

\(\igen{a,b}=\igen g\) for some \(g\in R\) and any such \(g\) is a gcd of \(a\) and \(b\text{.}\)
the gcd of \(a\) and \(b\) is unique up to multiplication by a unit.

Proof.

The existence of \(g\) is granted by definition in a PID. Now \(a,b\in \igen g\) gives that \(g\mid a\) and \(g\mid b\text{.}\) If \(g'\mid a\) and \(g'\mid b\) we have that \(a,b\in \igen{g'}\text{,}\) so \(\igen g=\igen{a,b}\subseteq \igen{g'}\) by minimality. This gives \(g\in\igen{g'}\text{,}\) hence \(g'\mid g\text{.}\)

Remark 10.18.

If \(R\) is not only a PID but a Euclidean domain with norm function \(N\text{,}\) then the Euclidean algorithm can be used to compute a gcd of any two nonzero \(a,b\in R\text{.}\)

Definition 10.19. Prime.

Suppose \(R\) is an integral domain. An element \(p \in R\) is a prime element if \(p \neq 0\) and the ideal \((p)\) is a prime ideal.

Definition 10.20. Irreducible.

Suppose \(R\) is an integral domain. An element \(r \in R\) is irreducible if \(r\neq 0\text{,}\) \(r\) is not a unit, and whenever \(r = xy\) with \(x,y \in R\) then either \(x\) or \(y\) is a unit.

Example 10.21. Examples of Primes and Irreducibles.

The prime elements of \(\Z\) are the prime integers and their negatives; they are also irreducible
Any element \(\a\in Z[i]\) with \(N(\a)\) a prime integer is irreducible e.g. \(\a=1+2i\) is irreducible
The element \(13=(2+3i)(2-3i)\) is not irreducible in \(\Z[i]\)
The polynomial \(x^2+x+[1]\in (\Z/2)[x]\) is irreducible; indeed if it factors nontrivially, it must factor as a product of two linear polynomials: \(x^2+x+[1]=(x+[a])(x+[b])\text{.}\) Then \(-[b]\) is a root for \(x^2+x+[1]\text{.}\) But neither \([0]\) nor \([1]\) are roots for this polynomial, a contradiction.

Exercise 10.22. In \(\Z[\sqrt{-5}]\text{,}\) \(2\) Prime but not Irreducible.

In the domain \(\Z[\sqrt{-5}]\) the element \(2\) is irreducible but not prime.

Solution.

Note that \(2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})\) and thus \((1+\sqrt{-5})(1-\sqrt{-5})\in (2)\text{.}\) However we claim that neither \(1+\sqrt{-5}\) nor \(1-\sqrt{-5}\) are in \((2)\text{.}\) If \(\a=1\pm\sqrt{-5}\in (2)\) then \(\a=2\beta\) for some \(\beta\in\Z[\sqrt{-5}]\) and so \(6=N(\a)=N(2)N(\beta)=4N(\beta)\text{,}\) a contradiction. Thus \(2\) is not prime.

Exercise 10.23.

In the ring \(\C[x,y]/\igen{x^2-y^3}\) the element \(y\) is irreducible but not prime.

Solution.

Since \(\C[x,y]/\igen{x^2-y^3}\cong \C[t^3, t^2]\subseteq \C[t]\) this ring is a domain. The element \(y\) is irreducible for degree reasons. The ideal \(\igen{y}\) contains \(x^2\) but doesn’t contain \(x\) so \(y\) is not prime.

Theorem 10.24. Irreducible is Prime in PID.

Let \(R\) be a domain and let \(r\in R\text{.}\)

If \(r\) is a prime element, then \(r\) is irreducible.
If \(R\) is a PID and \(r\) is irreducible, then \(r\) is a prime element.

Proof.

Suppose \(R\) is an integral domain and that \(r\) is prime. Then \(r \ne 0\) and \(r\) is not a unit. Suppose \(r = yz\text{.}\) Then \(yz \in \igen r\) and hence by definition either \(y \in \igen r\) or \(z \in \igen r\text{.}\) If \(y \in \igen r\text{,}\) we have \(y = rt\) for some \(t\) and so \(y = yzt\text{.}\) Since \(r \ne 0\text{,}\) \(y \ne 0\text{,}\) and \(R\) is an integral domain, we must have \(zt = 1\text{,}\) showing that \(z\) is a unit.

Assume \(R\) is a PID and that \(r\) is irreducible. Since \(r\) is not a unit, \(\igen r\) is a proper ideal and hence is contained in a maximal ideal \(M\) by [provisional cross-reference: cite] We show \(\igen r= M\) and hence \(\igen r\) is prime. Since \(R\) is a PID, \(M = \igen y\) for some \(y\text{.}\) So \(x = yt\) for some \(t\text{.}\) But \(x\) is irreducible and \(y\) is not a unit, which forces \(t\) to be a unit and hence \(\igen x = \igen y = M\text{.}\)

Theorem 10.25. Factorizations, Permutations, Associates.

Assume \(R\) is a PID \(r\in R\) is non zero and not a unit and \(r = p_1 \cdots p_n = q_1 \cdots q_m\) are two different irreducible factorization of \(r\text{.}\) Then \(n=m\) and there is a permutation \(\sigma\) such that, for all \(i\text{,}\) we have \(p_i\) and \(q_{\sigma(i)}\) are associates.

Proof.

Without loss of generality, assume \(n \leq m\text{.}\) We induct on \(m\text{.}\)

If \(m =1\) then \(n=1\) as well since \(n\leq m\) and \(n=0\) would yield \(r = p_1 \cdots p_n =1\text{,}\) a contradiction. If \(n=m=1\) we have \(p_1=q_1\) and there is nothing more to prove.

Assume \(m > 1\) and that irreducible factorizations with \(\leq m\) factors are unique up to reordering factors and taking associates.

Since \(R\) is a PID, irreducible elements are prime by [provisional cross-reference: cite]. Since \(p_1 \cdots p_n\in (p_n)\) we have that \(q_1 \cdots q_m\in (p_n)\) and since \((p_n)\) is a prime ideal it follows that \(q_i \in (p_n)\) for some \(i\text{.}\) Upon reordering, we may as well assume \(q_m\in (p_n)\text{.}\) Thus \(q_m = p_n u\) for some \(u \in R\text{.}\) Since \(q_m\) is irreducible and \(p_n\) is not a unit, \(u\) must be a unit. We get

\begin{equation*} p_1 \cdots p_n = q_1 \cdots q_{m-1} up_n \end{equation*}

and hence, since \(R\) is an integral domain, we may divide by \(p_n\) to obtain

\begin{equation*} p_1 \cdots p_{n-1} = q_1 \cdots q_{m-2} (q_{m-1}u). \end{equation*}

Notice that \((q_{m-1}u)=(q_{m-1})\) is a prime ideal so \(q_{m-1}u\) is irreducible by [provisional cross-reference: cite] .

By the inductive hypothesis we deduce that \(n-1=m-1\) hence \(n=m\) and also that \(p_1, \ldots, p_{n-1}\) are associates to \(q_1, \ldots, q_{m-2}, q_{m-1}u\) in some order. This together with \(p_n\) associate to \(q_m\) establishes the claim.

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