Let \(R\) be a ring.

An element \(x\in R\) is irreducible if \(x\) is not a unit and for all \(y, z\in R\), \(x = y z\) implies \(y\) or \(z\) is a unit.

An element \(x \in R\) is prime if \(x \ne 0\), \(x\) is not a unit and for all \(y,z\in R\), we have \(x | yz \implies x|y \text { or } x|z\). Note \(x\) is prime if and only if \(x R\) is a prime ideal.

In an integral domain, all primes are irreducbile. The converse is not always true. For example, take \(R =\mathbb{Z}[\sqrt{-5}]\). Then by using the norm, it can be deduced that the units of \(R\) are \(\pm 1\). We have \(6 = 2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\), where all the factors are irreducible but not prime.

An integral domain \(R\) is a unique factorization domain (UFD) if every nonzero nonunit of \(R\) can be expressed as a product of irreducibles and furthermore the factorization is unique up to order and associates. In a UFD, all irreducibles are prime.

Example: \(\mathbb{Z}\) is a UFD. All fields are trivially UFD’s.

Gauss' Theorem: If \(R\) is a UFD then the polynomial ring \(R[x]\) is a UFD.

Proof: Consult an abstract algebra textbook.


  1. Let \(R = F[x_1,...,x_n]\) where \(F\) is a field. By Gauss' Theorem \(R\) is a UFD. If \(p \in R\) is any irreducible polynomial over \(F\) then it is also prime, and the principal ideal \(p R\) is a prime ideal of \(R\).

  2. Let \(R = \mathbb{Z}\) or \(R = F[x]\) for some field \(F\). By using the Euclidean algorithm, it can be seen that every ideal of \(R\) is principal. (In \(\mathbb{Z}\), the nonzero prime ideals are generated by a prime, while in \(F[x]\), the nonzero prime ideals are generated by irreducible polynomials.) In these rings, it turns out that all nonzero prime ideals are also maximal, for reasons we shall see below.

A principal ideal domain (PID) is an integral domain in which all ideals are principal.

Proposition: Let \(R\) be a PID. Then every nonzero prime ideal is maximal.

Proof: Let \(I = x R\) be some nonzero prime ideal. Suppose \(I\) is strictly contained in some ideal \(y R\) of \(R\). Then \(x = y z\) for some \(z\in R\). Since \(I\) is prime, we must have \(y \in I\) or \(z \in I\). The former would imply \(y R\) is contained in \(I\), a contradiction. So we must have \(z \in I\). In other words, \(z = x t\) for some \(t \in R\), hence \(x = x y t \). implying that \(y t = 1\). Thus \(y\) is a unit so \(y R = R\).

Example: Going back to the example \(R = F[x_1,...,x_n]\) for some field \(F\), set \(M = \{p\in R|p(0) = 0\}\) (the set of polynomials with zero constant term). We have \(A/M \cong F\) thus \(M\) is maximal. Now if \(n \gt 1\), \(M\) is not principal, because any set of elements that generates \(M\) must contain at least \(n\) elements.

Ben Lynn blynn@cs.stanford.edu 💡