In a principal ideal domain, "prime" and "irreducible" are the same

Let \(R\) be a ring which is a PID, and let \(r \not = 0\) be an element of \(R\). Then \(r\) is irreducible if and only if \(r\) is prime.

In fact, it is easier to prove a stronger statement: that the following are equivalent. noteEvery proof known to the author is of this shape either implicitly or explicitly, but when it’s explicit, it should be clearer what is going on.

  1. \(r\) is irreducible.

  2. \(r\) is prime.

  3. The generated ideal \(\langle r \rangle\) is maximal in \(R\).

motivate the third bullet point


\(2 \Rightarrow 1\)

A proof that “prime implies irreducible” appears on the page for irreducibility.

\(3 \Rightarrow 2\)

We wish to show that if \(\langle r \rangle\) is maximal, then it is prime. (Indeed, \(r\) is prime if and only if its generated ideal is prime.)

An ideal \(I\) is maximal if and only if the quotient \(R/I\) is a field. (Proof.)

An ideal \(I\) is prime if and only if the quotient \(R/I\) is an integral domain. (Proof.)

All fields are integral domains. (A proof of this appears on the page on integral domains.)

Hence maximal ideals are prime.

\(1 \Rightarrow 3\)

Let \(r\) be irreducible; then in particular it is not invertible, so \(\langle r \rangle\) isn’t simply the whole ring.

To show that \(\langle r \rangle\) is maximal, we need to show that if it is contained in any larger ideal then that ideal is the whole ring.

Suppose \(\langle r \rangle\) is contained in the larger ideal \(J\), then. Because we are in a principal ideal domain, \(J = \langle a \rangle\), say, for some \(a\), and so \(r = a c\) for some \(c\). It will be enough to show that \(a\) is invertible, because then \(\langle a \rangle\) would be the entire ring.

But \(r\) is irreducible, so one of \(a\) and \(c\) is invertible; if \(a\) is invertible then we are done, so suppose \(c\) is invertible.

Then \(a = r c^{-1}\). We have supposed that \(J\) is indeed larger than \(\langle r \rangle\): that there is \(j \in J\) which is not in \(\langle r \rangle\). Since \(j \in J = \langle a \rangle\), we can find \(d\) (say) such that \(j = a d\); so \(j = r c^{-1} d\) and hence \(j \in \langle r \rangle\), which is a contradiction.