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.