In an integral domain $R$, we have the following characterisation of prime elements: $p \in R$ is prime $\iff$ $pR$ is a prime ideal $\iff$ $R/pR$ is an integral domain.

I was wondering whether there is a similar characterisation of irreducible elements and would appreciate any suggestions on this matter.

  • 2
    $\begingroup$ The only characterization I know is this: $x$ is irreducible iff $xR$ is a maximal ideal among the proper principal ideals of $R$. $\endgroup$ – Xam Apr 27 '17 at 19:17

There is no way to characterize irreducible elements purely in terms of ring-theoretic properties of the quotient ring. To see this, consider the domain $R=k[x]$ and the subring $S=k[x^2,x^3]\subset R$, where $k$ is a field. Note that $x^2$ is irreducible as an element of $S$, and the quotient $S/(x^2)$ is isomorphic to the ring $k[t]/(t^2)$ by sending $x^3$ to $t$. On the other hand, $x^2$ is not irreducible as an element of $R$, and $R/(x^2)$ is isomorphic to $k[t]/(t^2)$ by sending $x$ to $t$. So we have two domains, one with a reducible element and one with an irreducible element, such that the quotients by these elements are isomorphic. This means you can't tell whether an element is irreducible by just looking at the quotient ring (up to isomorphism).

On the other hand, you can characterize the ideals generated by irreducible elements as the ideals that are maximal among proper principal ideals. That is, $p\in R$ is irreducible iff $(p)$ is a proper ideal and there is no proper principal ideal which strictly contains $(p)$. (Proof: $(p)\subset (q)\subset R$ iff $q$ is a nontrivial factor of $p$.)

  • $\begingroup$ Nice answer, +1. $\endgroup$ – Xam Apr 27 '17 at 19:18
  • $\begingroup$ Very helpful, thanks Eric. $\endgroup$ – goblin Jun 26 '18 at 11:12

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