Prime element of a ring

by Patrick Stevens Jul 27 2016 updated Aug 21 2016

Despite the name, "prime" in ring theory refers not to elements which are "multiplicatively irreducible" but to those such that if they divide a product then they divide some term of the product.

[summary: A prime element of a ring is one such that, if it divides a product, then it divides (at least) one of the terms of the product.]

[summary(Technical): Let $~$(R, +, \times)$~$ be a ring which is an integral domain. We say $~$p \in R$~$ is prime if, whenever $~$p \mid ab$~$, it is the case that either $~$p \mid a$~$ or $~$p \mid b$~$ (or both).]

An element of an Integral domain is prime if it has the property that $~$p \mid ab$~$ implies $~$p \mid a$~$ or $~$p \mid b$~$. Equivalently, if its generated [ideal_ring_theory ideal] is [prime_ideal prime] in the sense that $~$ab \in \langle p \rangle$~$ implies either $~$a$~$ or $~$b$~$ is in $~$\langle p \rangle$~$.

Be aware that "prime" in ring theory does not correspond exactly to "prime" in number theory (the correct abstraction of which is irreducibility). It is the case that they are the same concept in the ring $~$\mathbb{Z}$~$ of integers (proof), but this is a nontrivial property that turns out to be equivalent to the Fundamental Theorem of Arithmetic ([alternative_condition_for_ufd proof]).



- Primes are always irreducible; a proof of this fact appears on the page on irreducibility, along with counterexamples to the converse.