In my commutative algebra lecture notes it says:

A non-zero element $p$ of a ring $R$ which is not a unit of $R$ is called a prime element if $p=ab$ implies $a$ is a unit or $b$ is a unit.

Is this not the definition of an irreducible element?? Everywhere else I've read that a non-zero, non-unit element $p$ is a prime element if $p|ab $ implies $p|a$ or $p|b$

Thanks :)

  • 1
    $\begingroup$ You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element. $\endgroup$
    – drhab
    Dec 21, 2014 at 11:58
  • $\begingroup$ You are correct. This is the definition of irreducible element. $\endgroup$
    – Crostul
    Dec 21, 2014 at 12:19
  • 1
    $\begingroup$ Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students. $\endgroup$
    – egreg
    Dec 21, 2014 at 14:12

1 Answer 1


An element in an integral domain is irreducible if it is neither a unit nor a product of non-units.

An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab \in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)

Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p \mid ab$ implies that $p \mid a$ or $p \mid b$. Say $p \mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.

The implication "irreducible implies prime" is false in general, for instance in the ring $A = \mathbf{Z}[\sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - \sqrt{-5}) (2 + \sqrt{-5})$ but does not divide either of the two elements $2 \pm \sqrt{-5}$.

The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 \times 3 = (2 - \sqrt{-5}) (2 + \sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = \mathbf{Z}[\sqrt{-5}]$ is not a unique factorization domain.

  • $\begingroup$ The nicest definition of a prime ideal $\mathfrak{p}$ of a commutative ring $A$ is to say that $A \backslash \mathfrak{p}$ ($A$ deprived from $\mathfrak{p}$) is stable by finite products. $\endgroup$
    – Olórin
    Dec 21, 2014 at 13:34
  • $\begingroup$ Equivalently in a domain, a nonunit $p\ne 0\,$ is irred if $\,p = ab\,\Rightarrow\,p\mid a\,$ or $\,p\mid b.\,$ Being so similar to the definition of a prime, it makes obvious the implication $\rm\, \color{#c00}{prime}\,\Rightarrow$ irred, $ $ viz. $$p = ab\,\Rightarrow\, p\mid ab\ \overset{\large p\ \rm\color{#c00}{prime}}\Rightarrow p\mid a\ \ {\rm or}\ \ p\mid b\,$$ $\endgroup$ Dec 21, 2014 at 16:47
  • $\begingroup$ I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring $\endgroup$
    – Sushil
    Feb 15, 2015 at 15:01

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