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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 :)

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    $\begingroup$ 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 '14 at 11:58
  • $\begingroup$ You are correct. This is the definition of irreducible element. $\endgroup$ – Crostul Dec 21 '14 at 12:19
  • $\begingroup$ 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 '14 at 14:12
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An element in an integral domain is irreductible if it is not 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.

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  • $\begingroup$ 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 '14 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. $$p = ab\,\Rightarrow\, p\mid ab\ \overset{\large p\ \rm\color{#c00}{prime}}\Rightarrow p\mid a\ \ {\rm or}\ \ p\mid b\,$$ $\endgroup$ – Bill Dubuque Dec 21 '14 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 '15 at 15:01

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