Ideal generated by a irreducible element

Is the ideal generated by an irreducible element always a prime ideal in a ring?

If so why?

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No, ideals generated by irreducible elements are prime if and only if the element is a prime element.

For a counterexample, consider $R=\mathbb{Z}(\sqrt{-5})$ and $2$.

The element $2$ is irreducible in $R$, but $(1+\sqrt{-5})(1-\sqrt{-5})\in (2)$, and neither $1+\sqrt{-5}$ nor $1-\sqrt{-5}$ are in $(2)$. So $(2)$ is not prime, even though $2$ is irreducible.

You may want to prove the following:

Proposition. Let $R$ be a commutative ring with identity, and let $a\in R$.

1. $(a)$ is a prime ideal if and only if $a$ is a prime element of $R$; that is, $a$ is not a unit, and if $a|bc$ in $R$, then either $a|b$ or $a|c$.
2. Note added Dec 2011: This certainly holds if $a$ is not a zero divisor; I don't know if it holds in general.

$(a)$ is maximal among principal ideals (that is, $(a)\neq R$, and if $(a)\subseteq (x)\subseteq R$, then either $(a)=(x)$ or $(x)=R$) if and only if $a$ is irreducible.

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But isn't $(2)$ maximal wrt principal ideals and $\mathbb{Z}(\sqrt{-5})/(2)$ an integral domain? –  Mohan Oct 14 '11 at 19:33
@user774025: Yes, $(2)$ is maximal among principal ideals (because $2$ is irreducible), but $\mathbb{Z}(\sqrt{-5})/(2)$ is not a principal ideal domain: the elements $1+\sqrt{-5}+(2)$ and $1-\sqrt{-5}+(2)$ are nonzero, but their product is zero in the quotient. $(2)$ is not a maximal ideal of $\mathbb{Z}[\sqrt{-5}]$, since $(2)\subset (2,1+\sqrt{-5})\subset \mathbb{Z}[\sqrt{-5}]$, and both inclusions are proper. –  Arturo Magidin Oct 14 '11 at 19:35
@Arturo: Could you please give a hint for your Proposition 2. $(\Rightarrow)$? Don't we need the assumption that $R$ is a domain? –  Leon Dec 9 '11 at 1:39
@Leon: Maybe... certainly that is sufficient, but I seem to recall that it was not needed if you did things cleverly enough... –  Arturo Magidin Dec 9 '11 at 6:41
@Leon: I've been (on and off) trying to see if I can recall the argument, find the source I think I had, or convince myself I'm wrong. Unsuccessfully. I should probably edit it. It also suffices for $a$ to not be a zero divisor... –  Arturo Magidin Dec 12 '11 at 7:08

HINT $\$ Since $\rm\:(p)\:$ is prime iff $\rm\:p\:$ is prime, your assumption is equivalent to saying that all atoms (irreducibles) are prime. This hypothesis, combined with atomicity (every nonzero nonunit factors into a product of atoms) is equivalent to a domain being a UFD. So it is a very strong hypothesis - one that fails in any non-UFD number ring (which are always atomic by way of norms).

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The fact that this is a "strong hypothesis" is the key fact to grasp. For many of us it is not obvious that this is the case until we have an example, and the realisation of this is also historically significant. So there is no shame in not knowing, and a lot to discover when you do realise ... –  Mark Bennet Oct 14 '11 at 20:12