Ideal generated by a irreducible element Is the ideal generated by an irreducible element always a prime ideal in a ring?
If so why?
 A: 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).
A: 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 $a\in R$.

*

*$(a)$ 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 $a|c$.


*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.
