in what follows all the rings are commutative, nontrivial, with unit. Recall the following definitions:

1) $\pi\in A$ is **prime** if $(\pi)$ is a nonzero prime ideal

2) $\pi\in A$ is **irreducible** if $\pi$ is nonzero, non invertible, and for all $a,b\in A$, $\pi=ab$ implies that $a$ or $b$ is a unit.

3) $\pi\in A$ is **indivisible** if $\pi$ is nonzero, non invertible, and for all $a,b\in A$, $\pi=ab$ implies that $\pi\mid a$ or $\pi\mid b$.

The following facts are known:

- if $A$ is a domain, prime$\Rightarrow$ irreducible.

This is not true anymore if $A$ has zero divisor (e.g. $A=\mathbb{Z}/6\mathbb{Z}$: $A$ has prime elements, but no irreducible elements)

irreducible $\Rightarrow$ indivisible

there exist indivisible elements which are not irreducible: $A=\mathbb{Z}/6\mathbb{Z}$, $\pi=3$ . However, this one is prime.

if $A$ is a domain, irreducible $\iff$ indivisible

if $A$ is noetherian, $A$ has indivisible elements

if $A$ is a noetherian domain which is not a field, $A$ has irreducible elements.

After this lengthy introduction, let me ask the following questions:

**Q1**: can we find an example of an indivisible element which is neither prime or irreducible ? If possible, I would like $A$ to be noetherian or, even better, finite.**Q2**: can we find an example of a**noetherian**ring which is not a field, which has no prime elements AND no irreducible elements ? (so necessarily, $A$ has zero divisors)**Q3**: can we find an example of a**noetherian domain**which is not a field which has no prime elements ?**Q4**: if the answer to Q3 is**NO**, can we find an example of a**domain**which has irreducible elements, but which has no prime elements ?

Thanks for your time.

Greg

musthave indivisble elements. $\endgroup$ – Arturo Magidin Jan 20 '15 at 20:50