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



Irreducible/prime/indivisible elements
To answer Q2: Every noetherian (commutative) ring that is not a finite product of fields admits irreducible elements. Since any finite product of $\ge 2$ fields admits prime elements, this shows that the answer to Q2 is no.


In an arbitrary commutative ring $A$, define $x\in A$ to be quasi-irreducible if $Ax$ is maximal among principal ideals $\neq A$. Thus if $A$ is noetherian and not a field, then it admits nonzero quasi-irreducible elements. (Note that 0 is quasi-irreducible iff $A$ is a field.) Moreover, irreducible implies quasi-irreducible.


The converse to holds for nonzero elements when $A$ is a domain, but also more generally when $A$ is connected, in the sense that it admits no nontrivial idempotent: (\*) if $A$ is connected and $x\in A$ with $x\neq 0$, $x$ quasi-irreducible, then $x$ is irreducible.


Indeed, suppose that $x\neq 0$ is quasi-irreducible. Suppose by contradiction that $x=yz$ with $y,z$ non-invertible. Since $x$ is quasi-irreducible, this implies that $y,z\in Ax$. Write $y=ax$ and $z=bx$. Hence $x=abx^2$. That is, $x(1-abx)=0$. Then $(abx)^2=ab(abx^2)=abx$. Thus $abx$ is idempotent. Since $A$ is connected, either this implies $abx=1$, so $x$ is invertible (contradiction), or $abx=0$, whence $abx^2=x=0$, another contradiction.


Note that the converse to (\*) does not hold in the non-connected case: for instance in a product of 2 fields, $(1,0)$ is quasi-irreducible but not irreducible. 


Let us now prove the main claim. Assume that $A\neq 0$ is an arbitrary noetherian ring. Then it can be written as a product $A=A\_1\times \dots \times A\_k$ with each $A\_i$ connected.


Suppose that one of the $A\_i$ is not a field. After reindexing, we can suppose $i=1$. Let $x\_1$ be a quasi-irreducible element in $A\_1$; by (\*) it is irreducible in $A\_1$. Then $(x\_1,1,\dots,1)$ is an irreducible element in $A$.


Finally, if all $A\_i$ are fields and $k\ge 2$, then $(0,1,\dots,1)$ is a prime element.


