# Integral domain without prime elements which is not a field

I was asked whether I knew a ring without prime elements which is not a field. The first thing I thought of was the Cartesian product of fields with component-wise addition and multiplication. But now I am looking for a ring which does not have any zero-divisors. I could not think of one.

So does anyone know an integral domain (commutative ring without zero divisors) which is not a field and has no prime elements?

My first idea was adjoining elements to a known ring such as $\mathbb{Z}[\alpha]$. But then the problem is that I always find some prime in $\mathbb{Z}$, which is also a prime in this new ring.

Take the ring of power series $R=k[[X,X^{1/2},X^{1/3},\ldots]]$.

If $f\in R$ is non-invertible, then $f = uX^{1/n}$ for some unit $u$ and positive integer $n$. But $X^{1/n}$ is not prime, so $f$ is not prime.

For a noetherian example, take the ring $k[[X^2,X^3]]$, or $k[X^2,X^3]_{(X^2,X^3)}$.

More generally, if $C$ is an irreducible curve with $y\in C$, then the local ring $R=\mathcal{O}_{C,y}$ will be a noetherian domain with a unique nonzero prime ideal $P$. If $y$ is a singularity, then $P$ will not be principal.

Another (kind of) natural example comes from the $$p$$-adic world: Consider e.g. $$\mathbb C_p$$, the completion of an algebraic closure of $$\mathbb Q_p$$. The $$p$$-adic absolute value $$\lvert \cdot \rvert_p$$ extends uniquely to this field, and the elements of value $$\le 1$$ form a ring usually called $$\mathcal{O}_{\mathbb C_p}$$.

This ring is local with unique maximal ideal $$m = \{x \in \mathcal{O}_{\mathbb C_p} : \lvert x \rvert_p < 1 \}$$ which is not finitely generated. The only other prime ideal of this ring is $$(0)$$.

More generally, this holds true for any valuation ring whose value group is of rank $$1$$ and dense in $$\mathbb R$$. E.g. one can take the valuation ring of any algebraic closure of $$\mathbb Q_p$$; or of the extension of $$\mathbb Q_p$$ by all $$p$$-power roots of unity $$\mathbb Q_p(\zeta_{p^\infty})$$.