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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.

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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.

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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 $\mathbb Q_p(\zeta_{p^\infty})$.

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