I am stuck at the following problem:

Show that the set of zero divisors in a valuation ring $V$ is a prime ideal.

Here is the definition of valuation ring and a theorem that might be connected to this exercise:

Definition: A commutative ring $V$ is called valuation ring if for all $a, b \in V$ holds either $a \mid b$ or $b \mid a$.

Theorem: A the ideals of a valuation ring are totally ordered with respect to inclusion and in particular every valuation ring contains exactly one maximal ideal.

Let $ZD(V)$ denote the set of zero divisors in $V$. I understand that $ZD(V)$ is an ideal, but I do not know how to show that it is a prime ideal. I tried to start in the standard way: Let $r_1, r_2 \in V$ and assume that $r_1r_2 \in ZD(V)$. We need to show that $r_1 \in ZD(V)$ or $r_2 \in ZD(V)$.

I understand that $r_1r_2 \in ZD(V)$ implies that there is a $k \in V$ sucht that $k \cdot r_1r_2 = 0$. Now we probably need to use the assumption that $V$ is a valuation ring, so we have either $r_1 \mid r_2$ or $r_2 \mid r_1$. However, when I try to make a case distinction from that I get stuck, also using $r_1 \mid k$ or $k \mid r_1$ did not help.

Could you give me a hint?

  • 1
    $\begingroup$ Assume $r_1|r_2$. Then $r_2 = a r_1$. So $k a r_1 r_1=0$. So $r_1$ is in $ZD(V)$. $\endgroup$
    – user1054
    Apr 28, 2021 at 15:12

1 Answer 1


The set of zero divisors in an arbitrary ring does not usually form an ideal.

Take, for example, any ring $A$ with a nontrivial idempotent $e$ (i.e. an element $e \notin \{0, 1\}$ such that $e^2 = e$). Then $e (1-e) = 0$, so $e$ and $1-e$ are zero divisors, but $(e, 1 - e) = A$, so the zero divisors of $A$ generate the unit ideal.

But whenever the set of zero divisors does form an ideal, it forms a prime ideal.

Indeed, let $A$ be a ring and $a,b \in A$ such that $ab$ is a zero divisor. Then $ab c = 0$ for some $c \not = 0$.

Case (i) $bc \not= 0$. Then $a$ is a zero divisor.

Case (ii) $bc = 0$. Then $b$ is a zero divisor.

A more natural way to phrase this is that the set of regular elements ( = non-zero divisors) forms a multiplicative set.

What you need to show, then, is that in a valuation ring the set of zero divisors does form an ideal.

This boils down to showing that if $a,b$ are zero divisors then (i) $a + b$ is a zero divisor and (ii) $r a$ is a zero divisor for any $r \in A$. Check that (ii) holds in any ring. So the focus in your problem is that zero divisors in a valuation ring are closed under addition.

EDIT with update to answer the comment of Batrachotoxin

In general the ideal generated by the zero divisors is not going to be a maximal ideal. Consider, for example, any valuation domain that is not a field. Meanwhile, in a field the zero divisors do generate the maximal ideal $(0)$.

It's also easy to produce examples of valuation rings which are not domains and in which the ideal generated by the zero divisors is maximal.

For example, consider the ring of dual numbers over a field $K$, $A := K[x]/(x^2)$.

A generic element of $A$ looks like $ax + b$, with $a,b \in K$. If $a,b \not= 0$ then $ax + b$ is a unit since it is the sum of a unit and a nilpotent element. Thus the ideal of non-zero divisors is principally generated by $(x)$, and this is the maximal ideal of $A$.

  • $\begingroup$ I just wanted to know if it is possible to show that in this case, the ideal be a maximal ideal as well? And if not, is there a concrete example for the same? $\endgroup$ Nov 21, 2021 at 20:11
  • $\begingroup$ @Batrachotoxin I updated the answer for you $\endgroup$ Nov 21, 2021 at 20:29
  • $\begingroup$ Thank you so very much! $\endgroup$ Dec 2, 2021 at 10:06

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