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Let $J$ denote the set of zero-divisors of a commutative ring $R$.

Since we automatically have $RJ \subseteq J$, hence $J$ is automatically halfway to being an ideal. Furthermore, its already "prime", meaning that:

  • $ab \in J \rightarrow a\in J \vee b \in J$
  • If $R$ is non-trivial, then $J$ is a proper subset of $R$.

In general, however, $J$ may not be closed under sums. For instance, in $\mathbb{Z}/6\mathbb{Z}$, we have that $2$ and $3$ are zero-divisors, but their sum $5,$ being a unit (since $5^2 = 1$), is not. On the other hand, if $R$ is furthermore an integral domain, then the zero-divisors of $R$ are certainly closed under addition, since $0$ is the only zero-divisor of $R$. Hence commutative rings whose zero-divisors are closed under addition generalize integral domains. In fact, they strictly generalize them; for example, the ring $\mathbb{Z}/4\mathbb{Z}$ has the property that its set of zero-divisors $\{0,2\}$ are an ideal, but this is not an integral domain.

Question.

  1. What do we call commutative rings with this property?

  2. Which commutative rings have this property? More precisely: what are the main (and/or most useful or interesting) characterizations of this property?

I would also be interested in having a name for those zero divisors $a \in R$ with the property that $a+J \subseteq J$.

Motivation. Given commutative ring $R$, we can define a binary relation $\sim_R$ on $R$ as follows: $a \sim b$ means that $a-b$ is a zero-divisor. This is automatically reflexive and symmetric. We have:

  • $a \sim b \rightarrow ca \sim cb$
  • $a \sim b \rightarrow c+a \sim c+b$

Now here's the cool thing: mimicking Artuo's proof here, we see that if a square matrix $M$ with entries in $R$ is symmetric, then the eigenvectors of $M$ whose eigenvalues are $\sim$-distinct must necessarily be orthogonal.

Unfortunately, the relation $\sim_R$ needn't be transitive for an arbitrary commutative rings $R$. Indeed, a bit of scribbling shows that $\sim_R$ is transitive iff the zero-divisors of $R$ form an ideal (equivalently, iff they're closed under sums.)

Hence the question.

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goblin GONE
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  • Consider a finite dimensional local algebra. – Pedro Jul 15 '15 at 05:17
  • $\mathbb Z/4\mathbb Z$ is a ring where zero-divisors are nilpotent, or $(0)$ has only one associated prime, or $(0)$ is primary ideal, and in all the rings having this property $ZD(R)$ is a minimal prime ideal. (It seems you missed an easy observation: if $ZD(R)$ is an ideal, then it must be a prime ideal.) – user26857 Jul 15 '15 at 14:43
  • @user26857, thanks. It seems that $\mathbb{Z}/p^n\mathbb{Z}$ also has the property you mention, for $p$ prime. I say this because, if I understand correctly, this is just the statement: "for all primes $p$ and all naturals $n \geq 1$ and all $r \geq 1$, there exists $k \geq 1$ such that $p^n \mid (rp)^k$." This is clearly true, of course. – goblin GONE Jul 15 '15 at 15:20
  • That's right. This is the case when $ZD(R)$ is a minimal prime ideal. – user26857 Jul 15 '15 at 16:08

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