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$\DeclareMathOperator\ann{ann}$Let $a$ and $b$ be two non-zero zero divisors of a commutative ring $R$ with 1 such that $\ann(a) \ne \ann(b)$.

is it always possible to find a sequence of non-zero elements $a_1,\dotsc,a_k \in R$ such that $a \in \ann(a_1)$, $a_1 \in \ann(a_2)$, …, $a_{k-1} \in \ann(a_k)$, and $a_k \in \ann(b)$?

Please share your thoughts or some references.

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If you allow some $a_i$ to be $0$, then the answer is obviously yes.

If there are no $0$-divisors in $R$, then the answer is vacuously yes. If there is a $0$-divisor $a$ in $R$, then taking $b = 1$ gives an example where no such sequence exists.

Suppose that $a$ and $b$ both have non-$0$ annihilators. Say $x \ne 0 \ne y$ satisfy $a x = 0$ and $b y = 0$. If $x y \ne 0$, then you may take $a_1 = x y$. Otherwise, take $a_1 = x$ and $a_2 = y$.

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  • $\begingroup$ Thanks. I have modified the question accordingly. So for me $a$ and $b$ are non-zero zero divisors. $\endgroup$
    – GA316
    Commented Mar 11, 2021 at 6:22
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Your question has been answered by LSpice, but in case you wonder about similar questions in the future, there is a sizable body of work on zero-divisor graphs of commutative rings.

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