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Armendariz ring: $R$ is said to be an Armendariz ring if whenever $f(x)= \displaystyle\sum_{i=0}^{n}a_ix^i$ and $g(x)=\displaystyle\sum_{j=0}^{m}b_jx^j \in R[x]$ satisfy $f(x)g(x)=0$, then $ a_ib_j=0$ for each $i,j$.

Tripotent ring: A ring $R$ is tripotent if its each element is tripotent i.e $x^3=x$ for all $x\in R$.

My attempt: For example $\mathbb{Z}_3$ is tripotent but $\mathbb{Z}_n$ is Armendariz for every positive integer $n$ by this paper.

Some interesting results on tripotent rings are:

(1) Every element of tripotent ring can be written as a sum of two commuting idempotents.

(2) If $R$ is a commutative ring in which every element is the sum of two idempotents then $R$ is a tripotent ring.

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    $\begingroup$ Every tripotent ring is Armendariz. This follows from the observation in the paper that Armendariz rings are closed under products and subrings, together with the fact that tripotent rings are subrings of products of copies of $\mathbb{F}_2$ and $\mathbb{F}_3$ (see math.stackexchange.com/questions/4950234/…). The same argument shows that $n$-potent rings are Armendariz. $\endgroup$ Commented Aug 3 at 8:19
  • $\begingroup$ what about idempotent Armendariz rings ( the ring in which $f(x)g(x)=0$ implies $a_ib_j$ is idempotent), are they also Armendariz rings? $\endgroup$
    – Chaudhary
    Commented Aug 3 at 8:22

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