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Is there a mistake in this paper or I am understanding it incorrectly? They defined an idempotent Armendariz ring.

Idempotent Armendariz ring: A ring $R$ is said to be idempotent Armendariz (id-Armendariz) if whenever polynomials $p(t)=\displaystyle\sum_{i=0}^{n}a_it^i$ and $q(t)=\displaystyle\sum_{j=0}^{m}b_jt^j$ in $R[t]$ satisfy $p(t)q(t)=0$, then $(a_ib_j)^2 =a_ib_j$ for each $i,j$. In theorem $2.12$ of this paper, they proved that $R$ is id-Armendariz if and only if $R[x]$ is id-Armendariz. I think this is not possible because the only idempotents of $R[x]$ are the idempotents of $R$. For example in some ring $R[x]$ the degree of $x$ will increase when we square the polynomial.

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  • $\begingroup$ Okay, I accept that. Now, I will try to find a counterexample to clear it more $\endgroup$
    – Chaudhary
    Commented Aug 14 at 14:08
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    $\begingroup$ I am editing the last line $\endgroup$
    – Chaudhary
    Commented Aug 14 at 14:10
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    $\begingroup$ After the edit, i'm not clear on why you think something is not possible. If $a_i$ and $b_j$ are elements of $R[x]$ are such that $a_ib_j$ is idempotent, what is wrong if it turns out that $a_i$ and $b_j$ are constant polynomials, i.e. elements of $R$? $\endgroup$
    – rschwieb
    Commented Aug 14 at 14:34
  • $\begingroup$ By the way, it might be good to avoid \mathcal in your posts. I say this because for some users (me included) there is a manual process for telling chrome the right way to render it, and if it is not rendered they all look like squares. It's a one-time thing, but I think for less savvy users it might never get resolved. $\endgroup$
    – rschwieb
    Commented Aug 14 at 14:37
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    $\begingroup$ Okay will avoid it next time $\endgroup$
    – Chaudhary
    Commented Aug 14 at 15:22

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