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Let $\{f_i(x)\}_{i\in I}$ be a subset of $R[x]$ where $R[x]$ is the polynomial ring of $R$(a commutative ring with identity). If the ring $R/\langle f_i^2(n)-f_i(n)\rangle_{i\in I, n\in A}$, for every subset $A $ of $ \mathbb{Z} $, has finitely many idempotent elements. How can we verify that $R[x]/\langle f_i^2(x)-f_i(x)\rangle_{i\in I}$ has finitely many idempotent elements.

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    $\begingroup$ $R=k$ a field, $A$ empty and $I$ infinite with $f_i$ all distinct seems to contradict your conclusion. This is very similar to mathoverflow.net/questions/225440/… , that was closed. $\endgroup$ Dec 15, 2015 at 17:19
  • $\begingroup$ @MattiaTalpo: It's not because the $f_i$ are all distinct that their images in the quotient are all distinct. Also keep in mind that the ideal $\langle f_i^2(x) - f_i(x) \rangle_{i \in I}$ is a principal ideal (even if $I$ is infinite) when $R=k$ is a field. $\endgroup$ Dec 16, 2015 at 8:20
  • $\begingroup$ @MattiaTalpo: In fact every quotient of $k[x]$ has finitely many idempotents, so your example is certainly incorrect. $\endgroup$ Dec 16, 2015 at 8:29
  • $\begingroup$ @TomDeMedts of course, you're right. Also, I had read the 'any subset $A$' not as 'every', but as 'some'... $\endgroup$ Dec 16, 2015 at 8:53

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