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Let $A$ be a commutative, associative, unital, finitely generated algebra over an algebraically closed field $k$. Denote by $N$ the nilradical of $A$, which is the set of all nilpotent elements of $A$ (or equivalently the intersection of all prime ideals of $A$).

Suppose that $A$ has no non-trivial idempotents. Is it true that $A/N$ also contains no non-trivial idempotents?

Aiming for a contradiction, let $e\in A/N$ be a non-trivial idempotent, then $e^2=e$ in $A/N$. In other words, $e^2-e\in N$. Then we know that there is some $n\ge 1$ such that $e^n(1-e)^n=0$. But then $\forall m\ge 1: \left((e(1-e))^m\right)^n=0$, hence $(e(1-e))^m\in N$. Does this necessarily imply that $e(1-e)=0$ in $N$, hence in $A$, contradicting that $A$ has no non-trivial idempotents?

Thank you.

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Yes, because idempotents lift modulo a nil ideal (or also this.) If there were a nontrivial idempotent of $A/N$, then it would lift to a nontrivial idempotent of $A$. (This is not hard to show.)

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