# If $A$ has no non-trivial idempotents, then neither does $A/N$

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.

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.)