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Let $A$ be a commutative ring, $\sqrt0$ is the nilradical and there exists an idempotent $e\in A/\sqrt0$, i.e. $e\neq0,1$ and $e^2-e=0$. Show that $A$ has an idempotent $\ne0,1$.

My attempt:

Denote a representative of $e$ in $A$ as $x$. Then $x^n(x-1)^n=0$ for some $n>0$. If I could show that $x^n$ and $(x-1)^n$ are coprime than I could use Chinese remainder theorem to decompose $A$ as $A/(x)^n\times A/(x-1)^n$ and then it is easy to prove that $(0,1)$ is an idempotent.

Any suggestions how to show that $x^n$ and $(x-1)^n$ are coprime? There are proofs of this fact on this website but the ones I found are for the case when $x$ is a formal variable of a ring, which are not applicable here.

user26857
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    There's a related post here about constructing the idempotent directly. It's in a question about nilpotent ideals but it works fine for nil ideals like $\sqrt{0}$. – rschwieb Apr 10 '19 at 13:50

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For any commutative ring with unity $A$ and ideals $\mathfrak{a}$ and $\mathfrak{b}$, we claim that $\mathfrak{a}+\mathfrak{b}=A\implies \mathfrak{a}^{k}+\mathfrak{b}^{l}=A$ for any positive integers $k$ and $l$.

We have $\sqrt{\mathfrak{a}^{k}+\mathfrak{b}^{l}}=\sqrt{\sqrt{\mathfrak{a}^{k}}+\sqrt{\mathfrak{b}^{l}}}=\sqrt{\sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}}=\sqrt{\mathfrak{a}+\mathfrak{b}}=A.$ Thus, $\mathfrak{a}^{k}+\mathfrak{b}^{l}=A.$ (All these use just basic properties of the radical of ideals)

In your case $(x)$ and $(x-1)$ are coprime and so, we conclude that $(x)^{n}$ and $(x-1)^{n}$ are coprime.

Indrayudh Roy
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