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I am trying to prove the following result:

If $R$ is a commutative ring with identity, $I$ is an ideal of $R$ such that $I^2=\{0\}$ and $a+I$ is an idempotent element in $R/I$ then there are some idempotent element of $R$ in the coset $a+I$.

What I tried to do?

Well, suppose that $a+I$ is idempotent in $R/I$, then $(a+I)(a+I)=a+I$, so, $a^2+I=a+I$ and then $a^2-a \in I$. We know also that $1-a+I$ is idempotent since $(1-a+I)(1-a+I)=1-2a+a^2+I = 1-2a+a+I=1-a+I$. And then, from this we know that $a(1-a)+I$ is idempotent.

We need to show that there exist some $m \in I$ such that $(a+m)^2=a+m$, but $(a+m)^2=a^2+2am+m^2=a^2+2am$ since $m^2=0$, so $m$ needs to satysfy $a^2+(2m-1)a-m=0.$

I dont know if this reasoning is right, neither if this is a way to go over it. Can someone give me a hint to finish it?

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marked as duplicate by user26857 abstract-algebra Dec 2 '16 at 15:48

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  • $\begingroup$ It might interest some that the ring does not have to be commutative, and that "$I$ a nilpotent ideal" can be relaxed to "$I$ a nil ideal" $\endgroup$ – rschwieb Dec 2 '16 at 13:50
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First, take $i$ such that $a=a^2+i$. Since we can only obtain formulas using $a$ and $i$, it is natural to try elements of the form $$a+\lambda a i+\mu i$$ where $\lambda,\mu$ are unknown (also, intuitively, given in terms of $a$ and $i$). Let's try to see when such an element is idempotent; \begin{align*} a+\lambda ai+\mu i&=(a+\lambda ai+\mu i)^2=a^2+2\lambda a^2i+2\mu ai=(a-i)+2\lambda(a-i)i+2\mu ai\\ &=a+2(\lambda+\mu)ai-i \end{align*} So we need to choose $\mu=-1$ and $\lambda$ such that $\lambda=2(\lambda+\mu)=2\lambda-2$, i.e., $\lambda=2$.

Thus, $a+2ai-i$ is idempotent, as the calculation above shows.

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