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Disclaimer: R is a unital ring, not necessarily commutative.

Let $I$ be a nil ideal and $S/I \subseteq R/I$ be a nil subset (every element is nilpotent) of $R/I$ where $S$ is a subset of $R$.

Show that $S+I$ is a nil subset of $R$.

Attempt: Let $s\in S, i\in I$. We have to show that $i+s$ is nilpotent. I can see that some power of $s$ must lie in $I$ but I don't see how this helps.

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I can see that some power of 𝑠 must lie in 𝐼 but I don't see how this helps.

You are like a millimeter away from the solution.

Using your observation, $(i+s)^n=(\text{things with at least one $i$})+s^n\in I$. But $I$ is a nil ideal... see the solution?

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  • $\begingroup$ Thanks. Very clear. Should drink more coffee so I can solve these trivialities myself! $\endgroup$
    – user661541
    Commented Nov 13, 2019 at 18:40
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    $\begingroup$ @EpsilonDelta Ugh, of course. That is the picture I had in my mind but it did not come out in a correct way did it. Thanks you $\endgroup$
    – rschwieb
    Commented Nov 14, 2019 at 11:47

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