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I have the following problem

Let $R$ be a ring, $A$ be a nil left ideal, $B$ be a nil ideal. Prove that $A+B$ is a left nil ideal. (Nil ideal is an ideal where every element is nilpotent.)

Generally, the sum of two nilpotent elements may not be nilpotent (in the case of non-commutative ring), so a set of nilpotent elements could not be an ideal, so I have the question

What condition could make a set of nilpotent elements to be an ideal?

I see Köthe conjecture at https://en.wikipedia.org/wiki/K%C3%B6the_conjecture is an open problem but my problem gives $A$ is left nil ideal and $B$ is a two-sided nil ideal, so then could my problem be easier?

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  • $\begingroup$ I posted a comment here (now deleted) suggesting that one might try adapting the usual proof for commutative rings. In particular, if $a \in A, b \in B$, one would hope that by choosing $n$ sufficiently large, the summands of $(a+b)^{n}$ would vanish, using the fact that $a, b, ab, ba$ are all nilpotent ($ab, ba \in B$, since $B$ is two-sided). I believe my previous comment undersold how ''exotic'' some of the summands might be; I'm now unsure whether or not it is enough to consider just $ab, ba$, since you might also might get powers of (e.g.) things like $b^{2}a$. Things would be... $\endgroup$ – Alex Wertheim Dec 30 '17 at 9:51
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    $\begingroup$ ...clearer with an explicit noncommutative generalization of the usual binomial formula. To that end, this MO thread might help: mathoverflow.net/questions/78813/…. It looks to me like everything ends up in terms of powers of $a, b$ and $[a, b] = ab-ba$, which are all nilpotent, so perhaps this does it! I admit I don't fully understand the MO thread, however. $\endgroup$ – Alex Wertheim Dec 30 '17 at 9:53
  • $\begingroup$ that's useful, thank you @AlexWertheim $\endgroup$ – Desunkid Dec 30 '17 at 11:27
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Let $a\in A$ and $b\in B$. We have \begin{align} (a+b)^1&=a+b_1\\ (a+b)^2&=a^2+ab+ba+b^2=a^2+b_2\\ (a+b)^3&=(a+b)(a^2+b_2)=a^3+ba^2+ab_2+bb_2=a^3+b_3 \end{align} and, by an easy induction, $(a+b)^n=a^n+b_n$ with $b_n\in B$.

If $a^m=0$, you get $$ (a+b)^m=a^m+b_m=b_m $$ and, since $b_m$ is nilpotent, $b_m^k=0$ for some $k$. Then $$ (a+b)^{mk}=((a+b)^m)^k=0 $$

A more “ring theoretical” proof. The left ideal $A+B$ is nil in $R/B$, because $$ (A+B)/B\cong A/(A\cap B) $$ Therefore, for every $a\in A$ and $b\in B$, there exists $m>0$ with $(a+b)^m\in B$. Since $(a+b)^m\in B$ and $B$ is nil, we are done.

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  • $\begingroup$ Beautiful! I don't know why I tried to be so explicit. Thanks for sharing this nice answer. $\endgroup$ – Alex Wertheim Dec 30 '17 at 12:01
  • $\begingroup$ Nice, thank you @egreg $\endgroup$ – Desunkid Dec 30 '17 at 12:24
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    $\begingroup$ @TrươngThừaChí You can compare the “computational“ proof with the corresponding “abstract” one. $\endgroup$ – egreg Dec 30 '17 at 13:17

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