TITLE: $A+B$ is Nil ideal
QUESTION [8 upvotes]: 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?

REPLY [10 votes]: 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.