Linear combinations of nilpotent elements This is meant to be a trivial question in a book that I'm reading, but after banging my head against it for some time and searching online I haven't found a solution.

Question: given a ring $R$ and two nilpotent elements $x$ and $y$, prove that for any $a,b \in R$ the element $ax + by$ is nilpotent. Hint: use the binomial expansion.

I'm having no trouble showing that $x+y$ is nilpotent, but when including the "coefficients" I'm having some trouble. I started by letting $n$ and $m$ be such that $x^n=0$ and $y^m=0$, and letting $k=\max\{n,m\}$. Then applying the binomial theorem gives:

$$ (ax+by)^k = \sum_{i=0}^k \binom{k}{i} \;(ax)^i \;(by)^{k-i} $$

If multiplication were commutative then this would be easy, as multiplication isn't necessarily commutative in a ring, I don't really know what to do. Both suggestions and complete answers are appreciated.
This community wiki solution is intended to clear the question from the unanswered queue.



As discovered in the comments, the user simply overlooked the requirement of commutativity, and in other comments learned how things can go wrong.

I am also going to point out that if the coefficients are distracting, then one can just prove independently that $xr$ and $x+y$ are nilpotent if $x,y$ are, and then the claim about linear combinations follows.
In general, if R is semicommutative then the set nil(R) of all nilpotent elements of R forms an ideal, so the result follows.