Let $R$ be a commutative ring. It is well-known that if $b \in R$ and $c \in R$ are two nilpotent elements with $b^k = 0$ and $c^\ell = 0$ (where $k$ and $\ell$ are positive integers), then $b+c$ is nilpotent again with $\left(b+c\right)^{k+\ell-1} = 0$.

I'm wondering if this has a converse of the following form:

Question 1.Let $R$ be a commutative $\mathbb{Q}$-algebra. Let $k$ and $\ell$ be two positive integers. Let $a \in R$ satisfy $a^{k+\ell-1} = 0$. Is it true that there exists a commutative ring $S$ such that $R$ is a subring of $S$, and such that $S$ has two elements $b$ and $c$ with $b^k = 0$ and $c^\ell = 0$ and $a = b+c$ ?

**Partial results:** I suspect that the answer is positive.

In order to prove a positive answer, it suffices to prove it for $\ell = 2$. This means splitting a nilpotent $a \in R$ with $a^n = 0$ into a sum $b + c$, where $b^{n-1} = 0$ and $c^2 = 0$. If such a splitting always exists, then by induction, we can split each nilpotent $a \in R$ with $a^{k+\ell-1} = 0$ into a sum $b + c_1 + c_2 + \cdots + c_{\ell-1}$ with $b^k = 0$ and $c_1^2 = c_2^2 = \cdots = c_{\ell-1}^2 = 0$; but then, we can set $c := c_1 + c_2 + \cdots + c_{\ell-1}$ and easily obtain $c^\ell = 0$.

I also know that the answer is positive when $k = \ell = 2$. Indeed, in this case, we have an element $a \in R$ with $a^3 = 0$, and we want to split it as a sum $a = b+c$ of two elements $b, c \in S$ satisfying $b^2 = c^2 = 0$. Here is one way to do this: Define a commutative ring $S$ to be $R \oplus \left(R / a\right)$, whose elements are added entrywise and multiplied by the rule $\left(p,\overline{q}\right)\left(u,\overline{v}\right) = \left(pu - qva^2/4, \overline{pv+qu}\right)$. We embed the ring $R$ into $S$ by equating each $r \in R$ with $\left(r,\overline{0}\right) \in S$. Now, we take $b = \left(a/2,\overline{1}\right)$ and $c = \left(a/2,\overline{-1}\right)$. It is then easy to see that $b^2 = \left(0,\overline{a}\right) = 0_S$ and $c^2 = \left(0,\overline{-a}\right) = 0_S$ and $b + c = \left(a,\overline{0}\right) = a$.

Could we do this without dividing by $2$ ? No, because the question clearly has a negative answer in characteristic $2$. Indeed, in characteristic $2$, if $b^2 = c^2 = 0$, then $\left(b+c\right)^2 = 0$, and thus $a$ cannot be written as $b + c$ unless $a^2 = 0$.

Question 2.What are the precise requirements needed on $R$ for Question 1 to have a positive answer for a given pair $\left(k,\ell\right)$ ? Presumably it should suffice for $\left(k+\ell-2\right)!$ to be invertible? Or maybe even $k+\ell-2$ ?

**Context.** This is motivated by the splitting principle in $\lambda$-ring theory, but I would be surprised if a proper connection exists. The Tschirnhaus transformation from the theory of polynomials looks vaguely related based on the $k = \ell = 2$ case.