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.

  • 3
    $\begingroup$ Seemingly it suffices to study the universal case. For example, is the map $\mathbb Z[T]/T^{k+\ell-1}\to\mathbb Z[X,Y]/(X^k,Y^\ell),T\mapsto X+Y$ faithfully flat? $\endgroup$
    – Z. M
    Jun 26 at 19:38
  • $\begingroup$ @Z.M: Can you walk me through the argument for why this would help? $\endgroup$ Jun 26 at 19:40
  • 3
    $\begingroup$ Base change this map along $\mathbb Z[T]/T^{k+\ell-1}\to R,T\mapsto a$, you get a universal candidate of $S$ (without assuming that $R\to S$ being injective), in the sense that every other $S$ factors uniquely through that. The faithful flatness is stronger than the injectivity. $\endgroup$
    – Z. M
    Jun 26 at 21:20

1 Answer 1


I can give a positive answer to question 1 and an answer to question 2.

Theorem: Let $R$ be a commutative algebra. Let $k$ and $\ell$ be two positive integers. Let $a\in R$ satisfy $a^{k+\ell-1}=0$. If $\binom{k+\ell-2}{k-1}$ is not a zero divisor in $R$, then 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 $a=b+c$.

Proof: Following a suggestion of Z.M., we take $S = R[ b,c]/ (b^k, c^\ell, a-b-c)$. The claim that $S$ has two elements $b$ and $c$ such that $b^k=0$ and $c^\ell=0$ and $a=b+c$ then holds automatically, so the main difficulty is verifying that $R \to S$ is injective.

To do this, we consider the $R$-module homomorphism $f \colon S \to R$ defined by $$f( r b^i c^j) = r a^{i+j} \binom{k+\ell-2-i-j}{k-1-i}$$ for $r\in R$ and nonnegative integers $i,j$. To see that $f$ is well-defined, we note that it is clearly a well-defined homomorphism $R[b,c]\to R$, where we take the binomial coefficient to vanish if the number on top is negative or the number on bottom is not between $0$ and the number on top. So it suffices to show that any multiple of $b^k, c^\ell$, or $a-b-c$ is sent to $0$. For $b^k$ and $c^\ell$ this follows from the aforementioned vanishing of binomial coefficients, and for $a-b-c$ it follows from

$$f ( a r b^i c^j ) = r a^{i+j+1} \binom{k+\ell-2-i-j}{k-1-i} = r a^{i+j+1} \left( \binom{k+\ell-3-i-j}{k-1-i} + \binom{k+\ell-3-i-j}{k-2-i}\right) = f( r b^{i+1} c^j) + f( r b^i c^{j+1}) = f( (b+c) r b^i c^j). $$

Now $f$ is a well-defined $R$-module homomorphism and sends $r\in R$ to $r\binom{k+\ell-2}{k-1}$. If $R\to S$ failed to be injective then it would send some $r\neq 0$ to $0$ which implies $f(r)=0$ which means $\binom{k+\ell-2}{k-1}$ would be a zero divisor, contradicting our assumption.

  • 1
    $\begingroup$ Great argument! So it suffices that $\left(k+\ell-2\right)!$ is a regular element of $R$ (that is, a non-zero-divisor). In other words, it suffices that every prime number $\leq k + \ell - 2$ is a regular element of $R$ (since a product of regular elements is regular). $\endgroup$ Jun 26 at 22:14
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    $\begingroup$ @darijgrinberg Yes, or even those primes $p$ such that adding $k-1$ to $\ell-1$ in base $p$ does not involve carrying. $\endgroup$
    – Will Sawin
    Jun 26 at 22:44
  • 2
    $\begingroup$ Even that regularity assumption is unnecessary in some cases. Consider for instance the ring $R=\mathbb{Z}/32\mathbb{Z}$. Taking $a=\overline{4}$, we have $a^3=0$. Then $R$ embeds in the ring $R[b]/(8b-16,b^2)$. The image of $a-b$ in the factor ring squares to zero. $\endgroup$ Jun 27 at 0:55

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