# Splitting a nilpotent into square-zeros by ring extension

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.

• 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?
– Z. M
Jun 26 at 19:38
• @Z.M: Can you walk me through the argument for why this would help? Jun 26 at 19:40
• 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.
– Z. M
Jun 26 at 21:20

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.

• 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). Jun 26 at 22:14
• @darijgrinberg Yes, or even those primes $p$ such that adding $k-1$ to $\ell-1$ in base $p$ does not involve carrying. Jun 26 at 22:44
• 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. Jun 27 at 0:55