In a ring with a $p$-derivation every $p$-power-torsion element is nilpotent

Question

Let $p$ be a prime. The definition of $p$-derivation on a ring (aka $\delta$-ring structure) can be read in [K, Definition 2.1.1]. In short, a $\delta$-ring is a commutative ring with unity $A$ plus a function $\delta:A\to A$ satisfying a bunch of axioms. If $(A,\delta)$ is a $\delta$-ring, then the map $\phi:A\to A$ given by $\phi(x)=x^p+p\delta(x)$ is a ring homomorphism. I am trying to solve the following:

[K, Exercise 2.5.4]. Prove that in any $\delta$-ring every $p$-power-torsion element is nilpotent.

Let $A$ be a $\delta$-ring and suppose $x\in A$ satisfies $p^nx=0$ for some $n\geq 1$. Then using the identity $\delta(p^n)=\frac{p^n-p^{np}}{p}$ [K, Example 2.2.2] and the product rule of $p$-derivations gives \begin{align*} 0&=\delta(p^nx)\\ &=p^{np}\delta(x)+x^p\delta(p^n)+p\delta(x)\delta(p^n)+p\delta(x)\delta(p^n)\\ &=p^{np}\delta(x)+x^p\frac{p^n-p^{np}}{p}+\delta(x)(p^n-p^{np})\\ &=p^{n-1}x^p-x^pp^{np-1}+p^n\delta(x)\\ &=p^{n-1}x^p+p^n\delta(x)\\ &=p^{n-1}(x^p+p\delta(x))\\ &=p^{n-1}\phi(x). \end{align*} How can I use this fact to derive nilpotency of $x$?

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References

[K] K. S. Kedlaya, Notes on Prismatic Cohomology https://kskedlaya.org/prismatic/frontmatter-1.html

Answer 1

The following proof is due to Devlin Mallory. By induction on $n$. For $n=0$ the result is trivial.

Let $n\geq 1$ and assume the result true for $n-1$. Suppose $p^nx=0$. We have \begin{align*} p^{n-1}x^{2p} &=p^{n-1}(\phi(x)-p\delta(x))^2\\ &=p^{n-1}(\phi(x)^2+(p\delta(x))^2-2\phi(x)\delta(x))\\ &=p^{n-1}(p\delta(x))^2 &\text{since }p^{n-1}\phi(x)=0\\ &=p^{n-1}p\delta(x)(\phi(x)-x^p)\\ &=\delta(x)(p^n\phi(x)-p^nx^p)\\ &=\delta(x)(0-0)=0, \end{align*} since $p^{n-1}\phi(x)=0$ (shown above) and $p^nx=0$. Applying the induction hypothesis, we get that $x^{2p}$ is nilpotent, thus $x$ is nilpotent.