Exponential of derivation is automorphism under suitable nilpotency assumptions

Weaker version: global powering and torsion assumptions
Suppose $$R$$ is a non-associative ring (i.e., a not necessarily associative ring) and $$d$$ is a derivation of $$R$$ satisfying the following:


 * 1) $$d$$ is nilpotent.
 * 2) The nilpotency of $$d$$ is at most one more than the powering threshold for $$R$$. In other words, there exists a natural number $$n$$ such that $$d^n = 0$$ and $$R$$ is powered for all primes strictly less than $$n$$.
 * 3) The binilpotency of $$d$$ is at most one more than the torsion-free threshold for $$R$$. In other words, there exists a natural number $$m$$ such that $$d^i(x) * d^j(y) = 0$$ for all $$x,y \in R, i + j \ge m$$, and $$R$$ is $$p$$-torsion-free for all primes $$p < m$$.

Then, $$d$$ is an exponentiable derivation. Explicitly, if $$d^n = 0$$, then the function:

$$\exp(d) := \operatorname{id} + \frac{d}{1!} + \frac{d^2}{2!} + \dots + \frac{d^{n-1}}{(n-1)!}$$

is an automorphism of $$R$$.

Stronger version: image powering and torsion assumptions
Suppose $$R$$ is a non-associative ring (i.e., a not necessarily associative ring) and $$d$$ is a derivation of $$R$$. Suppose that $$d$$ satisfies the following three conditions:


 * 1) $$d$$ is locally nilpotent.
 * 2) $$d$$ is an infinitely powered endomorphism of $$R$$, i.e., the  powering threshold for $$d$$ is $$\infty$$.
 * 3) $$d$$ is an infinitely bi-torsion-free endomorphism of $$R$$, i.e., the  bi-torsion-free threshold for $$d$$ is $$\infty$$.

Then, $$d$$ is an exponentiable derivation, i.e., the exponential of $$d$$ exists and is an automorphism of $$R$$.

Note on conditions
Note that conditions (1) and (2) are necessary for us to even be able to talk of the exponential of a derivation. Condition (3) is what helps us prove that the exponential must be an automorphism. It's possible that there are weaker versions of (3) that still provide the same guarantee, so the above form sufficient but not necessary conditions for a derivation to be exponentiable.

Related facts

 * Logarithm of automorphism is derivation under suitable nilpotency assumptions
 * Exponential of nilpotent derivation with divided powers is automorphism

Facts used

 * 1) uses::Exponential of nilpotent derivation with divided powers is automorphism
 * 2) uses::Binomial formula for powers of a derivation

Proof with global powering and torsion assumptions
Given: $$R$$ is a non-associative ring (i.e., a not necessarily associative ring) and $$d$$ is a derivation of $$R$$ satisfying the following:


 * 1) $$d$$ is nilpotent.
 * 2) The nilpotency of $$d$$ is at most one more than the powering threshold for $$R$$. In other words, there exists a natural number $$n$$ such that $$d^n = 0$$ and $$R$$ is powered for all primes strictly less than $$n$$.
 * 3) The binilpotency of $$d$$ is at most one more than the torsion-free threshold for $$R$$. In other words, there exists a natural number $$m$$ such that $$d^i(x) * d^j(y) = 0$$ for all $$x,y \in R, i + j \ge m$$, and $$R$$ is $$p$$-torsion-free for all primes $$p < m$$.

To prove: $$d$$ is an exponentiable derivation. Explicitly, if $$d^n = 0$$, then the function:

$$\exp(d) := \operatorname{id} + \frac{d}{1!} + \frac{d^2}{2!} + \dots + \frac{d^{n-1}}{(n-1)!}$$

is an automorphism of $$R$$.

Proof: We will show that the following is a nilpotent derivation with divided powers:

$$d^{(i)} := \left\lbrace \begin{array}{rl} \frac{d^i}{i!}, & 0 \le i < n \\ 0, & i \ge n\\\end{array}\right.$$

Once we have shown this, Fact (1) will complete the proof.

Note that once we have established that the above is a derivation, it is clearly nilpotent. So, we only need to check that it is a derivation.we need to check the divided power conditions and the Leibniz rule conditions. Let's check these.

Divided power conditions: We need to show that for all nonnegative integers $$i,j$$,

$$d^{(i)}d^{(j)} = \frac{i + j}{i}d^{(i + j)}$$.

Leibniz rule conditions: We need to show that, for all nonnegative integers $$k$$:

$$d^{(k)}(x * y) = \sum_{i + j = k} d^{(i)}(x) * d^{(j)}(y)$$

We make three cases: