ℚ-algebras with locally nilpotent derivations

Posted: March 15, 2024 in Derivation, Noncommutative Ring Theory Notes
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All rings in this post are commutative with identity. For the basics on derivations of rings see this post and this post.

Let \delta be a derivation of a ring R. Since \delta is additive, \delta(na)=n\delta(a) for all integers n and all a \in R. So every derivation of a ring is a \mathbb{Z}-derivation. If R is \mathbb{Q}-algebra and r=\frac{m}{n} \in \mathbb{Q}, where m,n are integers and n \ne 0, then n\delta(ra)=\delta(nra)=\delta(ma)=m\delta(a) and so \delta(ra)=r\delta(a). Thus every derivation of a \mathbb{Q}-algebra is a \mathbb{Q}-derivation.

Definition. We say that a derivation \delta of a ring R is locally nilpotent if for every a \in R, there exists a positive integer n such that \delta^n(a)=0.

Example. Let R be the polynomial ring \mathbb{C}[x]. Then the derivation \delta:=\frac{d}{dx} is locally nilpotent because if p(x) \in R has degree n, then \delta^{n+1}(p(x))=0.

The following Theorem characterizes all \mathbb{Q}-algebras R for which there exists a locally nilpotent derivation \delta and a \in R such that \delta(a)=1. The polynomial ring in the above Example gives one of those algebras since, in that example, \delta(x)=1. It turns out that any such algebra is a polynomial ring!

Theorem. Let \delta be a locally nilpotent derivation of a \mathbb{Q}-algebra R, and let K:=\ker \delta. If there exists x \in R such that \delta(x)=1, then x is transcendental over K and R=K[x].

Proof. Suppose, to the contrary, that x is algebraic over K, and let n be the smallest positive integer such that \alpha_nx^n+\alpha_{n-1}x^{n-1}+ \cdots + \alpha_1x+\alpha_0=0 for some \alpha_i \in K, \ \alpha_n \ne 0. Then, since by the product rule, \delta(x^i)=ix^{i-1}\delta(x)=ix^{i-1}, we have

0=\delta( \alpha_nx^n+\alpha_{n-1}x^{n-1}+ \cdots + \alpha_1x+\alpha_0)= \alpha_n\delta(x^n)+\alpha_{n-1}\delta(x^{n-1})+ \cdots + \alpha_1\delta(x)

=n\alpha_n x^{n-1}+(n-1)\alpha_{n-1}x^{n-1}+ \cdots + \alpha_1,

which contradicts the minimality of n. So x is transcendental over K. We now show that R=K[x]. For a \in R, let \nu(a) be the smallest positive integer m such that \delta^m(a)=0. The proof is by induction over \nu(a), \ a \in R. If \nu(a)=1, then \delta(a)=0 and so a \in K \subset K[x]. Suppose now that a \in R and m:=\nu(a) \ge 2. Let

\displaystyle y:=\sum_{n=0}^{m-1}\frac{(-1)^n\delta^n(a)x^n}{n!}=a-bx,

where

\displaystyle b=\sum_{n=1}^{m-1}\frac{(-1)^{n-1}\delta^n(a)x^{n-1}}{n!}.

So a=y+bx and so we are done if we prove that b,y \in K[x].

Claim 1: y \in K.

Proof. We have

\displaystyle \delta(y)=\sum_{n=0}^{m-1}\frac{(-1)^n}{n!}\delta(\delta^n(a)x^n)=\sum_{n=0}^{m-1}\frac{(-1)^n}{n!}(\delta^{n+1}(a)x^n+\delta^n(a)\delta(x^n))

\displaystyle =\sum_{n=0}^{m-1}\frac{(-1)^n}{n!}(\delta^{n+1}(a)x^n+n\delta^n(a)x^{n-1})=\sum_{n=0}^{m-1}\frac{(-1)^n\delta^{n+1}(a)x^n}{n!}+\sum_{n=1}^{m-1}\frac{(-1)^n\delta^n(a)x^{n-1}}{(n-1)!}

\displaystyle =\sum_{n=0}^{m-2}\frac{(-1)^n\delta^{n+1}(a)x^n}{n!}+\sum_{n=0}^{m-2}\frac{(-1)^{n+1}\delta^{n+1}(a)x^n}{n!}=0

and so y \in K.

Claim 2: b \in K[x].

Proof. By the Leibniz formula,

\displaystyle \delta^{m-1}(b)=\sum_{n=1}^{m-1}\frac{(-1)^{n-1}}{n!}\delta^{m-1}(\delta^n(a)x^{n-1})=\sum_{n=1}^{m-1}\frac{(-1)^{n-1}}{n!}\sum_{k=0}^{m-1}\binom{m-1}{k}\delta^{n+k}(a)\delta^{m-k-1}(x^{n-1}).

Now notice that \delta^{n+k}(a)\delta^{m-k-1}(x^{n-1})=0 for all k,n because if n+k \ge m, then \delta^{n+k}(a)=0, and if n+k < m, then \delta^{m-k-1}(x^{n-1})=0. Thus \delta^{m-1}(b)=0 and so \nu(b) < \nu(a)=m. Hence, by our induction hypothesis, b \in K[x]. \ \Box

Exercise. Let \delta be a locally nilpotent derivation of a \mathbb{Q}-algebra R, and let c \in R. Let K:=\ker \delta, and define the map f: R \to R by

\displaystyle f(a):=\sum_{n=0}^{\infty}\frac{\delta^n(a)c^n}{n!},

for all a \in R. Show that f is a K-algebra homomorphism.

Note. The above Theorem is Theorem 2.8 in here. The proof I’ve given is essentially the same as the proof given in there; I just made the proof easier to follow.

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