# Locally nilpotent derivations *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* An action of the additive group of a field on an affine variety is the same thing as a locally nilpotent derivation of its coordinate ring. The derivation is the infinitesimal action, and the action is its exponential. Such derivations are the main tool for telling affine space apart from varieties that resemble it: affine space has many of them, and a variety that is too rigid to carry them cannot be affine space. This lesson develops what the next lessons use. It proves the correspondence between derivations and actions, the additivity of the order of an element, which makes the kernel closed under taking factors, the description of these derivations on a polynomial ring in one variable, and two ways of replacing a derivation by a homogeneous one: taking the extreme component for a grading, and passing to the associated graded ring of a filtration. Throughout, \(k\) is a field of characteristic zero, rings are commutative \(k\)-algebras with identity, and homomorphisms are \(k\)-algebra homomorphisms. We use localization and polynomial rings over a domain, as in [Spectra of rings](course:AG-CA/spectra-of-rings) and [Localization, local properties and support](course:AG-CA/localization-local-properties-and-support). Basic references are [Kaliman–Makar-Limanov] and [OpenAI-cancellation]. ## 1. Derivations and local nilpotence A **derivation** of a ring \(A\) is a \(k\)-linear map \(D:A\to A\) with \(D(fg)=fD(g)+gD(f)\). It kills \(k\), because \(D(1)=D(1\cdot1)=2D(1)\). Induction on \(n\) gives the **Leibniz formula** (1.1): \[ \begin{aligned} D^n(fg)&=\sum_{i=0}^{n}\binom ni D^i(f)\,D^{n-i}(g). \end{aligned} \] A derivation is determined by its values on a set of generators of \(A\), since the product rule computes it on every monomial in the generators. **Definition 1.1.** A derivation \(D\) of \(A\) is **locally nilpotent** if for every \(f\in A\) there is \(n\ge1\) with \(D^n(f)=0\). Its **kernel** \(\ker D\) is a subring of \(A\) containing \(k\). **Lemma 1.2.** If \(A\) is generated by elements each killed by some power of \(D\), then \(D\) is locally nilpotent. **Proof.** Let \(N\) be the set of elements killed by some power of \(D\). It is closed under sums. If \(D^m(f)=0\) and \(D^n(g)=0\), then every term of (1.1) for \(D^{m+n-1}(fg)\) has \(i\ge m\) or \(m+n-1-i\ge n\), so \(D^{m+n-1}(fg)=0\). Thus \(N\) is a subring containing the generators and \(k\), so \(N=A\). \(\square\) **Example 1.3.** On \(k[x_1,\ldots,x_n]\), each partial derivative \(\partial/\partial x_i\) is locally nilpotent. More generally, a **triangular** derivation, one with \(D(x_1)\in k\) and \(D(x_i)\in k[x_1,\ldots,x_{i-1}]\) for \(i\ge2\), is locally nilpotent: by induction on \(i\), \(D\) maps \(k[x_1,\ldots,x_{i-1}]\) into itself and is locally nilpotent there, so \(D^{m+1}(x_i)=D^m(D(x_i))=0\) for large \(m\), and Lemma 1.2 applies. The derivation \(x\,\partial/\partial x\) of \(k[x]\) is not locally nilpotent, since every power sends \(x\) to \(x\). ## 2. Exponentials and actions of the additive group Let \(D\) be locally nilpotent and \(t\) a variable. For \(f\in A\) the sum \[ \exp(tD)(f)=\sum_{n\ge0}\frac{t^n}{n!}\,D^n(f)\in A[t] \] is finite. Extend \(D\) to polynomial rings \(A[s]\), \(A[t]\), \(A[s,t]\) by letting it kill the variables; the extension is locally nilpotent by Lemma 1.2. **Proposition 2.1.** Let \(D\) be a locally nilpotent derivation of \(A\). 1. \(\exp(tD):A\to A[t]\) is a ring homomorphism, and setting \(t=0\) gives the identity. 2. In \(A[s,t]\), \(\exp(sD)\bigl(\exp(tD)(f)\bigr)\) equals \(\exp((s+t)D)(f)\) for every \(f\in A\), where \(\exp(sD)\) is applied to the coefficients. 3. For a new variable \(w\), the map \(\Phi=\exp(wD):A[w]\to A[w]\), \(\Phi(f)=\sum_n w^nD^n(f)/n!\) with \(D(w)=0\), is a ring automorphism fixing \(w\), with inverse \(\exp(-wD)\). **Proof.** (1) By (1.1), \[ \begin{aligned} &\exp(tD)(f)\,\exp(tD)(g)\\ &\quad =\sum_{i,j}\frac{t^{i+j}}{i!\,j!}D^i(f)D^j(g)\\ &\quad =\sum_n\frac{t^n}{n!}\sum_{i}\binom niD^i(f)D^{n-i}(g)\\ &\quad =\exp(tD)(fg). \end{aligned} \] Additivity and \(\exp(tD)(1)=1\) are clear. (2) The left side is \(\sum_{m,n}s^mt^nD^{m+n}(f)/(m!\,n!)\), which is \(\sum_l D^l(f)(s+t)^l/l!\) by the binomial theorem. (3) The extended derivation of \(A[w]\) is locally nilpotent, so part (1) gives a ring homomorphism \(A[w]\to A[w][t]\); composing with \(t\mapsto w\) gives \(\Phi\), a ring homomorphism fixing \(w\). Since \(D\) kills \(w\), the same computation as in (2), with \(s=-w\) and \(t=w\), shows that \(\exp(-wD)(\Phi(f))\) equals \(\sum_lD^l(f)(w-w)^l/l!\), which is \(f\). Exchanging the roles of \(w\) and \(-w\) gives the other composite. \(\square\) **Definition 2.2.** An **action of the additive group** on \(\operatorname{Spec}A\) is a ring homomorphism \(\rho:A\to A[t]\), written \(\rho(f)=\sum_n\delta_n(f)t^n\), such that \(\delta_0\) is the identity and, for every \(f\), the identity (2.1) holds in \(A[s,t]\): \[ \begin{aligned} &\sum_{n}\rho_s\bigl(\delta_n(f)\bigr)\,t^n\\ &\quad=\sum_n\delta_n(f)\,(s+t)^n, \end{aligned} \] where \(\rho_s\) is \(\rho\) with the variable renamed \(s\). Geometrically, \(\rho\) is the comorphism of a morphism \(\mathbb A^1\times\operatorname{Spec}A\to\operatorname{Spec}A\), and (2.1) is the action law. **Proposition 2.3.** Every action of the additive group on \(\operatorname{Spec}A\) is \(\exp(tD)\) for a unique locally nilpotent derivation \(D\), namely \(D=\delta_1\). Conversely, every \(\exp(tD)\) is such an action. **Proof.** The converse is Proposition 2.1. Given \(\rho\), the coefficient of \(s^mt^n\) in (2.1) gives \[ \delta_m\delta_n=\binom{m+n}m\delta_{m+n}. \] With \(m=1\), \(\delta_1\delta_{n-1}=n\,\delta_n\), so induction gives \(\delta_n=\delta_1^n/n!\); here characteristic zero is used. The coefficient of \(t\) in \(\rho(fg)=\rho(f)\rho(g)\) is \(\delta_1(fg)=f\delta_1(g)+g\delta_1(f)\), so \(\delta_1\) is a derivation. Since \(\rho(f)\) is a polynomial in \(t\), \(\delta_1^n(f)=n!\,\delta_n(f)=0\) for large \(n\): the derivation \(\delta_1\) is locally nilpotent, and \(\rho=\exp(t\delta_1)\). Uniqueness holds because \(D\) is the coefficient of \(t\) in \(\exp(tD)\). \(\square\) The invariants of the action, the elements with \(\rho(f)=f\), are exactly the elements of \(\ker D\). ## 3. The order of an element **Definition 3.1.** Let \(D\) be a locally nilpotent derivation of a domain \(A\). For \(0\ne f\in A\), the **order** of \(f\) is \[ \nu_D(f)=\max\{n\ge0: D^n(f)\ne0\}. \] Equivalently, \(\nu_D(f)\) is the degree in \(t\) of the polynomial \(\exp(tD)(f)\): its coefficient of \(t^n\) is \(D^n(f)/n!\), which vanishes exactly when \(D^n(f)\) does. **Proposition 3.2.** Let \(D\) be a locally nilpotent derivation of a domain \(A\), and let \(f,g\in A\) be nonzero. 1. \(\nu_D(fg)=\nu_D(f)+\nu_D(g)\). 2. If \(D(f)\ne0\), then \(\nu_D(D(f))=\nu_D(f)-1\). 3. The kernel is **factorially closed**: if \(fg\ne0\) lies in \(\ker D\), then \(f,g\in\ker D\). In particular every unit of \(A\) lies in \(\ker D\). 4. If \(D(f)\ne0\), then \(D(f)\notin fA\). **Proof.** (1) The map \(\exp(tD)\) is multiplicative, and degrees in \(t\) add for nonzero polynomials over a domain. (2) is the definition. (3) If \(fg\in\ker D\) is nonzero, then \(\nu_D(f)+\nu_D(g)=0\), so both orders are zero. A unit \(e\) satisfies \(e\cdot e^{-1}=1\in\ker D\). (4) If \(D(f)=fq\), then \(q\ne0\) and (1), (2) give \(\nu_D(f)-1=\nu_D(f)+\nu_D(q)\ge\nu_D(f)\), which is impossible. \(\square\) **Corollary 3.3.** Let \(S\subset\ker D\) be a multiplicative set not containing \(0\). Then \(D(f/s)=D(f)/s\) defines a locally nilpotent derivation of \(S^{-1}A\), and \(D^n(f/s)=D^n(f)/s\). **Proof.** The quotient rule gives \(D(f/s)=(sD(f)-fD(s))/s^2\), which is \(D(f)/s\) because \(D(s)=0\); iterating gives the formula, and local nilpotence follows. \(\square\) ## 4. Derivations of a polynomial ring in one variable **Proposition 4.1.** Let \(F\) be a field of characteristic zero and \(\delta\) a locally nilpotent \(F\)-linear derivation of \(F[t]\). Then \(\delta(t)\in F\). **Proof.** Write \(\delta(t)=P\), so that \(\delta=P\,d/dt\). Suppose \(m=\deg P\ge1\), with leading coefficient \(\gamma\). We show by induction that \(\delta^n(t)\) has degree \(1+n(m-1)\) and leading coefficient \(\gamma^n\prod_{j=0}^{n-1}\bigl(1+j(m-1)\bigr)\). For \(n=1\) this is the definition. If \(Q=\delta^n(t)\) has degree \(e=1+n(m-1)\) and leading coefficient \(c\), then \(\delta(Q)=PQ'\) has degree \(m+e-1=1+(n+1)(m-1)\) and leading coefficient \(\gamma\,c\,e\). Every factor \(1+j(m-1)\) is a positive integer, hence nonzero in \(F\), so no power of \(\delta\) kills \(t\). This contradicts local nilpotence. So \(P\) is constant. \(\square\) ## 5. Homogeneous components for a grading Let \(A=\bigoplus_{i\in\mathbb Z}A_i\) be a \(\mathbb Z\)-graded ring, generated by finitely many homogeneous elements \(a_1,\ldots,a_r\). For a derivation \(D\) and \(j\in\mathbb Z\), let \(D_j(f)\) be the component of degree \(i+j\) of \(D(f)\) for \(f\in A_i\), and extend \(D_j\) additively. **Proposition 5.1.** Each \(D_j\) is a derivation, homogeneous of degree shift \(j\). Only finitely many \(D_j\) are nonzero, and \(D=\sum_jD_j\). If \(D\) is locally nilpotent and \(\lambda\) is the largest \(j\) with \(D_j\ne0\), then \(D_\lambda\) is locally nilpotent; the same holds for the smallest such \(j\). If \(A\) carries a second grading for which the \(a_m\) are homogeneous and \(D\) is homogeneous, then every \(D_j\) is homogeneous for the second grading with the same shift as \(D\). **Proof.** For \(f\in A_i\) and \(g\in A_l\), the component of degree \(i+l+j\) of \(D(fg)=fD(g)+gD(f)\) is \(fD_j(g)+gD_j(f)\), since multiplication by a homogeneous element shifts degrees. So \(D_j\) is a derivation. Its value on \(a_m\) is a component of \(D(a_m)\), and \(D(a_m)\) has finitely many nonzero components; a derivation vanishing on all generators is zero. Hence finitely many \(D_j\) are nonzero, and \(D=\sum_jD_j\) because this holds on homogeneous elements. Let \(f\in A_i\). Expanding \((\sum_jD_j)^n\), each composite \(D_{j_1}\cdots D_{j_n}(f)\) is homogeneous of degree \(i+j_1+\cdots+j_n\le i+n\lambda\), with equality only if every \(j_k=\lambda\). So the component of degree \(i+n\lambda\) of \(D^n(f)\) is \(D_\lambda^n(f)\), which therefore vanishes as soon as \(D^n(f)\) does. Every element is a finite sum of homogeneous ones, so \(D_\lambda\) is locally nilpotent. The smallest shift is treated in the same way. Finally, taking a component for the first grading does not change the degree for the second grading of a bihomogeneous element. \(\square\) ## 6. Filtrations and the associated graded ring Let \(A\) be a domain with a **degree function** \(\deg:A\setminus\{0\}\to\mathbb Z\) such that \(\deg(fg)=\deg f+\deg g\), that \(\deg(f+g)\le\max\{\deg f,\deg g\}\) when \(f+g\ne0\), and \(\deg c=0\) for \(c\in k^*\). Put \(\mathcal F_dA=\{f:\deg f\le d\}\cup\{0\}\), and let \[ \operatorname{gr}A=\bigoplus_{d\in\mathbb Z}\mathcal F_dA/\mathcal F_{d-1}A . \] Write \([f]_d\) for the class of \(f\in\mathcal F_dA\), and \(\operatorname{in}(f)=[f]_{\deg f}\) for the **initial form** of \(f\ne0\). Since degrees add, \(\operatorname{in}(f)\operatorname{in}(g)=\operatorname{in}(fg)\ne0\); every nonzero homogeneous element of \(\operatorname{gr}A\) is an initial form, so \(\operatorname{gr}A\) is a domain. **Proposition 6.1.** Let \(D\) be a locally nilpotent derivation of \(A\), and let \(j\in\mathbb Z\) satisfy \(D(\mathcal F_dA)\subset\mathcal F_{d+j}A\) for every \(d\). Then \[ D_0([f]_d)=[D(f)]_{d+j} \] defines a locally nilpotent derivation of \(\operatorname{gr}A\), homogeneous of degree shift \(j\). If \(D(h)=0\) and \(h\ne0\), then \(D_0(\operatorname{in}h)=0\). **Proof.** The map is well defined because \(D(\mathcal F_{d-1}A)\subset\mathcal F_{d+j-1}A\), and the product rule for \(D\) passes to classes. By induction, \(D_0^n([f]_d)=[D^n(f)]_{d+nj}\), which vanishes once \(D^n(f)=0\). Homogeneous classes span, so \(D_0\) is locally nilpotent. The last assertion is the case \(n=1\). \(\square\) The next statement chooses \(j\) and guarantees \(D_0\ne0\). Suppose \(A\) is generated by \(h_1,\ldots,h_r\), give a monomial \(h_1^{e_1}\cdots h_r^{e_r}\) the **weight** \(\sum e_i\deg h_i\), and say that the generators are **exact** if every nonzero \(f\in A\) is a polynomial in the \(h_i\) all of whose monomials have weight at most \(\deg f\). **Proposition 6.2.** Suppose the generators \(h_1,\ldots,h_r\) are exact, and let \(D\ne0\) be locally nilpotent. Put \(j=\max_i\bigl(\deg D(h_i)-\deg h_i\bigr)\), the maximum over the \(i\) with \(D(h_i)\ne0\). Then \(D(\mathcal F_dA)\subset\mathcal F_{d+j}A\) for all \(d\), and the derivation \(D_0\) of Proposition 6.1 is nonzero. **Proof.** Represent \(f\ne0\) as a polynomial in the \(h_i\) with monomials of weight at most \(\deg f\). By the product rule, \(D\) of a monomial of weight \(\omega\) is a sum of terms of degree at most \(\omega+j\), because differentiating a factor \(h_i\) raises its degree by at most \(j\). Hence \(\deg D(f)\le\deg f+j\). For an index \(i\) attaining the maximum, \(D_0(\operatorname{in}h_i)=[D(h_i)]_{\deg h_i+j}\), which is \(\operatorname{in}(D(h_i))\ne0\). \(\square\) ## 7. Exercises **Exercise 7.1 (easy).** Show that \(D=x_2\,\partial/\partial x_1+x_3\,\partial/\partial x_2\) is a locally nilpotent derivation of \(k[x_1,x_2,x_3]\), and compute \(\exp(tD)(x_1)\). **Exercise 7.2 (easy).** Show that every locally nilpotent derivation of \(k[x,x^{-1}]\) is zero. **Exercise 7.3 (medium).** Let \(D\) be locally nilpotent on a domain \(A\), and suppose \(D(a)=1\) for some \(a\in A\). Show that \(A=(\ker D)[a]\) and that \(a\) is transcendental over \(\ker D\). **Exercise 7.4 (easy).** Let \(D\) be a derivation of a \(\mathbb Z\)-graded ring, homogeneous of some degree shift. Show that \(\ker D\) is a graded subring: the homogeneous components of an element of the kernel lie in the kernel. ## 8. Solutions **7.1.** It is triangular in the order \(x_3,x_2,x_1\): \(D(x_3)=0\), \(D(x_2)=x_3\), \(D(x_1)=x_2\). So it is locally nilpotent by Example 1.3. Here \(D(x_1)=x_2\), \(D^2(x_1)=x_3\), \(D^3(x_1)=0\), so \(\exp(tD)(x_1)=x_1+tx_2+\tfrac{t^2}2x_3\). **7.2.** The ring is a domain generated by the units \(x\) and \(x^{-1}\). By Proposition 3.2(3), a locally nilpotent derivation kills both, hence is zero. **7.3.** Put \(B=\ker D\). We show \(f\in B[a]\) by induction on \(n=\nu_D(f)\). If \(n=0\), then \(f\in B\). If \(n>0\), then \(g=D^n(f)\) lies in \(B\), because \(D^{n+1}(f)=0\). Since \(D(a)=1\), we have \(D^n(a^n)=n!\) and \(D^n(ga^n)=g\,D^n(a^n)=n!\,g\). So \(f'=f-ga^n/n!\) satisfies \(D^n(f')=0\): either \(f'=0\) or \(\nu_D(f')