Let \(A[\mathbf{x}] = A[x_{1}, \cdots, x_{n}\)] be the polynomial ring in n variables over an integral domain \(A\), \(D\) an \(A\)-derivation of \(A[\mathbf{x}]\) and denote \[L^{D}_{ij} := D(x_{i}) x_{j} - D(x_{j}) x_{i}, \quad \text{ for each } i, j \in \{1, \cdots, n\}.\]
Theorem 1 [Nowicki 1, Conjecture 6.9.10, Kuroda's Proof]
Assume \(k[\mathbf y]=k[y_1,\ldots,y_n]\) is a polynomial ring in \(n\) variables over a field \(k\) of characteristic zero. If \(\Delta_{n}\) is the \(k[\mathbf{y}]\)-derivation of \(k[\mathbf{y}][\mathbf{x}]\) defined by \(\Delta_{n}(x_{i}) = y_{i}\) for \(i = 1, \cdots, n\), then
\[ \ker\Delta_n= k[\mathbf y]\big[L^{\Delta_n}_{ij}\big] = k[\mathbf y]\big[y_i x_j-y_jx_i:1\le i < j \le n\big]. \]
Equivalently, the ring of constants of \(\Delta_n\) is generated by the \(2\times 2\)-minors \(y_i x_j-y_jx_i,1\le i < j \le n\) over \(k[\mathbf y]\).
Proof
Induction step
We prove the conjecture by induction on \(n\). The assertion is clear when \(n=1\). Assume that \(n\geq2\), and for each \(l\leq n\), let \(S_l\) be the set of \(L_{i,j}:=L_{i,j}^{\Delta_n}\) with \(1\leq i<j\leq l\). By the induction hypothesis, \(\ker\Delta_{n-1}\) is generated by \(S_{n-1}\) over \(k[\mathbf y']:=k[y_1,\ldots,y_{n-1}]\), since \(L_{i,j}^{\Delta_{n-1}}=L_{i,j}^{\Delta_n}\) for each \(i,j\). The \(k[\mathbf y']\)-derivation \(\Delta_{n-1}\) naturally extends to a \(k[\mathbf y]\)-derivation \((\Delta_{n-1})_{k[\mathbf y]}\) of \(k[\mathbf y][\mathbf x']:=k[\mathbf y][x_1,\ldots,x_{n-1}]\). Then \((\Delta_{n-1})_{k[\mathbf y]}=\Delta_n|_{k[\mathbf y][\mathbf x']}\), so \(\ker(\Delta_{n-1})_{k[\mathbf y]}=k[\mathbf y][\mathbf x']\cap\ker\Delta_n\). Moreover, \(\ker(\Delta_{n-1})_{k[\mathbf y]}=k[\mathbf y]\otimes_{k[\mathbf y']}\ker\Delta_{n-1}\), since \(k[\mathbf y]\) is flat over \(k[\mathbf y']\). Thus,
\[ k[\mathbf y][\mathbf x']\cap\ker\Delta_n=k[\mathbf y][S_{n-1}]. \tag{1} \]
The \(\Gamma\)-grading
Let \(\mathbf e_1,\ldots,\mathbf e_n\) be the coordinate unit vectors of \(\mathbb Z^n\), let \(M\) be the \(\mathbb Z\)-submodule of \((\mathbb Z^n)^2\) generated by \((\mathbf e_j-\mathbf e_i,\mathbf e_i-\mathbf e_j)\) for \(1\leq i<j\leq n\), and set \(\Gamma=(\mathbb Z^n)^2/M\). We define \(\Gamma\)-gradings on \(k[\mathbf y][\mathbf x]\) and \(k[\mathbf y^{\pm1}][\mathbf x]:=k[\mathbf y][\mathbf x][(y_1\cdots y_n)^{-1}]\) as follows. Recall that a \(k\)-algebra \(R\) is \(\Gamma\)-graded if there are \(k\)-vector subspaces \(R_\gamma\), \(\gamma\in\Gamma\), such that \(R=\bigoplus_{\gamma\in\Gamma}R_\gamma\) and \(R_\gamma R_\mu\subset R_{\gamma+\mu}\) for all \(\gamma,\mu\in\Gamma\). Let \(\mathbb Z_{\geq0}\) denote the nonnegative integers, and write \(\mathbf y^a=y_1^{a_1}\cdots y_n^{a_n}\) and \(\mathbf x^b=x_1^{b_1}\cdots x_n^{b_n}\) for \(a=(a_1,\ldots,a_n)\) and \(b=(b_1,\ldots,b_n)\). For each \(\gamma\in\Gamma\), define \(k[\mathbf y][\mathbf x]_\gamma\) (resp. \(k[\mathbf y^{\pm1}][\mathbf x]_\gamma\)) to be the \(k\)-vector space generated by \(\mathbf y^a\mathbf x^b\) with \(a,b\in(\mathbb Z_{\geq0})^n\) (resp. \(a\in\mathbb Z^n\) and \(b\in(\mathbb Z_{\geq0})^n\)) whose image in \(\Gamma\) is \(\gamma\). Note that \(\Delta_n(k[\mathbf y][\mathbf x]_\gamma)\subset k[\mathbf y][\mathbf x]_{\gamma-\delta}\) for each \(\gamma\in\Gamma\), where \(\delta\) is the image of \((-\mathbf e_n,\mathbf e_n)\) in \(\Gamma\). Hence \(\ker\Delta_n=\bigoplus_{\gamma\in\Gamma}(k[\mathbf y][\mathbf x]_\gamma\cap\ker\Delta_n)\).
Reduction to a homogeneous element
Thus, it is enough to show that every \(0\neq\Phi\in k[\mathbf y][\mathbf x]_\gamma\cap\ker\Delta_n\) belongs to \(k[\mathbf y][S_n]\). Choose \(a=(a_1,\ldots,a_n)\in\mathbb Z^n\) and \(l\in\mathbb Z_{\geq0}\) such that the image of \((a,l\mathbf e_n)\) in \(\Gamma\) is \(\gamma\). Let \(m\) be the \(x_n\)-degree of \(\Phi\), with \(0\leq m\leq l\), and let \(\phi\in k[\mathbf y][\mathbf x']\) be the coefficient of \(x_n^m\) in \(\Phi\). Then \(\phi\in k[\mathbf y][\mathbf x]_\mu\), where \(\mu\) is the image of \((a,(l-m)\mathbf e_n)\) in \(\Gamma\). Furthermore, \(0=\Delta_n(\Phi)=\Delta_n(\phi)x_n^m+m\phi y_nx_n^{m-1}+\Delta_n(\Phi-\phi x_n^m)\). Since the last two terms have \(x_n\)-degree at most \(m-1\), we have \(\Delta_n(\phi)=0\). Hence \(\phi\in k[\mathbf y][S_{n-1}]\) by (1).
Write \(\phi=\sum_{b,\mathbf u}r'_{b,\mathbf u}\mathbf y^b\widehat{\mathbf y}^{-\mathbf u}L^{\mathbf u}\), where \(b\in(\mathbb Z_{\geq0})^n\), \(\mathbf u=(u_{i,j})_{1\leq i<j\leq n-1}\) with \(u_{i,j}\in\mathbb Z_{\geq0}\), and \(r'_{b,\mathbf u}\in k\). Here \(\widehat{\mathbf y}^{-\mathbf u}:=\prod_{1\leq i<j\leq n-1}(y_iy_j)^{-u_{i,j}}\) and \(L^{\mathbf u}:=\prod_{1\leq i<j\leq n-1}L_{i,j}^{u_{i,j}}\). We may assume that \(r'_{b,\mathbf u}=0\) whenever \(\mathbf y^b\widehat{\mathbf y}^{-\mathbf u}\notin k[\mathbf y]\). Let \(\eta(b,\mathbf u)\) be the image of \((b-|\mathbf u|\mathbf e_n,|\mathbf u|\mathbf e_n)\) in \(\Gamma\), where \(|\mathbf u|:=\sum_{1\leq i<j\leq n-1}u_{i,j}\). Then \(\mathbf y^b\widehat{\mathbf y}^{-\mathbf u}L^{\mathbf u}\in k[\mathbf y^{\pm1}][\mathbf x]_{\eta(b,\mathbf u)}\), since \((y_iy_j)^{-1}L_{i,j}\in k[\mathbf y^{\pm1}][\mathbf x]_\delta\) for each \(i,j\).
Since \(\phi\in k[\mathbf y][\mathbf x]_\mu\) and \(\mu\) is the image of \((a,(l-m)\mathbf e_n)\), we may assume that \(r'_{b,\mathbf u}=0\) unless \(|\mathbf u|=l-m\) and \(b=a+(l-m)\mathbf e_n\). For each \(\mathbf u\) with \(r_{\mathbf u}:=r'_{a+(l-m)\mathbf e_n,\mathbf u}\neq0\), write \(\mathbf y^ay_n^{l-m}\widehat{\mathbf y}^{-\mathbf u}=y_1^{\rho_1(\mathbf u)}\cdots y_{n-1}^{\rho_{n-1}(\mathbf u)}y_n^s\), where \(\rho_i(\mathbf u)\in\mathbb Z_{\geq0}\) for \(i=1,\ldots,n-1\) and \(s=a_n+l-m\). Then \(\phi=y_n^s\sum_{\mathbf u}r_{\mathbf u}y_1^{\rho_1(\mathbf u)}\cdots y_{n-1}^{\rho_{n-1}(\mathbf u)}L^{\mathbf u}\). Since \(|\mathbf u|=l-m\), it follows that
\[ \sum_{i=1}^{n-1}\rho_i(\mathbf u)=\sum_{i=1}^{n-1}a_i-2(l-m). \tag{2} \]
The minimal-degree argument
Now we show that \(\Phi\in k[\mathbf y][S_n]\) by contradiction. Replacing \(\Phi\) if necessary, we may assume that \(m\) is minimal among the \(x_n\)-degrees of elements of \(\ker\Delta_n\setminus k[\mathbf y][S_n]\). To obtain a contradiction, it suffices to prove
\[ m\geq2l-\sum_{i=1}^{n-1}a_i. \tag{3} \]
Indeed, (3) and (2) imply \(\sum_{i=1}^{n-1}\rho_i(\mathbf u)\geq m\). Hence, for each \(\mathbf u\), there exist integers \(0\leq\rho_i'(\mathbf u)\leq\rho_i(\mathbf u)\) such that \(\sum_{i=1}^{n-1}\rho_i'(\mathbf u)=m\). Define
\[ \Phi':=y_n^s\sum_{\mathbf u}r_{\mathbf u}L^{\mathbf u}\prod_{i=1}^{n-1}y_i^{\rho_i(\mathbf u)-\rho_i'(\mathbf u)}L_{i,n}^{\rho_i'(\mathbf u)} =y_n^s\sum_{\mathbf u}r_{\mathbf u}L^{\mathbf u}\prod_{i=1}^{n-1}y_i^{\rho_i(\mathbf u)-\rho_i'(\mathbf u)}(y_ix_n-y_nx_i)^{\rho_i'(\mathbf u)}. \]
Then \(\Phi'\in k[\mathbf y][S_n]\) has \(x_n\)-degree \(m\), and the coefficient of \(x_n^m\) is \(\phi\). Therefore, the \(x_n\)-degree of \(\Phi-\Phi'\) is less than \(m\). Since \(\Phi-\Phi'\in\ker\Delta_n\setminus k[\mathbf y][S_n]\), this contradicts the minimality of \(m\).
Proof of the degree bound
We first record a standard fact about locally nilpotent derivations.
Lemma (Factorial closure)
Let \(D\) be a locally nilpotent derivation of an integral domain \(R\) of characteristic zero. Then \(\ker D\) is factorially closed in \(R\): if \(0\neq fg\in\ker D\), then \(f,g\in\ker D\).
Proof
For \(0\neq h\in R\), define \(\deg_D(h):=\max\{r\geq0:D^r(h)\neq0\}\), which is finite because \(D\) is locally nilpotent. If \(a=\deg_D(f)\) and \(b=\deg_D(g)\), then the Leibniz formula gives \(D^{a+b}(fg)=\binom{a+b}{a}D^a(f)D^b(g)\neq0\), while \(D^{a+b+1}(fg)=0\). Hence \(\deg_D(fg)=a+b\). If \(fg\in\ker D\), then \(\deg_D(fg)=0\), so \(a=b=0\), and therefore \(f,g\in\ker D\). \(\square\)
It remains to establish (3) for every nonzero homogeneous element \(\Phi\in\ker\Delta_n\). Suppose, to the contrary, that (3) fails, and choose such a \(\Phi\) with \(m\) minimal. Then \(t:=2l-\sum_{i=1}^{n-1}a_i-m>0\), and by (2), \(\sum_{i=1}^{n-1}\rho_i(\mathbf u)=m-t\) for each \(\mathbf u\). Hence, the \(x_n\)-degree of
\[ \Phi_1:=\sum_{\mathbf u}r_{\mathbf u}L^{\mathbf u}\prod_{i=1}^{n-1}L_{i,n}^{\rho_i(\mathbf u)} =\sum_{\mathbf u}r_{\mathbf u}L^{\mathbf u}\prod_{i=1}^{n-1}(y_ix_n-y_nx_i)^{\rho_i(\mathbf u)} \]
is \(m-t\). The coefficient of \(x_n^{m-t}\) in \(y_n^s\Phi_1\) is \(\phi\), so the coefficient of \(x_n^m\) in \(y_n^s\Phi_1L_{1,n}^t\) is equal to that in \(y_1^t\Phi\). Consequently, the \(x_n\)-degree \(m'\) of \(\Phi_2:=y_1^t\Phi-y_n^s\Phi_1L_{1,n}^t\) is less than \(m\). We claim that \(\Phi_2=0\). Indeed, if \(\gamma'\) is the image of \((a+t\mathbf e_1,l\mathbf e_n)\) in \(\Gamma\) and \((a_1',\ldots,a_n'):=a+t\mathbf e_1\), then \(\Phi_2\in k[\mathbf y][\mathbf x]_{\gamma'}\cap\ker\Delta_n\), and \(2l-\sum_{i=1}^{n-1}a_i'=2l-\sum_{i=1}^{n-1}a_i-t=m>m'\). Thus \(\Phi_2=0\) by the minimality of \(m\).
Hence \(y_1^t\Phi=y_n^s\Phi_1L_{1,n}^t\). Since neither \(y_n\) nor \(L_{1,n}\) is divisible by \(y_1\), it follows that \(\Phi_1\) is divisible by \(y_1\). By the lemma, \(\ker\Delta_n\) is factorially closed. Since \(\Delta_n\) is locally nilpotent, \(\Delta_n(\Phi_1)=0\), \(\Phi_1\neq0\), and \(\Delta_n(x_n)\neq0\), the polynomial \(\Phi_1\) is not divisible by \(x_n\). Setting \(x_n=0\) therefore gives the nonzero polynomial
\[ \sum_{\mathbf u}r_{\mathbf u}L^{\mathbf u}\prod_{i=1}^{n-1}(-y_nx_i)^{\rho_i(\mathbf u)}=(-y_n)^{m-t}\Psi, \qquad \Psi:=\sum_{\mathbf u}r_{\mathbf u}L^{\mathbf u}\prod_{i=1}^{n-1}x_i^{\rho_i(\mathbf u)}. \]
Thus \(\Psi\neq0\), and \(\Psi\) is divisible by \(y_1\), since \(\Phi_1\) is. Define \(\sigma\in\operatorname{Aut}_k k[\mathbf y][\mathbf x]\) by \(\sigma(x_i)=y_i\) and \(\sigma(y_i)=x_i\) for \(i=1,\ldots,n\). Then \(\sigma(\Psi)\) is divisible by \(x_1\). On the other hand, \(\sigma(L_{i,j})=L_{j,i}\) and \(\sigma(x_i)=y_i\) belong to \(\ker\Delta_n\) for each \(i,j\), so \(\sigma(\Psi)\in\ker\Delta_n\). Since \(\ker\Delta_n\) is factorially closed and \(x_1\notin\ker\Delta_n\), this is impossible for the nonzero polynomial \(\sigma(\Psi)\). Therefore (3) holds. Hence \(\Phi\in k[\mathbf y][S_n]\), completing the proof of the conjecture.
A differential approach \({\rm DCF}_0\)
There is a useful way to read Nowicki's theorem inside differential algebra. Let \((\mathcal U,\partial)\models {\rm DCF}_0\), with constant field \(C=\ker\partial\), and consider a tuple satisfying \(\partial y_i=0\) and \(\partial x_i=y_i\) for \(i=1,\ldots,n\). The derivation induced on the polynomial ring \(k[\mathbf y][\mathbf x]\) is exactly Kuroda's derivation \(\Delta_n\). Thus \(\ker\Delta_n\) is the ring of polynomial first integrals of the differential system \(\mathbf y'=0\), \(\mathbf x'=\mathbf y\).
The basic generators appear immediately from this viewpoint. Since \(\partial y_i=0\) and \(\partial x_i=y_i\), we have \(\partial L_{ij}=\partial(y_ix_j-y_jx_i)=y_iy_j-y_jy_i=0\). Hence every \(L_{ij}\) is a differential constant. Geometrically, the system carries the definable action of the additive constant group \((C,+)\) given by \(c\cdot(\mathbf x,\mathbf y)=(\mathbf x+c\mathbf y,\mathbf y)\). The quantities \(y_i\) and \(L_{ij}\) are invariant under this action, and for \(\mathbf y\neq0\) they separate its orbits. In this sense, the minors \(L_{ij}\) provide concrete algebraic coordinates for the quotient by the one-dimensional constant flow.
The rational version is particularly transparent. On the chart \(y_1\neq0\), put \(t=x_1/y_1\). Then \(\partial t=1\), while \(L_{1j}=y_1x_j-y_jx_1\) gives \(x_j=y_jt+L_{1j}/y_1\). Thus the whole differential field is obtained from its constants by adjoining a single element \(t\) with \(t'=1\). Consequently, at the level of rational functions, the first integrals are generated by the \(y_i\) and the \(L_{ij}\):
\[ \operatorname{Frac}(k[\mathbf y][\mathbf x])^{\Delta_n}=k(\mathbf y,L_{ij}). \]
From this perspective, the real content of Nowicki's theorem is not the rational description but the passage back to the polynomial ring. It asserts that no hidden denominators are needed:
\[ k[\mathbf y][\mathbf x]\cap k(\mathbf y,L_{ij})=k[\mathbf y][L_{ij}]. \]
Equivalently, the polynomial first integrals are already generated by the obvious differential constants. This is precisely where Kuroda's argument goes beyond the elementary \(DCF_0\) picture: the differential-field calculation identifies the rational quotient almost immediately, whereas the \(\Gamma\)-grading, the minimal-degree argument, and factorial closure are what eliminate denominators and recover the coordinate ring itself.
There is also a model-theoretic way to phrase the same geometry. For fixed \(\mathbf y\in C^n\), consider the definable equivalence relation \(\mathbf x\sim\mathbf z\) if \(\mathbf z=\mathbf x+c\mathbf y\) for some \(c\in C\). On the locus \(\mathbf y\neq0\), the tuple \((\mathbf y,(L_{ij})_{i<j})\) gives a concrete code for the corresponding orbit. Thus Nowicki's generators may be viewed as polynomial representatives of the natural quotient invariants of a definable \((C,+)\)-action. The theorem says that, in this special linear situation, the model-theoretically natural invariants and the algebraically generated polynomial constants coincide exactly.
References
- Andrzej Nowicki, Polynomial Derivations and Their Rings of Constants, Uniwersytet Mikołaja Kopernika, Toruń, 1994.
- Shigeru Kuroda, "A Simple Proof of Nowicki's Conjecture on the Kernel of an Elementary Derivation," Tokyo Journal of Mathematics 32 (2009), no. 1, 247-251.