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]

Assume that \(k[\mathbf{y}] = k[y_{1}, \cdots, y_{n}]\) is the 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 \(\ker \Delta_{n}\) is generated by \(L^{\Delta_{n}}_{ij}\) for \(1 ≤ i < j ≤ n\) over \(k[\mathbf{y}]\).

Proof of Theorem 1

We prove the conjecture by induction on \(n\).

Step \(1\). The assertion is clear when \(n = 1\).

Step \(n-1\). By the assumption on induction, \(\ker \Delta_{n-1}\) is generated by \(S_{n-1}\) over \(k[\mathbf{y}^{\prime}] := k[y_{1}, \cdots , y_{n-1}]\), since \(L^{\Delta_{n-1}}_{i,j} = L^{\Delta_{n}}_{i,j}\) for each \(i, j\).

Step \(n\). Split the theorem into several lemmas.

Lemma 1

The \(k[\mathbf{y}^{\prime}]\)-derivation \(\Delta_{n-1}\) that naturally extends to a \(k[\mathbf{y}]\)-derivation \((\Delta_{n-1})_{k[\mathbf{y}]}\) of \(k[\mathbf{y}][\mathbf{x}^{\prime}] := k[\mathbf{y}][x_{1}, \cdots, x_{n-1}]\) satisfies

\[\begin{equation} k[\mathbf{y}][\mathbf{x}^{\prime}] \cap \ker \Delta_{n} = k[\mathbf{y}][S_{n-1}]. \end{equation}\]

Proof of Lemma 1

As discussed in Section 1, then \((\Delta_{n-1})k[\mathbf{y}] = \Delta_{n}|_{k[\mathbf{y}][\mathbf{x}^{\prime}]}\), so we have \(\ker(\Delta_{n-1})k[\mathbf{y}] = k[\mathbf{y}][\mathbf{x}^{\prime}] \cap \ker \Delta_{n}\). Moreover, \(\ker(\Delta_{n-1})k[\mathbf{y}] = k[\mathbf{y}] \otimes_{k[\mathbf{y}^{\prime}]} \ker \Delta_{n-1}\), since \(k[\mathbf{y}]\) is flat over \(k[\mathbf{y}^{\prime}]\). \(\square\)

Let \(\mathbf{e}_{1}, \cdots , \mathbf{e}_{n}\) be the coordinate unit vectors of \(\mathbb{R}^{n}\), \(M\) 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 \(\Gamma = (\mathbb{Z}^{n})^{2}/M\).

Theorem 2

For each \(\gamma \in \Gamma\), we define \(k[\mathbf{y}][\mathbf{x}]_{\gamma}\) to be the \(k\)-vector space generated by \(\mathbf{y}^{a}\mathbf{x}^{b}:= y^{a_{1}} \cdots y^{a_{n}} x^{b_{1}} \cdots x^{b_{n}}\) for \(a = (a_{1}, \cdots , a_{n})\) and \(b = (b_{1}, \cdots , b_{n})\) in \((\mathbb{Z}_{\geq 0})^{n}\) such that the image of \((a, b)\) in \(\Gamma\) is equal to \(\gamma\). Then

\[\ker \Delta_{n} = \bigoplus_{\gamma \in \Gamma} k[\mathbf{y}] [ \mathbf{x}]_{\gamma} \cap \ker \Delta_{n}\]

Proof of Theorem 2

Note that \(\Delta_{n}(k[\mathbf{y}][\mathbf{x}]_{\gamma})\) is contained in \(k[\mathbf{y}][\mathbf{x}]_{\gamma - \delta}\) for each \(\gamma \in \Gamma\), where \(\delta\) is the image of \((-\mathbb{e}_{n}, \mathbb{e}_{n})\) in \(\Gamma\). \(\square\)

Hence, we are reduced to showing that each \(0 \neq \varPhi \in k[\mathbf{y}][\mathbf{x}]_{\gamma} \cap \ker \Delta_{n}\) belongs to \(k[\mathbf{y}][S_{n}]\) for \(\gamma \in \Gamma\).

We may find \(a = (a_{1}, \cdots , a_{n}) \in \mathbb{Z}^{n}\) and \(l \in \mathbb{Z}_{\geq 0}\) such that the image of \((a, l\mathbf{e}_{n})\) in \(\Gamma\) is equal to \(\gamma\). Let \(m\) be the \(x_{n}\)-degree of \(\varPhi\), where \(0 \leq m \leq l\), and \(\phi \in k[\mathbf{y}][\mathbf{x}^{\prime}]\) the coefficient of \(x_{n}^{m}\) in \(\varPhi\). By replacing \(\varPhi\) if necessary, we may assume that \(m\) is the minimum among the \(x_{n}\)-degrees of elements of \(\ker \Delta_{n}\backslash k[\mathbf{y}][S_{n}]\).

Lemma 2

Write \(\phi\) in expression of sum of \(y_{1}\cdots y_{n}\)

\[\phi = y_{n}^{s}\sum_{\mathbf{u}} r_{\mathbf{u}} y_{1}^{\rho_{1}(\mathbf{u})} \cdots y_{n-1}^{\rho_{n-1}(\mathbf{u})} L^{\mathbf{u}}\]

where \(L^{\mathbf{u}} = \prod_{1 \leq i<j\leq n-1} L_{i,j}^{u_{i,j}}\) for each \(\mathbf{u}\). There holds the identity

\[\begin{equation} \sum_{i=1}^{n-1} \rho_{i} (\mathbf{u}) = \sum_{i=1}^{n-1}a_{i} - 2(l-m) \end{equation}\]

Lemma 3

There holds the inequality

\[\begin{equation} m \geq 2l - \sum_{i=1}^{n-1} a_{i}. \end{equation}\]

Proof of Lemma 2

then, \(\phi\) belongs to \(k[\mathbf{y}][\mathbf{x}]_{\mu}\), where \(\mu\) is the image of \((a, (l - m)\mathbf{e}_{n})\) in \(\Gamma\). Furthermore, \(0 = \Delta_{n}(\varPhi) = \Delta_{n}(\varphi)x^{m}_{n} + m\phi y_{n}x_{n}^{m-1} + \Delta_{n}(\varPhi - \phi x^{m}_{n})\), and the \(x_{n}\)-degrees of \(m\phi y_{n}x_{n}^{m-1}\) and \(\Delta_{n}(\varPhi - \phi x^{m}_{n})\) are at most \(m - 1\). Hence, \(\Delta_{n}(\phi) = 0\). Thus, \(\phi\) belongs to \(k[\mathbf{y}][S_{n-1}]\) by (1). Write \(\phi = \sum_{b,\mathbf{u}} r^{\prime}_{b,\mathbf{u}}y^{b}\hat{y}^{-\mathbf{u}}L^{\mathbf{u}}\), where the sum is taken over \(b \in (\mathbb{Z}_{\geq 0})^{n}\) and \(u = (u_{i,j})_{i,j}\) with \(u_{i,j} \in \mathbb{Z}_{\geq 0}\) for \(1 \leq i < j \leq n - 1\), \(r^{\prime}_{b,\mathbf{u}} \in k\) for each \(b\) and \(\mathbf{u}\), and

\(\hat{y}^{-u} = \prod_{1\leq i < j \leq n-1} (y_{i}y_{j})^{-u_{i,j}}, \quad L^{\mathbf{u}} = \prod_{1 \leq i<j\leq n-1} L_{i,j}^{u_{i,j}}\) for each \(\mathbf{u}\).

We may assume that \(r_{b,\mathbf{u}}^{\prime} = 0\) if \(\mathbf{y}^{b}\hat{\mathbf{y}}^{-\mathbf{u}}\) is not in \(k[\mathbf{y}]\). Let \(\eta(b, \mathbf{u})\) be the image of \((b - |\mathbf{u}|\mathbf{e}_{n}, |\mathbf{u}|\mathbf{e}_{n})\) in \(\Gamma\), where \(|\mathbf{u}| = \sum_{i,j} u_{i,j}\). Then, \(\mathbf{y}^{b}\hat{\mathbf{y}}^{-\mathbf{u}}L^{\mathbf{u}}\) belongs to \(k[\mathbf{y}^{\pm 1}][\mathbf{x}]_{\eta(b,\mathbf{u})}\) for each \(b\) and \(\mathbf{u}\) where \(k[\mathbf{y}^{\pm 1}][\mathbf{x}] := k[\mathbf{y}][\mathbf{x}][(y_{1} \cdots y_{n})^{-1}]\), since \((y_{i}y_{j})^{-1} L_{i,j}\) belongs to \(k[\mathbf{y}^{\pm 1}][\mathbf{x}]_{\delta}\) for each \(i, j\). Since \(\phi\) is in \(k[\mathbf{y}][\mathbf{x}]_{\mu}\), and \(\mu\) is the image of \((a, (l - m)\mathbf{e}_{n})\), we may assume that \(r_{b,\mathbf{u}}^{\prime} = 0\) unless \(|u| = l - m\) and \(b = a + (l - m)\mathbf{e}_{n}\). For each \(\mathbf{u}\) with \(r_{\mathbf{u}} := r^{\prime}_{a + (l-m) \mathbf{e}_{n},\mathbf{u}}\neq 0\), write \(\mathbf{y}^{a} y_{n}^{l-m} \hat{\mathbf{y}}^{-\mathbf{u}} = y_{1}^{\rho_{1}(\mathbf{u})} \cdots y_{n-1}^{\rho_{n-1}(\mathbf{u})} y_{n}^{s}\), where \(\rho_{i}(u) \in \mathbb{Z}_{\geq 0}\) for \(i = 1, \cdots , n - 1\), and \(s = a_{n} + l - m\). Then, we have \(\phi = y_{n}^{s}\sum_{\mathbf{u}} r_{\mathbf{u}} y_{1}^{\rho_{1}(\mathbf{u})} \cdots y_{n-1}^{\rho_{n-1}(\mathbf{u})} L^{\mathbf{u}}\). Since \(|u| = l - m\), it follows that

\[\begin{equation} \sum_{i=1}^{n-1} \rho_{i} (\mathbf{u}) = \sum_{i=1}^{n-1}a_{i} - 2(l-m) \end{equation}\]

Proof of Lemma 3

Take \(\varPhi\) which does not satisfy (3) so that \(m\) would be the minimum among the \(x_{n}\)-degrees of such polynomials. Then, \(t:= 2l - \sum_{i=1}^{n-1}a_{i} -m\) is positive, and \(\sum_{i=1}^{n-1} \rho_{i}(\mathbf{u}) = m - t\) for each \(\mathbf{u}\) by (2). Hence, the \(x_{n}\)-degree of

\[\varPhi_{1}:= \sum_{\mathbf{u}} \sum_{\mathbf{u}} r_{\mathbf{u}} L^{\mathbf{u}} \prod_{i=1}^{n-1} L_{n,i}^{\rho_{i}(\mathbf{u})} = \sum_{\mathbf{u}} r_{\mathbf{u}} L^{\mathbf{u}} \prod_{i=1}^{n-1} (y_{i}x_{n} - y_{n}x_{i})^{\rho_{i}(\mathbf{u})}\]

is \(m - t\). The coefficient of \(x_{n}^{n-t}\) in \(y_{n}^{s}\varPhi_{1}\) is equal to \(\phi\), so the coefficient of \(x_{n}^{m}\) in \(y_{n}^{s}\varPhi_{1}L_{n,1}^{t}\) is equal to that in \(y_{1}^{t}\varPhi\). Consequently, the \(x_{n}\)-degree \(m^{\prime}\) of \(\varPhi_{2}:= y_{1}^{t}\varPhi - y_{n}^{s}\varPhi_{1}L_{n,1}^{t}\) is less than \(m\). We claim that \(\varPhi_{2} = 0\). In fact, if \(\gamma^{\prime}\) is the image of \((a + t\mathbf{e}_{1}, l\mathbf{e}_{n})\) in \(\Gamma\), and \((a^{\prime}_{1}, \cdots , a_{n}^{\prime} ) := a + t\mathbf{e}_{1}\), then \(\varPhi_{2}\) belongs to \(k[\mathbf{y}][\mathbf{x}]_{\gamma^{\prime}} \cap \ker \Delta_{n}\), and

\[2l - \sum_{i=1}^{n-1} a^{\prime}_{i} = 2l - \sum_{i=1}^{n-1} a_{i} - t = m > m^{\prime}\]

This implies that \(\varPhi_{2} = 0\) by the minimality of \(m\). Hence, \(y_{1}^{t} \varPhi= y_{n}^{s}\varPhi_{1}L_{n,1}^{t}\). Thus, \(\varPhi_{1}\) is divisible by \(y_{1}\), since neither are \(y_{n}\) and \(L_{n,1}\). Recall that the kernel of a locally nilpotent derivation \(D\) of an integral domain \(R\) containing \(Q\) is factorially closed in \(R\), that is, \(D(f g) = 0\) implies \(D(f ) = D(g) = 0\) for each \(f, g \in R \ \{0\}\) (cf. [2, Proposition 1.3.32 (iii)]). Note that \(\Delta_{n}\) is locally nilpotent, \(\Delta_{n}(\varPhi_{1}) = 0\), \(\varPhi_{1} \neq 0\) and \(\Delta_{n}(x_{n}) \neq 0\). Hence, \(\varPhi_{1}\) is not divisible by \(x_{n}\). By substituting zero for \(x_{n}\), we obtain from \(\varPhi_{1}\) a nonzero polynomial

\[\sum_{\mathbf{u}}r_{\mathbf{u}} L^{\mathbf{u}} \prod_{i=1}^{n-1} (- y_{n} x_{i})^{\rho_{i}(\mathbf{u})} = ( - y_{n} )^{m-i} \varPsi, \text{ where } \varPsi = \sum_{\mathbf{u}} r_{\mathbf{u}} L^{\mathbf{u}} \prod_{i=1}^{n-1} x_{i}^{\rho_{i}(\mathbf{u})}.\]

Then, \(\varPhi \neq 0\), and \(\varPhi\) is divisible by \(y_{1}\), since so is \(\phi_{1}\). Define \(\sigma \in \text{Aut}_{k} k[\mathbf{y}][\mathbf{x}]\) by \(\sigma(x_{i}) = y_{i}\) and \(\sigma(y_{i}) = x_{i}\) for \(i = 1, \cdots, n\). Then, \(\sigma(\varPhi)\) is divisible by \(x_{1}\). On the other hand, \(\sigma(L_{i,j}) = L_{j,i}\) and \(\sigma(x_{i}) = y_{i}\) are in \(\ker \Delta_{n}\) for each \(i, j\), so \(\sigma(\varPsi)\) belongs to \(\ker \Delta_{n}\). Thus, we have \(\sigma(\varPsi) = 0\), because \(x_{1}\) is not in \(\ker \Delta_[n]\) and \(\ker \Delta_{n}\) is factorially closed in \(k[\mathbf{y}][\mathbf{x}]\). This contradicts that \(\varPsi \neq 0\). \(\square\)

Proof of Theorem 1

In fact, (3) implies that \(\sum_{i=1}^{n-1} \rho_{i}(\mathbf{u}) \geq m\) by (2), so we have \(\sum_{i=1}^{n-1}\rho_{i}^{\prime}(\mathbf{u}) = m\) for some integers \(0 \leq \rho^{\prime}_{i}(\mathbf{u}) \leq \rho_{i}(\mathbf{u})\) for \(i = 1, \cdots, n - 1\) for each \(u\). Then,

\[\varPhi^{\prime} := y_{n}^{s} \sum_{\mathbf{u}} r_{\mathbf{u}} L^{\mathbf{u}} \prod_{i=1}^{n-1} y_{i}^{\rho_{i}(\mathbf{u}) - \rho_{i}^{\prime}(\mathbf{u})} L_{n,i}^{\rho_{i}^{\prime}(\mathbf{u})} = y_{n}^{s} \sum_{\mathbf{u}} r_{\mathbf{u}} \prod_{i=1}^{n-1} y_{i}^{\rho_{i}(\mathbf{u}) - \rho_{i}^{\prime}(\mathbf{u}) } (y_{i}x_{n} - x_{n}y_{i})^{\rho_{i}^{\prime}(\mathbf{u}) }\]

is an element of \(k[\mathbf{y}][S_{n}]\) having \(x_{n}\)-degree \(m\), in which the coefficient of \(x_{n}^{m}\) is equal to \(\phi\). Hence, the \(x_{n}\)-degree of \(\varPhi - \varPhi^{\prime}\) is less than \(m\). Since \(\varPhi - \varPhi^{\prime}\) is an element of \(\ker \Delta_{n} \backslash k[\mathbf{y}][S_{n}]\), this contradicts the minimality of \(m\). \(\square\)