Proof. We write a proof for the case \(n = 2\). The proof for \(n > 2\) is similar. Write monomials \(x^m:=x_1^{m_1}x_2^{m_2}\) for \(m=(m_1,m_2)\in\mathbb{Z}^2\). The bracket on monomials is \[\{x^u,x^v\}= \lambda_{12}\,(u_1v_2-u_2v_1)\,x^{u+v}\qquad(u,v\in\mathbb{Z}^2).\]

Suppose \(S\subseteq A\) is a Poisson subalgebra of finite codimension and \(S\neq A\). Let \[M:=\{m\in\mathbb{Z}^2:\;x^m\notin S\}.\] Finite codimension implies \(M\) is finite. Let \[T:=\mathbb{Z}^2\setminus M=\{m\in\mathbb{Z}^2:\;x^m\in S\},\] so \(T\) is cofinite and closed under addition (because \(S\) is a subalgebra).

Choose an integer \(N>0\) such that \((N,0)\in T\) and \((0,N)\in T\) (this is possible since \(M\) is finite). Put \[u:=(N,0),\qquad v:=(0,N).\] Fix any \(m\in\mathbb{Z}^2\). Since \(M\) is finite, there are only finitely many integers \(n\) for which either \(nu\in M\) or \(m-nu\in M\). Hence, we can choose a non-zero integer \(n\) such that both \(nu\in T\) and \(m-nu\in T\). Because \(S\) is Poisson-closed, \(x^{nu},x^{m-nu}\in S\) implies \[\{x^{nu},x^{m-nu}\}=\lambda_{12}\big((nN)(m_2)-0\cdot(m_1-nN)\big)\,x^{m}=\lambda_{12} nN m_2\,x^m\in S.\] As \(\operatorname{char}\Bbbk=0\) and \(\lambda_{12}\neq0\), the scalar \(nN m_2\) is zero only if \(m_2=0\). Thus for every \(m\) with \(m_2\neq0\) we get \(x^m\in S\).

If \(m_2=0\), repeat the same argument exchanging the roles of \(u\) and \(v\): choose \(n\) with \(nv\in T\) and \(m-nv\in T\), then \[\{x^{nv},x^{m-nv}\}= \lambda_{12}\,nN m_1\,x^m\in S,\] and again, the coefficient is nonzero unless \(m_1=0\).

Finally, the remaining case \(m=(0,0)\) corresponds to the unit \(x^0=1\in S\) since \(S\) is a subalgebra. Therefore every monomial \(x^m\) lies in \(S\), so \(M=\varnothing\) and \(S=A\), contradicting \(S\neq A\). ◻

Proof. (1) This follows from the fact that every good negatively-\({\mathbb{N}}\)-filtration of \(B\) induces a good negatively-\({\mathbb{N}}\)-filtration on \(A\) by restriction, see [26].

(2) This is [26].

(3) When \(\Bbbk\) is algebraically closed, \(A\otimes B\) is a domain if and only if both \(A\) and \(B\) are domains [29]. Then \(u_{firm}\) agrees with \(u_{good}\). So the assertion follows from part (2) or [26].

(4) Clear. ◻

Proof. (1) Let \(f\colon A\longrightarrow A\otimes R\) be an injective Poisson algebra morphism and let \(\overline{A}\) be the image of \(f\). Then \(\overline{A}\) is a Poisson subalgebra of \(A\otimes R\). Then we have \[\begin{align} \overline{A}& =u_{good}(\overline{A}) \qquad\qquad\qquad \qquad {\text{as A\cong \overline{A} is u_{good}-maximal}}\\ &\subseteq u_{good}(A\otimes R) \qquad\qquad \qquad {\text{by Lemma \ref{xxlem2463}(1)}}\\ &\subseteq u_{good}(A)\otimes u_{firm}(R) \;\qquad\;\; {\text{by Lemma \ref{xxlem2463}(2)}}\\ &= u_{good}(A)\otimes \Bbbk \qquad \; \qquad \qquad {\text{as R is u_{firm}-minimal}}\\ &=A\otimes \Bbbk \qquad \qquad \; \qquad\qquad\quad {\text{as A is u_{good}-maximal.}} \end{align}\]

(2) Let \(\phi\) be in \(\mathop{\mathrm{Aut}}_{Poi.alg}(A\otimes R\mid C)\) and let \(f=\phi\mid_{A\otimes \Bbbk}\colon A\otimes \Bbbk \longrightarrow A\otimes R\) be the restriction of \(\phi\) to \(A\otimes \Bbbk\) that is identified with \(A\). By part (1), \({\rm{im}}(f)\subseteq A\otimes \Bbbk\). So we can consider \(f\) as a Poisson algebra endomorphism of \(A\). Since \(\phi\) is invertible, we can apply part (1) to the inverse of \(\phi\). Hence \(f\) is bijective. This implies that there is a map from \(\mathop{\mathrm{Aut}}_{Poi.alg}(A\otimes R\mid C)\to \mathop{\mathrm{Aut}}_{Poi.alg}(A\mid C)\) when \(C\) is identified with \(C\otimes \Bbbk\). Now it is routine to check that the short sequence is exact.

(3) By the proof of part (1), every injective Poisson algebra morphism \(f: A\to A\otimes R\), induces an injective map \(\pi\colon A\stackrel{f}{\cong} \overline{A}\hookrightarrow A\otimes \Bbbk\cong A\). Since \(A\) is Dixmier, \(\pi\) is a bijection, which implies that \({\rm{im}}(f)=\overline{A}=A\otimes \Bbbk\) as required. Therefore, \(A\) is \(R\)-Dixmier. ◻

Proof. The equivalence of (1) and (2) is due to Vancliff [30], see also [3]. Since \(\Bbbk[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]\) is \(\mathbb{Z}^{n}\)-graded Poisson algebra with \({\rm deg}\, (x_{i})=e_{i}=(0, \cdots, 1, \cdots, 0)\) for \(i=1, \cdots, n\), each Poisson-central element in \(L_{\Lambda}(\Bbbk)\) is a sum of scalar multiples of monomials in \(x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}\), each of which is Poisson-central. Without loss of generality, we may assume that each Poisson-central element of \(L_{\Lambda}(\Bbbk)\) is of the form: \(x=\prod_{i=1}^{n} x_{i}^{a_{i}}\) with \(a_{i}\in \mathbb{Z}\). Then we shall have \(\{x, x_{j}\}=0\) for \(j=1, \cdots, n\). As a result, we have \[\sum_{i=1}^{n} a_{i}\lambda_{ij} =0\] for \(j=1, \cdots, n\). Now the equivalence of (2) and (3) follows. ◻

Proof. \((1) \Rightarrow (2)\): Suppose that \(\mathcal{Q} \subseteq A\) is a Poisson subalgebra of the simple Poisson torus \(A=\Bbbk_{\Lambda}[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]\) such that \(A\stackrel{\phi}{\cong} \mathcal{Q}\) as Poisson algebras. Denote the composition of the embedding of \(\mathcal{Q}\) into \(A\) with the isomorphism \(\phi\) by \[\psi: A\stackrel{\phi}{\longrightarrow} \mathcal{Q} \hookrightarrow A.\] Obviously, \(\psi\) is an (injective) Poisson endomorphism of \(A\). By (1), we know that \(\psi\) is an automorphism, which implies that \(\mathcal{Q}=A\).

\((2) \Rightarrow (1)\): Let \(\phi \colon A \longrightarrow A\) be a Poisson endomorphism of \(A\). Since \(A\) is a simple Poisson torus, we know that \(\phi\) is injective. As a result, \(A\) is isomorphic to \(\mathcal{Q}\colon =\phi(A)\) as a Poisson algebra, and \(\mathcal{Q}\subseteq A\) is a Poisson subalgebra of \(A\). By (2), we know that \(\mathcal{Q}=A\), which implies that \(\phi\) is an automorphism.

\((2) \Rightarrow (3)\): Suppose \(\Lambda^{\prime}=(\lambda_{ij}^{\prime})\in M_{n}(\Bbbk)\) is a skew-symmetric matrix such that \(A=\Bbbk_{\Lambda^{\prime}}[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\) and there exist an invertible integer matrix \(G=(g_{ij}) \in {\rm GL}_{n}(\mathbb{Z})\) together with positive integers \(p_{1}, \cdots, p_{n}\) such that \[\lambda_{ij}^{\prime}=\sum_{1\leq k,t\leq n} p_{k}p_{t} g_{ki}g_{tj} \lambda_{kt}^{\prime}, \quad \forall 1\leq i, j\leq n.\] Note that we have \(\{z_{i}, z_{j}\}=\lambda_{ij}^{\prime} z_{i}z_{j}\) for \(1\leq i, j\leq n\). Let \(\mathcal{Q}\) be the Poisson subalgebra of \(A\) generated by \(y_{1}^{\pm 1}\colon =z_{1}^{\pm p_{1}}, \cdots, y_{n}^{\pm 1}\colon =z_{n}^{\pm p_{n}}\). Since \(z_{1}, \cdots, z_{n}\) are algebraically independent and \(p_{i}\geq 1\), we know that \(\mathcal{Q}\) is indeed a Poisson torus of rank \(n\) generated by \(y_{1}^{\pm 1}, \cdots, y_{n}^{\pm 1}\) subject to the following Poisson bracket: \[\{y_{i}, y_{j}\}=\{z_{i}^{p_{i}}, z_{j}^{p_{j}}\}=p_{i}p_{j} \lambda_{ij}^{\prime} y_{i}y_{j}\] for \(1\leq i, j\leq n\). Let us define an algebra morphism \[\phi \colon A=\Bbbk[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\longrightarrow \mathcal{Q}=\Bbbk[y_{1}^{\pm 1}, \cdots, y_{n}^{\pm 1}]\] by setting \[\phi(z_{i})=\prod_{k=1}^{n} y_{k}^{g_{ki}}\] for \(i=1, \cdots, n\). It is easy to check that \[\begin{align} \{\phi(z_{i}), \phi(z_{j})\} &=&\{\prod_{k=1}^{n} y_{k}^{g_{ki}}, \prod_{t=1}^{n} y_{t}^{g_{tj}}\}\\ &=& (\sum_{k, t=1}^{n} p_{k}p_{t} g_{ki} g_{tj} \lambda_{kt}^{\prime})(\prod_{k=1}^{n} y_{k}^{g_{ki}}) (\prod_{t=1}^{n} y_{t}^{g_{tj}})\\ &=& \lambda_{ij}^{\prime} (\prod_{k=1}^{n} y_{k}^{g_{ki}}) (\prod_{t=1}^{n} y_{t}^{g_{tj}})\\ &=& \lambda_{ij}^{\prime} \phi(z_{i}) \phi(z_{j})\\ &=& \phi(\lambda_{ij}^{\prime} z_{i} z_{j})\\ &=& \phi(\{z_{i}, z_{j}\}) \end{align}\] for any \(1\leq i, j\leq n\). Therefore, \(\phi\) is a Poisson algebra morphism. Since \(G=(g_{ij})\in {\rm GL}_{n}(\mathbb{Z})\) is an invertible integer matrix, \(\phi\) is indeed a Poisson isomorphism. By (2), we know that \[\Bbbk[z_{1}^{\pm p_{1}}, \cdots, z_{n}^{\pm p_{n}}]=\mathcal{Q}=A=\Bbbk[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}].\] As a result, we must have that \(p_{1}=\cdots =p_{n}=1\).

\((3) \Rightarrow (2)\): Suppose that \(\mathcal{Q}\subseteq A=L_{\Lambda}(\Bbbk)=\Bbbk_{\Lambda}[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]\) is a Poisson subalgebra of \(A\) and \(A\) is isomorphic to \(\mathcal{Q}\) as a Poisson algebra. Therefore, there is an isomorphism of Poisson algebras: \[\phi \colon A=\Bbbk[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}] \longrightarrow \mathcal{Q}\subseteq A.\] Since \(A=\Bbbk[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]\) is generated by \(x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}\) and \(\phi(x_{i})\) is a monomial in \(\Bbbk[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]\) for each \(i=1, \cdots, n\), we know that \(\mathcal{Q}\) is generated by the monomials \(y_{i}^{\pm 1}\colon =[\phi(x_{i})]^{\pm 1}=x^{\pm b_{i}}=(\prod_{k=1}^{n} x_{k}^{b_{ki}})^{\pm 1}\) for some \(b_{i}=(b_{1i}, \cdots, b_{ni}) \in \mathbb{Z}^{n}\) with \(i=1, \cdots, n\). As a result, the subset \(\mathbb{S}=\{b\in \mathbb{Z}^{n}\mid x^{b}\in \mathcal{Q}\}\) is actually a subgroup of \(\mathbb{Z}^{n}\). Since \(\phi\) is an isomorphism of algebras, we must have that \(\mathbb{S}\) is a free abelian group of the same rank as \(\mathbb{Z}^{n}\). Now we can choose a basis \(\{c_{1}, \cdots, c_{n}\}\) of \(\mathbb{Z}^{n}\) such that \(\{p_{1} c_{1}, \cdots, p_{n} c_{n}\}\) is a basis of \(\mathbb{S}\) (whose rank is equal to \(n\)) for some positive integers \(p_{1}\mid p_{2}\mid \cdots \mid p_{n}\). Setting \(z_{i}^{\pm 1}\colon =x^{\pm c_{i}}=(\prod_{k=1}^{n} x_{k}^{c_{ki}})^{\pm 1}\) for \(i=1, \cdots, n\), we know that \(A\) is the same as the Poisson torus \(L_{\Lambda^{\prime}}(\Bbbk)=\Bbbk_{\Lambda^{\prime}}[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\), which is generated by \(z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}\) subject to the following Poisson bracket \[\{z_{i}, z_{j}\}=\lambda^{\prime}_{ij}z_{i}z_{j}\] where \(\Lambda^{\prime}=(\lambda^{\prime}_{ij})\) with \(\lambda_{ij}^{\prime} :=\sum_{1\leq k,t\leq n} c_{ki}c_{tj} \lambda_{kt}\) for \(1\leq i, j\leq n\). As a result, \(\mathcal{Q}\) is indeed the Poisson torus \(\Bbbk[(z_{1}^{p_{1}})^{\pm 1}, \cdots, (z_{n}^{p_{n}})^{\pm 1}]\) generated by \(z_{1}^{\pm p_{1}}, \cdots, z_{n}^{\pm p_{n}}\) subject to the following Poisson bracket: \[\{z_{i}^{p_{i}}, z_{j}^{p_{j}}\}=p_{i}p_{j}\lambda_{ij}^{\prime}z_{i}^{p_{i}}z_{j}^{p_{j}},\quad \forall 1\leq i, j\leq n.\] Since the Poisson torus \(L_{\Lambda^{\prime}}(\Bbbk)=\Bbbk_{\Lambda^{\prime}}[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\) (the same as \(A\)) is also isomorphic to the Poisson torus \(\mathcal{Q}=\Bbbk[z_{1}^{\pm p_{1}}, \cdots, z_{n}^{\pm p_{n}}]\) as a Poisson algebra via \(\phi\colon A \longrightarrow \mathcal{Q}\), there exists an invertible integer matrix \(G=(g_{ij})\in {\rm GL}_{n}(\mathbb{Z})\) such that \(\phi(z_{i})=\alpha_{i} \prod_{k=1}^{n} z_{k}^{p_{k} g_{ki}}\) where \(\alpha_{i} \in \Bbbk^{\times}\) for \(i=1, \cdots, n\). Therefore, we have that \[\lambda_{ij}^{\prime}=\sum_{k, t=1}^{n} p_{k}p_{t} g_{ki} g_{tj} \lambda_{kt}^{\prime}\] for any \(1\leq i, j\leq n\). Since \(A=L_{\Lambda^{\prime}}(\Bbbk)=\Bbbk_{\Lambda^{\prime}}[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\) and \(G\) is in \({\rm GL}_{n}(\mathbb{Z})\), we must have that \(p_{1}=\cdots=p_{n}=1\) using (3). Therefore, we have that \(\mathbb{S}=\mathbb{Z}^{n}\), which implies that \(\mathcal{Q}=\Bbbk[x^{\pm c_{1}}, \cdots, x^{\pm c_{n}}]=\Bbbk[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]=A\). ◻

Proof of Theorem 8.. Note that \(\Lambda =\lambda M\) for some \(\lambda \in \Bbbk^{\times}\) and a skew-symmetric matrix \(M=(m_{ij})\in M_{n}(\mathbb{Z})\) with integer entries. Recall that \(L_{\Lambda}(\Bbbk)\) is the Poisson torus generated by \(x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}\) subject to the following Poisson bracket: \[\{x_{i}, x_{j}\}=\lambda_{ij} x_{i}x_{j}=\lambda m_{ij}x_{i}x_{j}\] for \(1\leq i\leq j\leq n\). Since the Poisson torus \(L_{\Lambda}(\Bbbk)\) is Poisson simple, by Lemma 2, we know that the matrix \(\Lambda\) is non-singular, which is equivalent to the fact that the matrix \(M=(m_{ij})\) is non-singular. That is, \({\rm det}(M)\neq 0\).

To show that \(L_{\Lambda}(\Bbbk)\) is Dixmier, we prove that Condition (3) of Lemma 3 holds. Suppose there exist a skew-symmetric matrix \(\Lambda^{\prime}=(\lambda_{ij}^{\prime})\in M_{n}(\Bbbk)\) such that \(L_{\Lambda}(\Bbbk)=\Bbbk_{\Lambda^{\prime}}[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\) and an invertible integer matrix \(G=(g_{ij})\in {\rm GL}_{n}(\mathbb{Z})\) together with positive integers \(p_{1}, \cdots, p_{n}\) such that

\[\label{F2462461} \lambda_{ij}^{\prime} =\sum_{k, t=1}^{n} p_{k} p_{t} g_{ki} g_{tj} \lambda_{kt}^{\prime}\tag{4}\]

for \(1\leq i, j\leq n\). We shall prove that \(p_{1}=\cdots =p_{n}=1\).

Since \(\Bbbk_{\Lambda}[x_{1}^{\pm 1}, \cdots, x_{n}^{\pm 1}]=\Bbbk_{\Lambda^{\prime}}[z_{1}^{\pm 1}, \cdots, z_{n}^{\pm 1}]\) as Poisons algebras, there exists an invertible integer matrix \(T=(t_{ij})\in {\rm GL}_{n}(\mathbb{Z})\) such that \[z_{i}=\prod_{k=1}^{n} x_{k}^{t_{ki}},\] for \(i=1, \cdots, n\) and so from \(\{z_i , z_j \} = \{ \prod_{k=1}^{n} x_{k}^{t_{ki}}, \prod_{l=1}^{n} x_{l}^{t_{lj}} \}\), we have that \(\Lambda^{\prime} =T^{t} \Lambda T\), where \(T^t\) denotes the matrix transpose of \(T\). As a result, we have \(\Lambda^{\prime}=T^{t} (\lambda M) T=\lambda (T^{t} M T)\). Setting \(M^{\prime}=(m_{ij}^{\prime})=T^{t} M T\), then we have \(\Lambda^{\prime}=\lambda M^{\prime}\) or equivalently \(\lambda_{ij}^{\prime}=\lambda m_{ij}^{\prime}\) for \(1\leq i, j\leq n\). Since \({\rm det}(M)\neq 0\), we have that \({\rm det}(M^{\prime})={\rm det}(T^{t} M T)=({\rm det}(T))^{2} {\rm det} (M)\neq 0\). This also implies that \(\det(\Lambda') \neq 0.\) Now formula (4 ) gives us \[\Lambda' = G^t D \Lambda' D G, \quad \text{where } D = \mathrm{Diag}[p_1, \dots, p_n].\] and consequently \(1 = {\det(G)}^2 (\prod_{i=1}^{n} p_{i})^{2}\). Because \(G\) has entries in \(\mathbb{Z}\), we conclude that \(\prod_{i=1}^{n} p_i = 1,\) and therefore \[p_1 = \cdots = p_n = 1.\] By Lemma 3, we know that every Poisson endomorphism of \(L_{\Lambda}(\Bbbk)\) is an automorphism. Thus, we have proved \(L_{\Lambda}(\Bbbk)\) is Dixmier. Since \(L_{\Lambda}(\Bbbk)\) is \(u_{\mathrm{good}}\)-maximal (see [26]) and \(R\) is \(u_{\mathrm{firm}}\)-minimal, it follows from Theorem 7(3) that \(L_{\Lambda}(\Bbbk)\) is \(R\)-Dixmier. ◻

Proof. (1,3,4) See the proof of Lemma 1.

(2) We modify the proof of [26] as below. Since \(A\otimes \Bbbk\) is a Poisson subalgebra of \(A\otimes B\), we have \(u_{good,w}(A)\otimes \Bbbk \subseteq u_{good,w}(A\otimes B)\) by part (1). Similarly, \(\Bbbk\otimes u_{good,w}(B)\subseteq u_{good,w}(A\otimes B)\). Therefore \(u_{good,w}(A)\otimes u_{good,w}(B)\subseteq u_{good,w}(A\otimes B)\).

For \(u_{good,w}(A\otimes B)\subseteq u_{good,w}(A) \otimes u_{firm,w}(B)\), let \({\mathbf{F}}\) be a good negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(A\) and \({\mathbf{G}}\) be a firm negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(B\). By [26], \({\mathbf{H}}:={\mathbf{F}}\otimes {\mathbf{G}}\) is a good negatively-\({\mathbb{N}}\)-filtration of \(A\otimes B\) with \(H_0=F_0(A)\otimes G_0(B)\). By the definition of the product of Poisson algebras, \[\{A_{i}\otimes B_j, A_{l}\otimes B_{k}\} \subseteq \{A_{i}, A_{l}\}\otimes B_j B_{k} +A_{i}A_{l}\otimes\{B_j,B_{k}\} \subseteq A_{i+l-w}\otimes B_{j+k} + A_{i+l}\otimes B_{j+k-w}\] which implies that \({\mathbf{H}}\) is a \(w\)-filtration. So \[\begin{align} u_{good,w}(A\otimes B)&=\bigcap_{{\mathbf{H}} \in \Phi_{good,w}(A\otimes B)} H_0\\ &\subseteq \bigcap_{{\mathbf{H}}={\mathbf{F}}\otimes {\mathbf{G}}, {\mathbf{F}}\in \Phi_{good,w}(A), {\mathbf{G}}\in \Phi_{firm,w}(B)} F_0(A)\otimes G_0(B)\\ &=[\bigcap_{{\mathbf{F}}\in \Phi_{good,w}(A)} F_0(A) ]\otimes [\bigcap_{{\mathbf{G}}\in \Phi_{firm,w}(B)}G_0(B) ] \\ &=u_{good,w}(A)\otimes u_{firm,w}(B). \end{align}\] ◻

Proof. Assertion (4) is [4], where we put \(\epsilon = 1\).
Assertions (1)–(3) basically follow from [4], but let us explain part (1).

Let \(K\) be the fraction field of \(B\). By [4], \[^{w}\Gamma(K):= \bigcap_{\nu} F^{\nu}_0(K)\] where \(\nu\) runs over all \(w\)-valuations on \(K\) and \(F^{\nu}_0(K) = \{ a \in K \vert v(a) \geq 0\}\). By [4], \(^{w}\Gamma(K)\supseteq B\). For every negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(B\), say \({\mathbf{F}}\), it defines a \(w\)-valuation of \(B\) by [4]. By [4], it extends to a \(w\)-valuation of \(K\). Then equation \(^{w}\Gamma(K)\supseteq B\) implies that \(F_0(B)=B\). So \({\mathbf{F}}\) is trivial. Consequently, \(u_{good, w}(B)=B\). ◻

Proof. Since \(\Bbbk\) is algebraically closed, we have \(u_{good,w}=u_{firm,w}\). Let \(A\) and \(A'\) be in \({\mathcal{P}}_{n}\) and let \(f\colon A\longrightarrow A'\otimes R\) be an injective Poisson algebra morphism where \(R\in {\mathcal{F}}_{n-1}\). We have: \[\begin{align} A&=u_{good, n-1}(A) \qquad\; \qquad \; \qquad\; \qquad {\text{by Lemma \ref{xxlem24614}(1)}}\\ &\xrightarrow{f} u_{good,n-1}(A'\otimes R) \qquad\;\qquad\;\qquad {\text{by Lemma \ref{xxlem24611}(1)}}\\ &= u_{good,n-1}(A')\otimes u_{good, n-1}(R) \qquad {\text{by Lemma \ref{xxlem24611}(3)}}\\ &=A'\otimes \Bbbk. \end{align}\] Thus, \(f\) can be regarded as an injective map from \(A\) to \(A' \otimes \Bbbk \cong A'\). By Lemma 5(4), the image of \(f\) is \(A'\otimes \Bbbk\). So the assertion follows. ◻

Proof. The proofs of all four parts are similar to the proof of [26], so we only prove part (1). Parts (3) and (4) also follow from [26].

(1) Let \(A\) be in \({\mathcal{P}}_n\) and \(R_1, R_2\) be in \({\mathcal{F}}_{n-1}\) of the same Gelfand-Kirillov dimension. Suppose \(A\otimes R_1 \cong B\otimes R_2\) for another Poisson algebra \(B\). Since \(A\otimes R_1\) is a domain, so is \(B\). Applying \(u_{good, n-1}\) to the isomorphism \(A\otimes R_1 \cong B\otimes R_2\) and using Lemma 4 and the fact \(u_{firm, n-1}(R_1)=u_{firm,n-1}(R_2)=\Bbbk\), we obtain that \[A\cong A\otimes \Bbbk \subseteq u_{good,n-1}(A\otimes R_1) \cong u_{good, n-1}(B\otimes R_2) \subseteq B\otimes \Bbbk\cong B.\] Now the proof of [26] applies. ◻

Proof. Suppose \(\rho\colon A\to A\otimes H\) is a Poisson Hopf coaction of \(H\) on \(A\). By the counit axiom, \(\rho\) is an injective map. By the proof of Theorem 10, the image of \(\rho\) is in \(A\otimes \Bbbk\) inside \(A\otimes H\). This means that the coaction \(\rho\) is trivial. ◻

Proof. Let \(I\) be a nonzero Poisson ideal of \(A\otimes B\). Let \(0\neq c\in I\) and write \(c\) as \(\sum_{i=1}^{n} a_i \otimes b_i\) where \(a_i\in A\) and \(b_i\in B\). Suppose \(n\) is minimal among all possible nonzero \(c\) in \(I\). We claim that \(n=1\). Suppose to the contrary that \(n>1\). Then we may assume that \(\{a_1,\cdots, a_n\}\) (resp. \(\{b_1,\cdots, b_n\}\)) are linearly independent.

Case 1: \(\{a_1,a_2\}\neq 0\). Then \(I\) contains the element \(\{c, a_1\otimes 1\}=\sum_{i=1}^{n} \{a_i,a_1\}\otimes b_i=\sum_{i=2}^{n} \{a_i,a_1\}\otimes b_i\), which is nonzero since \(\{b_1,\cdots, b_n\}\) are linearly independent, and \(\{a_2,a_1\}\neq 0\). This contradicts the minimality of \(n\).

It remains to consider the following case:

Case 2: \(\{a_1,a_2\}=0\). Note that, for every \(f \in A\), we have that \((f \otimes 1 ) c = \sum_{i=1}^{n} a_i f \otimes b_i \in I\), which is nonzero, for any nonzero \(f\). By Case 1, we may further assume that \(\{a_1 f, a_2 f\}=0\). So \[0=\{a_1f, a_2 f\} =\{a_1, f\} a_2 f+\{f, a_2\} a_1 f+\{a_1,a_2\}f^2=\{a_1, f\} a_2 f+\{f, a_2\} a_1 f.\] Hence \(\{a_1,f\}a_2=\{a_2,f\}a_1\). This implies that \[\{a_1 a_2^{-1},f\}=\{a_1,f\} a_2^{-1}-\{a_2, f\} a_2^{-2} a_1=(\{a_1, f\} a_2-\{a_2, f\} a_1) a_2^{-2}=0.\] Since \(f\) is arbitrary, \(a_1 a_2^{-1}\) is in the center of \(Q(A)\) which is \(\Bbbk\) by hypothesis. So \(a_{1}\) and \(a_{2}\) are linearly dependent, which is a contradiction. ◻

Proof of Lemma 7. Let \(I\) be a nonzero Poisson ideal of \(S\otimes A\). By Lemma 6, there is an element \(0\neq s\otimes a\in I\) where \(s\in S\) and \(a\in A\). Consider \(J_{a}:=\{s\in S\mid s\otimes a\in I \}\). Using the equality \(\{ s'\otimes 1 , s \otimes a\} = \{ s', s\} \otimes a\), for every \(s, s' \in S\), it is easy to see that \(J_{a}\) is a nonzero Poisson ideal of \(S\). Since \(S\) is Poisson simple, \(J_{a}=S\) and \(1\otimes a\in I\).

(1) By the above, we may assume that \(1\otimes a\in I\). Thus \(I\cap (\Bbbk\otimes A)\) is not zero. Since \(A\) is Poisson simple, so is \(\Bbbk \otimes A\) and consequently \(1\otimes 1 \in \Bbbk\otimes A = I\cap (\Bbbk \otimes A) \subseteq I\). So \(I=S\otimes A\).

(2) Let \(I\) be a nonzero graded Poisson ideal of \(S\otimes A\). By the first paragraph, \(1\otimes a\in I\) for some \(0\neq a\in A\). Since \(I\cap (\Bbbk\otimes A)\) is not zero and \(A\) is projectively simple, \(I\cap (\Bbbk \otimes A)\) contains \(1\otimes A_{\geq h}\) for some \(h>0\). So The assertion follows. ◻

Proof. Let \({\mathbf{G}}'\) be the unique nontrivial negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(P\) provided by (ii). Define a filtration \({\mathbf{H}}'\) of \(A\otimes P\) by \(H'_i(A\otimes P): =A\otimes G'_i(P)\) for all \(i\). Then it is a nontrivial negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(A\otimes P\). It remains to show the uniqueness.

Let \({\mathbf{H}}\) be any nontrivial negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(A\otimes P\). By restriction, it induces a negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(A\), denoted by \({\mathbf{F}}\) (resp. a negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(P\), denoted by \({\mathbf{G}})\). Since \(u_{good, w}(A)=A\), \({\mathbf{F}}\) is trivial, that is \(F_i(A)=A\) for all \(i\leq 0\) and \(F_i(A)=0\) for all \(i>0\). Since \(P\) has the unique nontrivial negatively-\({\mathbb{N}}\)-\(w\)-filtration, \({\mathbf{G}}\) is either trivial or equal to \({\mathbf{G}}'\). If \({\mathbf{G}}\) is trivial, we obtain that \(H_0(A\otimes P)=A\otimes P\) which implies that \({\mathbf{H}}\) is trivial, yielding a contradiction. So it forces that \({\mathbf{G}}={\mathbf{G}}'\). Since \(A\otimes \Bbbk \subseteq H_0(A\otimes P)\), localizing elements in \(A\setminus \{0\}\), the filtration \({\mathbf{H}}\) induces a negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(K\otimes P\) where \(K=Q(A)\), which is denoted still by \({\mathbf{H}}\). Similarly, \({\mathbf{H}}'\) induces a negatively-\({\mathbb{N}}\)-\(w\)-filtration of \(K\otimes P\), which is still denoted by \({\mathbf{H}}'\). For each \(i\leq 0\), \(H'_{i}(K\otimes P)=K\otimes G'_i(P) \subseteq H_i(K\otimes P)\). By [4], there is a natural graded Poisson algebra morphism \[\phi: \mathop{\mathrm{gr}}_{{\mathbf{H}}'} K\otimes P (\cong K\otimes \mathop{\mathrm{gr}}_{{\mathbf{G}}'} P) \to \mathop{\mathrm{gr}}_{\mathbf{H}} K\otimes P.\]

The kernel of \(\phi\) is a graded Poisson ideal, and so by Lemma 7(2), if \(\ker(\phi)\) is nonzero, then there exists an integer \(m \leq 0\) such that \(\ker(\phi)\) contains all elements \(1 \otimes p\) for every \(p + {\mathbf{G}}'_{i + 1}(P) \in {\mathbf{G}}'_{i }(P)/ {\mathbf{G}}'_{i + 1}(P)\) with \(i \leq m\). Since \(\phi\) maps \(1 \otimes p\) to \(1 \otimes p\) for all \(p + {\mathbf{G}}'_{i + 1}(P) \in {\mathbf{G}}'_{i }(P)/ {\mathbf{G}}'_{i + 1}(P)\) and \(i < 0\), this implies that \({\mathbf{G}}'_{i}(P) = {\mathbf{G}}'_{i+1}(P)\) for all \(i \leq m\), which is a contradiction, since \(\operatorname{gr}_{{\mathbf{G}}'} P\) is a projectively simple and therefore infinite dimensional. Thus \(\ker(\phi) = 0\) and \(\phi\) is injective. By [4], \({\mathbf{H}} = {\mathbf{H}}'\) on \(K \otimes P\). Restricted to \(A \otimes P\), we obtain \({\mathbf{H}} = {\mathbf{H}}'\) as required. ◻

Proof. Set \(w=\deg \Omega -3\). Since \({\rm deg} (\Omega)\) is at least \(4\), \(w\) is greater than or equal to \(1\). By Lemma 5(3), \(P_{\Omega-\xi}\) has a unique nontrivial negatively-\({\mathbb{N}}\)-\(w\)-filtration such that the associated graded ring is isomorphic to \(P_{\Omega}\). Since \(\Omega\) has an isolated singularity, \(P_{\Omega}\) is projectively simple, see [4] Lemma 1.6. So \(P:= P_{\Omega-\xi}\) satisfies the hypotheses (i-iv) of Lemma 8 As a result, \(K\otimes P_{\Omega-\xi}\) (as a Poisson \(\Bbbk\)-algebra) has a unique nontrivial negatively-\({\mathbb{N}}\)-\(w\)-filtration, say \({\mathbf{F}}\), such that the associated graded ring is isomorphic to \(K\otimes P_{\Omega}\). Now let \[\phi : K \otimes P_{\Omega-\xi} \longrightarrow K \otimes P_{\Omega-\xi}\] be an injective Poisson \(K\)-algebra endomorphism. Then it is easy to see that \[\phi(\mathbf{F}) := \{\, F_i + \phi(F_i(K\otimes P_{\Omega-\xi})) \,\}\] is a nontrivial negatively–\({\mathbb{N}}\)–\(w\)–filtration of \(K\otimes P_{\Omega-\xi}\). This implies that \[\mathbf{F} = \phi(\mathbf{F}).\] That is, for each \(i\), \[F_i + \phi(F_i(K\otimes P_{\Omega-\xi})) = F_i,\] and consequently \[\phi(F_i(K\otimes P_{\Omega-\xi})) \subseteq F_i.\] This shows that we obtain a \(w\)-graded Poisson \(\Bbbk\)-algebra morphism \[\mathop{\mathrm{gr}}\phi : \mathop{\mathrm{gr}}_{\mathbf{F}}(K\otimes P_{\Omega-\xi}) \longrightarrow \mathop{\mathrm{gr}}_{\mathbf{F}}(K\otimes P_{\Omega-\xi})\] sending \(x + F_{i+1} \in F_i/F_{i+1}\) to \(\phi(x) + F_{i+1}\).

Note that \(\mathop{\mathrm{gr}}\phi\) is injective. Indeed, if \[G_i := \{\, x \in F_i \mid \phi(x) \in F_{i+1} \,\},\] then the collection \(\{\, G_i + F_{i+1} \,\}\) is a negatively–\({\mathbb{N}}\)–\(w\)–filtration, which cannot be trivial since \(K \otimes P_{\Omega-\xi}\) is not finite-dimensional. By uniqueness of \(\mathbf{F}\), we must have \[G_i = F_{i+1},\] which shows that \(\mathop{\mathrm{gr}}\phi\) is injective.

Recall that each \(F_i\) is of the form \[F_i = K \otimes G'_i(P_{\Omega-\xi}),\] where \(G'\) is the unique nontrivial negatively–\({\mathbb{N}}\)–\(w\)–filtration on \(P_{\Omega-\xi}\). This implies that each \(F_i/F_{i+1}\) is a finite-dimensional \(K\)-module. Because \(\phi\) is a \(K\)-algebra morphism, for each \(i\), and for all \(k \in K\) and \(x + F_{i+1} \in F_i/F_{i+1}\), \[\mathop{\mathrm{gr}}\phi(k(x+F_{i+1})) = \mathop{\mathrm{gr}}\phi(kx+F_{i+1}) = \phi(kx)+F_{i+1} = (k\phi(x))+F_{i+1} = k (\,\mathop{\mathrm{gr}}\phi(x+F_{i+1})).\] Hence the restriction of \(\mathop{\mathrm{gr}}\phi\) to \(F_i/F_{i+1}\) is an injective \(K\)-linear map; since \(F_i/F_{i+1}\) is finite-dimensional over \(K\), this restriction is also surjective for each \(i\).

Suppose that \(\phi\) is not surjective. Then we may choose a maximal \(i \le 0\) such that there exists \[x \in F_i \setminus F_{i+1}\] which \(x\) is not in the image of \(\phi\). Since the restriction of \(\mathop{\mathrm{gr}}\phi\) to \(F_i/F_{i+1}\) is surjective, we have \[F_{i+1} + \phi(F_i) = F_i.\] Thus there exist \(y \in F_{i+1}\) and \(t \in \operatorname{im}(\phi)\) such that \[x = y + t.\] Now \(y\) must lie in \(\operatorname{im}(\phi)\); otherwise, we contradict the choice of \(i\). But then \(x = y + t\) is also in \(\operatorname{im}(\phi)\), which is a contradiction.

Therefore, \(\phi\) is surjective. ◻

Proof. We break our proof into three steps. In the first two steps, we consider the case when \(\xi=0\) and the last step deals with the case when \(\xi\neq 0\).

Step 1: Recall that \(P_{\Omega}\) has the standard grading \[P_{\Omega} = \bigoplus_{i \in \Bbb Z} P_i,\] as explained before Lemma 5. On the other hand, \[K \otimes P_{\Omega} = \bigoplus_{i \in \Bbb Z} K \otimes P_i\] is also a graded Poisson algebra. The goal of this step is to show that the cardinality of \(\mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(K \otimes P_{\Omega} \mid K),\) the subgroup of \(\mathop{\mathrm{Aut}}_{\mathrm{Poi.alg}}(K \otimes P_{\Omega} \mid K)\) consisting of grading-preserving automorphisms, is bounded above by \(42\, d(d-3)^2\).

First, we show that this cardinality is bounded by the cardinality of \(\mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(P'_{\Omega}),\) where \(P^{\prime}_{\Omega}:=\Bbbk^{\prime}[x,y,z]/(\Omega)\) and \(\Bbbk'\) denotes the Poisson center of \(K\). Note that \(\Bbbk \subseteq \Bbbk^{\prime} \subseteq K\), and that \(\Bbbk^{\prime}\) is a field with trivial Poisson bracket.

Now let \[\phi \in \mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(K \otimes P_{\Omega} \mid K).\] Since \(\{\lambda \otimes 1,\; \lambda' \otimes x\} = 0\), for any \(\lambda \in K\), \(\lambda' \in \Bbbk'\), and \(x \in P_{\Omega}\), and since \(\phi\) fixes \(K\) pointwise, it follows that \(\{\lambda \otimes 1,\; \phi(\lambda' \otimes x)\} = 0.\) Writing \[\phi(\lambda' \otimes x) = \sum_{i = 1}^{n} \mu_i \otimes x_i,\] where \(n\) is a finite integer, each \(\mu_i \in K\) and the elements \(x_i \in P_{\Omega}\) are linearly independent, we conclude that each \(\mu_i\) belongs to \(\Bbbk'\). Consequently, \(\phi(\lambda' \otimes x) \in \Bbbk' \otimes P_{\Omega}.\)

Therefore, restriction induces an injective group homomorphism \[\mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(K \otimes P_{\Omega} \mid K) \longrightarrow \mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(\Bbbk' \otimes P_{\Omega} \mid \Bbbk').\] Note that \[\Bbbk' \otimes P_{\Omega} \cong \Bbbk'[x,y,z]/(\Omega) = P'_{\Omega},\] and clearly \[\mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(\Bbbk' \otimes P_{\Omega} \mid \Bbbk') = \mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(P'_{\Omega}).\]

By the final part of the proof of [4], the order of the group \[\mathop{\mathrm{Aut}}_{\mathrm{Poi.alg}}(P'_{\Omega})\] is bounded above by \(42\, d(d-3)^2\). Consequently, the cardinality of \[\mathop{\mathrm{Aut}}_{\mathrm{gr.Poi.alg}}(K \otimes P_{\Omega} \mid K)\] is also bounded above by \(42\, d(d-3)^2\), which proves the claim in this step.
Step 2: We now consider \(\mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega}\mid K)\). Let \(\sigma\in \mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega}\mid K)\). Since \(K\otimes P_{\Omega}\) has the unique negatively-\({\mathbb{N}}\)-\(w\)-filtration \({\mathbf{F}}\), \(\sigma\) preserves \({\mathbf{F}}\). So \(\sigma\) induces a Poisson automorphism \(\sigma': \mathop{\mathrm{gr}}_{\mathbf{F}} K\otimes P_{\Omega} \to \mathop{\mathrm{gr}}_{\mathbf{F}} K\otimes P_{\Omega}\) which preserves the degree. Bearing in mind that \(\mathop{\mathrm{gr}}_{\mathbf{F}} K\otimes P_{\Omega} = K\otimes P_{\Omega}\), we have that \(\sigma'\) is in \(\mathop{\mathrm{Aut}}_{gr.Poi.alg}(K\otimes P_{\Omega}\mid K)\). Replacing \(\sigma\) by \(\sigma (\sigma')^{-1}\) we may assume that \(\sigma'\) is the identity. We claim that \(\sigma\) is the identity in this case. Since \(\sigma'\) is the identity, \(\sigma(x)=x+a\), \(\sigma(y)=y+b\) and \(\sigma(z)=z+c\) where \(a,b,c\in K\). (Note that in \(K \otimes P_{\Omega}\) we identify \(K\) with \(K \otimes \Bbbk\), and \(\Bbbk\otimes P_{\Omega}\) with \(P_{\Omega}\).) Observe that in \(K\otimes \Bbbk[x,y,z]\), \[\Omega(x+a, y+b,z+c) =\Omega(x,y,z)+a \Omega_{x} +b \Omega_{y} +c \Omega_z + ldt\] where \(ltd\) is a sum of terms of Adams degree less or equal to \(d-2\). Hence in \(K\otimes P_{\Omega}\), \[\begin{align} 0&=\sigma(0)=\sigma(\Omega) =\Omega (x+a, y+b,z+c)\\ &=\Omega(x,y,z)+a \Omega_{x} +b \Omega_{y} +c \Omega_z + ldt =a \Omega_{x} +b \Omega_{y} +c \Omega_z + ldt \end{align}\] which implies that \(a \Omega_{x} +b \Omega_{y} +c \Omega_z=0\) when restricted to degree \(d-1\). Since \(\Omega\) has i.s., \(\Omega_x, \Omega_y, \Omega_z\) are \(\Bbbk\)-linearly independent. Consider these as elements in \(K\otimes P_{\Omega}\), these are \(K\)-linearly independent, which implies that \(a=b=c=0\) as required. Therefore, every Poisson algebra automorphism in \(\mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega}\mid K)\) preserves the grading. Thus, the assertion for \(\xi = 0\) follows from Step 1.

Step 3: Let \(\xi\neq 0\). We will construct a group homomorphism from \(\mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega-\xi}\mid K)\to \mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega}\mid K)\). Let \(\sigma\in \mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega-\xi}\mid K)\). Since \(K\otimes P_{\Omega -\xi}\) has the unique negatively-\({\mathbb{N}}\)-\(w\)-filtration \({\mathbf{F}}\) [Lemma 8], \(\sigma\) preserves \({\mathbf{F}}\). So \(\sigma\) induces a Poisson automorphism \(\sigma': \mathop{\mathrm{gr}}_{\mathbf{F}} K\otimes P_{\Omega-\xi} \to \mathop{\mathrm{gr}}_{\mathbf{F}} K\otimes P_{\Omega-\xi}\) which preserves the degree. Bearing in mind that \(\mathop{\mathrm{gr}}_{\mathbf{F}} K\otimes P_{\Omega-\xi } = K\otimes P_{\Omega}\), we have that \(\sigma'\) is in \(\mathop{\mathrm{Aut}}_{gr.Poi.alg}(K\otimes P_{\Omega}\mid K)\).

Using the fact that \({\mathbf{F}}\) is unique, one can easily show that \(\sigma\to \sigma'\) is a group homomorphism. We claim that this is an injective map. Suppose \(\sigma'\) is the identity, by using the filtration \({\mathbf{F}}\), we have \(\sigma(x)=x+a\), \(\sigma(y)=y+b\) and \(\sigma(z)=z+c\) where \(a,b,c\in K\). Note that in \(K\otimes \Bbbk[x,y,z]\), \[\Omega(x+a, y+b,z+c)-\xi =\Omega(x,y,z)-\xi +a \Omega_{x} +b \Omega_{y} +c \Omega_z + ldt\] where \(ltd\) is a sum of terms of degree less than or equal to \(d-2\). Hence in \(K\otimes P_{\Omega}\), \[\begin{align} 0&=\sigma(0)=\sigma(\Omega-\xi) =\Omega (x+a, y+b,z+c)\\ &=\Omega(x,y,z)-\xi+a \Omega_{x} +b \Omega_{y} +c \Omega_z + ldt =a \Omega_{x} +b \Omega_{y} +c \Omega_z + ldt \end{align}\] which implies that \(a \Omega_{x} +b \Omega_{y} +c \Omega_z=0\) in the associated graded ring \(\mathop{\mathrm{gr}}_{\mathbf{F}} (K\otimes P_{\Omega-\xi})\). Since \(\Omega\) has i.s., \(\Omega_x, \Omega_y, \Omega_z\) are \(\Bbbk\)-linearly independent. Consider these as elements in \(K\otimes P_{\Omega}\), these are \(K\)-linearly independent, which implies that \(a=b=c=0\). So \(\sigma\) is the identity and the map \(\sigma\to \sigma'\) is injective. The assertion follows from Step 1. ◻

Proof of Theorem 1. (1) We use induction on \(d\). If \(d=1\), the assertion follows from [4]. Now we assume \(d\geq 2\). Without loss of generality, we may assume that \(\deg \Omega_1> \deg \Omega_2> \cdots >\deg \Omega_d=:m\). Let \(A\) be the Poisson domain \(\bigotimes_{i=1}^{d-1} P_{\Omega_i-\xi_i}\). Then \(P=A\otimes P_{\Omega_d-\xi_d}\). By Lemma 5(1), \(u_{good, m-3}(P_{\Omega_i-\xi_i})=P_{\Omega_i-\xi_i}\) for all \(i\leq d-1\). By Lemma 4 (3), \(u_{good, m-3}(A)=A\). By Lemma 5(3), \(u_{good,m - 3}(P_{\Omega_d-\xi_d})=\Bbbk\). Let \(\phi\) be an injective Poisson algebra endomorphism of \(P=A\otimes P_{\Omega_d-\xi_d}\). Applying \(u_{good, m -3}\) to \(\phi\) and using Lemma 5(2), we obtain an injective Poisson algebra morphism \[f\colon =u_{good, m -3}(\phi)\colon A\otimes \Bbbk \longrightarrow A\otimes \Bbbk.\] By the induction hypothesis, \(f\) is a bijective morphism. Replacing \(\phi\) by \((f^{-1}\otimes Id) \phi\), we may assume that \(f\) is the identity, namely, the restriction of \(\phi\) to \(A\) is the identity. Localizing elements in \(A\setminus \{0\}\), we obtain an injective Poisson algebra endomorphism \[\psi\colon K\otimes P_{\Omega_{d}-\xi_d} \longrightarrow K\otimes P_{\Omega_d-\xi_d}\] such that the restriction of \(\psi\) on \(A\otimes P_{\Omega_d-\xi_d}\) is \(\phi\). Since \(f\) is identity, \(\psi\) is \(K\)-morphism and now by Lemma 9, \(\psi\) is bijective. Since \(\psi\) is the identity on \(K\), by Lemma 10, there is an integer \(m\) such that \(\psi^{m}\) is the identity. This implies that \(\phi^{m}\) is the identity. Therefore, \(\phi\) is a bijection as required.

(2) This follows from the next corollary.

(3) This follows from induction, Lemma 7 and [36]. ◻

Proof. (1) Without loss of generality, we may assume \(C\) is a Poisson field. In particular, by [26], we may assume that \(C\) is \(u_{\mathrm{good},\,w}\)-maximal for all \(w \geq 0\). We use induction on \(d\). If \(d=0\), there is nothing to prove. If \(d=1\), the statement follows from Lemma 10. Now we assume \(d\geq 2\). Similar to the proof of Theorem 1 and assume that \(\deg \Omega_1> \deg \Omega_2> \cdots >\deg \Omega_d=\colon m\). Let \(A\) be the Poisson domain \(C \otimes \bigotimes_{i=1}^{d-1} P_{\Omega_i-\xi_i}\). Then \(C\otimes P = A\otimes P_{\Omega_d-\xi_d}\). So \(A\) is \(u_{good, m- 3}\)-maximal and \(P_{\Omega_d-\xi_d}\) is \(u_{good, m - 3}\)-minimal. By Theorem 9(2), there is a short exact sequence of groups \[1\longrightarrow \mathop{\mathrm{Aut}}_{Poi.alg}(A\otimes P_{\Omega_d-\xi_d}\mid A) \longrightarrow \mathop{\mathrm{Aut}}(A\otimes P_{\Omega_d-\xi_d}\mid C)\longrightarrow \mathop{\mathrm{Aut}}_{Poi.alg}(A\mid C)\to 1.\] Let \(K\) be the fractional field of \(A\), then \(\mathop{\mathrm{Aut}}_{Poi.alg}(A\otimes P_{\Omega_d-\xi_d}\mid A)\) is a subgroup of \(\mathop{\mathrm{Aut}}_{Poi.alg}(K\otimes P_{\Omega_d-\xi_d}\mid K)\) where latter is bounded by \(42 \deg \Omega_d (\deg \Omega_d-3)^2\) by Lemma 10. Now the assertion follows from the induction and the above exact sequence.

(2) Note that \(L_{\Lambda}(\Bbbk)\) is \(u_{good}\)-maximal (see [26]) and \(P\) is \(u_{firm}\)-minimal (follows from Theorem 10). The assertion now follows from Theorem 7(2), where we take \(C = \Bbbk\). ◻

Proof of Theorem 3. Suppose that \(A\) is the Poisson algebra \(L_{\Lambda}(\Bbbk) \otimes P\) where \(P=\bigotimes _{i=1}^{d} P_{\Omega_i-\xi_i}\). First, we prove that \(A\) is Dixmier. We use induction on \(d\). If \(d=0\), the assertion follows from Theorem 8, with this remark that \(\Bbbk\) is algebraically closed and every commutative domain is a firm domain. Now we assume \(d\geq 1\). Without loss of generality, we may assume that \(\deg \Omega_1> \deg \Omega_2> \cdots >\deg \Omega_d=\colon m\). Let \(A'\) be the Poisson domain \(L_{\Lambda}(\Bbbk) \otimes \prod_{i=1}^{d-1} P_{\Omega_i-\xi_i}\). Then \(A =A'\otimes P_{\Omega_d-\xi_d}\). By Lemma 5(1), \(u_{good, m - 3}(P_{\Omega_i- \xi_i})=P_{\Omega_i-\xi_i}\) for all \(i\leq d-1\). Therefore using Lemma 4 (2) and [26], \(u_{good, m- 3}(A')=A'\). We also use Lemma 5(3) to obtain \(u_{\mathrm{good},\,m-3}(P_{\Omega_d-\xi_d})=\Bbbk\).

Let \(\phi\) be an injective Poisson algebra endomrophism of \(A = A' \otimes P_{\Omega_d-\xi_d}\). Applying \(u_{good, m-3}\) to \(\phi\) and using Lemma 4(2), we obtain an injective Poisson algebra morphism \[f\colon =u_{good, m - 3}(\phi)\colon A'\otimes \Bbbk \longrightarrow A'\otimes \Bbbk.\] By the induction hypothesis, \(f\) is a bijective morphism. Replacing \(\phi\) by \((f^{-1}\otimes Id) \phi\), we may assume that \(f\) is the identity, namely, the restriction of \(\phi\) to \(A'\) is the identity. Inverting all elements in \(A'\setminus \{0\}\), we obtain an injective Poisson algebra endomorphism \[\psi: K\otimes P_{\Omega_{d}-\xi_d} \longrightarrow K\otimes P_{\Omega_d-\xi_d}\] such that the restriction of \(\psi\) on \(A'\otimes P_{\Omega_d-\xi_d}\) is \(\phi\). Since \(\psi\) is the identity on \(K\), by Lemma 9, \(\psi\) is bijective and consequently by Lemma 10, there is an integer \(m\) such that \(\psi^{m}\) is the identity. This implies that \(\phi^{m}\) is the identity. Therefore, \(\phi\) is a bijection as required.

Now, to conclude Theorem 3, it is enough to show that \(A\) and \(P\) are \({\mathcal{F}}_{w}\)-Dixmier whenever \[w \leq \min\{\deg \Omega_i\} - 1.\] Indeed, as defined earlier, \({\mathcal{F}}_{w}\) is the class of Poisson firm domains \(S\) such that \[u_{\mathrm{firm},\,w}(S) = \Bbbk.\] Since \(\Bbbk\) is algebraically closed, every commutative domain is a firm domain. Therefore, by definition, we have \({\mathcal{G}}_{w} \subseteq {\mathcal{F}}_{w}.\) So let us see that \(A\) is \({\mathcal{F}}_{w}\)-Dixmier if \(w\leq \min\{\deg \Omega_i\}-1 = m - 4\). By the above proof, one sees that \(A\) is \(u_{good, w}\)-maximal. Then every \(R\) in \({\mathcal{F}}_{w}\) is \(u_{good, w}\)-minimal. The assertion follows from Theorem 9(3). So this takes care of the case when \(A=L_{\Lambda}(\Bbbk) \otimes \bigotimes_{i=1}^{d} P_{\Omega_i-\xi_i}\).

When \(P= \bigotimes_{i=1}^{d} P_{\Omega_i-\xi_i}\), by Theorem 1, \(P\) is Dixmier. Note that \(P\) is \(u_{good, w}\)-maximal and that \(R\in {\mathcal{F}}_{w}\) is \(u_{good, w}\)-minimal. By Theorem 9(3), \(P\) is \(R\)-Dixmier. ◻

Proof. Let \(x\in Q(A)\setminus A\) that is integral over \(A\). Then \(x\) is a root of a monic polynomial \(f(t)\) with coefficients in \(A\). So, for each \(i\), \(x\in Q(A_i)\) is a root of the same monic polynomial \(f(t)\) whose coefficients are in \(A\subseteq A_i\). Since each \(A_i\) is integrally closed, \(x\in A_i\) for all \(i\). So \(x\in A\) as required. ◻

Proof. Take an injective Poisson algebra map \(f \colon A \to K\), where \(K\) is a Poisson field. By definition, \(\mathop{\mathrm{c}}_{f}(A)\) is the intersection of a family of integrally closed Poisson domains. Since the intersection of Poisson subalgebras is again a Poisson algebra, it follows from Lemma 11 that \(\mathop{\mathrm{c}}_{f}(A)\) is an integrally closed Poisson domain. Therefore, using Part (1) of Definition 8, we have \(f^{-1}(\mathop{\mathrm{c}}_{f}(A)) \in \Phi(Q(A))\); moreover, \(f^{-1}(\mathop{\mathrm{c}}_{f}(A))\) contains \(A\).

Similarly, if \(g \colon A \to K'\) is another injective Poisson algebra homomorphism, where \(K'\) is a Poisson field, then \(g^{-1}(\mathop{\mathrm{c}}_{g}(A)) \in \Phi(Q(A))\), and \(g^{-1}(\mathop{\mathrm{c}}_{g}(A))\) contains \(A\). We have \[f(A) \subseteq f\bigl(g^{-1}(\mathop{\mathrm{c}}_{g}(A))\bigr) \subseteq f(Q(A)) \subseteq K.\] Since \(g^{-1}(\mathop{\mathrm{c}}_{g}(A))\) is a Poisson algebra, so is \(f\bigl(g^{-1}(\mathop{\mathrm{c}}_{g}(A))\bigr)\). Consequently, by the definition of \(\mathop{\mathrm{c}}_{f}(A)\), we obtain \[\mathop{\mathrm{c}}_{f}(A) \subseteq f\bigl(g^{-1}(\mathop{\mathrm{c}}_{g}(A))\bigr).\] This implies that \[f^{-1}(\mathop{\mathrm{c}}_{f}(A)) \subseteq g^{-1}(\mathop{\mathrm{c}}_{g}(A)).\] Finally, using the definition of \(\mathop{\mathrm{c}}(A)\) in Part (2) of Definition 8 and the fact that \(g \colon A \to K'\) was an arbitrary injective Poisson algebra homomorphism into a Poisson field, we conclude that \[\mathop{\mathrm{c}}(A) = f^{-1}(\mathop{\mathrm{c}}_{f}(A)).\] ◻

Proof. By Lemma 12, \(\mathop{\mathrm{c}}(A)=\mathop{\mathrm{c}}_{f}(A)\) where \(f\) is the inclusion \(A\to Q(A)\). Since \(B\) is integrally closed, \(B\in \Phi(Q(A))\). Consequently, \(\mathop{\mathrm{c}}(A)\subseteq B\). Now let \(C\) be any element in \(\Phi(Q(A))\) containing \(A\). Since \(B\) is integral over \(A\), \(B\subseteq C\). So \(B\) is the minimal element in \(\Phi( Q(A))\) containing \(A\). Therefore \(\mathop{\mathrm{c}}(A)=B\). ◻

Proof. (1) Let \(\phi\colon A\longrightarrow A'\) be a Poisson algebra isomorphism. A Poisson algebra map \(f\colon A\longrightarrow K\), where \(K\) is a Poisson field, is injective if and only if \(f'\colon =f\circ \phi^{-1}\colon A'\longrightarrow K\) is injective. By definition, \(\mathop{\mathrm{c}}_{f}(A)=\mathop{\mathrm{c}}_{f'}(A')\). Hence, \(\phi\) induces an isomorphism from \(f^{-1}(\mathop{\mathrm{c}}_{f}(A)) \to (f')^{-1}(\mathop{\mathrm{c}}_{f'}(A))\). By Lemma 12, \(\phi\) induces an isomorphism from \(\mathop{\mathrm{c}}(A)\to \mathop{\mathrm{c}}(A')\).

(2) The assertion follows from the following diagram \[\begin{CD} 0@>>> A @>>> \mathop{\mathrm{c}}(A) @>>> \mathop{\mathrm{CompM}}(A) @>>> 0\\ @. @V \stackrel{\phi} {\cong} VV @VV \stackrel{\phi} {\cong} V @VVV @.\\ 0@>>> A' @>>> \mathop{\mathrm{c}}(A') @>>> \mathop{\mathrm{CompM}}(A') @>>> 0.\\ \end{CD}\]

(3) The assertions follow from part (1). ◻

Proof. We have an inclusion map \(f\colon A\hookrightarrow B\hookrightarrow Q(B)=: K\). Let \(f'\) be the inclusion of \(B\) into \(K\). Then both \(\mathop{\mathrm{c}}_{f}(A)\) and \(\mathop{\mathrm{c}}_{f'}(B)\) are Poisson subalgebras of \(K\) by the definition. Also by the definition, \(\mathop{\mathrm{c}}_{f}(A)\) is a Poisson subalgebra of \(\mathop{\mathrm{c}}_{f'}(B)\) both of which are considered as Poisson subalgebras of \(K\). By Lemma 12, \(\mathop{\mathrm{c}}(A)=f^{-1}(\mathop{\mathrm{c}}_{f}(A))=\mathop{\mathrm{c}}_{f}(A)\), since \(f\) is the inclusion. Similarly, \(\mathop{\mathrm{c}}(B)= (f')^{-1}(\mathop{\mathrm{c}}_{f'}(B))=\mathop{\mathrm{c}}_{f'}(B)\). Therefore, we obtain the left commuting square of diagram 7 . By the definition, the top and bottom rows are short exact sequences. Therefore, we have an induced map \(\mathop{\mathrm{CompM}}(A)\longrightarrow \mathop{\mathrm{CompM}}(B)\) that makes the right square commutative. ◻

Proof. (1) Let \(\sigma: A\longrightarrow A\) be an injective Poisson endomorphism of \(A\). By Lemma 15(1), we have a commutative diagram \[\notag\begin{CD} 0@>>> A @>>> B=\mathop{\mathrm{c}}(A) @>>> \mathop{\mathrm{CompM}}(A) @>>> 0\\ @. @V\sigma VV @V\sigma' VV @VVg V @.\\ 0@>>> A @>>> B=\mathop{\mathrm{c}}(A) @>f >> \mathop{\mathrm{CompM}}(A) @>>> 0\\ \end{CD}\] where \(\sigma'\) denotes the induced Poisson endomorphism of \(B\). Since \(\sigma'\) and \(f\) are surjective, \(g\) is surjective. By hypothesis, \(\mathop{\mathrm{CompM}}(A)\) is finite-dimensional. So \(g\) is bijective. By the \(5\)-lemma, \(\sigma\) is an automorphism as required. Therefore, Theorem 5 holds.

(2) Let \(f\colon A\longrightarrow A\otimes R\) be an injective Poisson algebra morphism. By Lemma 15(1), there is an induced Poisson algebra morphism \(f'\colon \mathop{\mathrm{c}}(A)\longrightarrow \mathop{\mathrm{c}}(A\otimes R)\). Since both \(B\) and \(B\otimes R\) are integrally closed with \(A\subseteq B\subseteq Q(A)\) and \(A\otimes R\subseteq B\otimes R\subseteq Q(A\otimes R)\), we have that \(B=\mathop{\mathrm{c}}(A)\) and \(B\otimes R=\mathop{\mathrm{c}}(A\otimes R)\) by Lemma 13. Thus \(f'\) is the induced injective Poisson algebra morphism \(f'\colon \mathop{\mathrm{c}}(A)=B \longrightarrow \mathop{\mathrm{c}}(A\otimes R)=B\otimes R\). Since \(B\) is \(R\)-Dixmier, we have \({\rm{im}}(f')= B\otimes \Bbbk\). This implies that \({\rm{im}}(f)\subseteq (A\otimes R) \cap (B\otimes \Bbbk)= A\otimes \Bbbk\). By part (1), \(A\) is Dixmier. Hence \({\rm{im}}(f)=A\otimes \Bbbk\).

(3) This is a special case of part (2), indeed \(B \otimes \Bbbk[x_1, \cdots, x_n] \cong B [x_1, \cdots, x_n[\) is an integrally closed domain.

(4) Let \(\pi: R\to \Bbbk\) be an augmentation of \(R\). Suppose \(A\xrightarrow{\phi} A\otimes R\xrightarrow{Id_A\otimes \pi} A\) is the identity. Using Lemma 15 and the fact that \(\mathop{\mathrm{c}}(A) = B\) and \(\mathop{\mathrm{c}}(A\otimes R) = B \otimes R\), the map \(\phi\) induces a Poisson algebra morphism \(\sigma: B\to B\otimes R\). So we have a commutative diagram \[\begin{CD} A@>\phi >> A\otimes R @>>> A\\ @VVV @VVV @VVV @.\\ B@>\sigma>> B\otimes R @>>> B. \end{CD}\] Since the composition of the top row is the identity, the composition of the bottom row is also the identity. Since \(B\) is augmented \(R\)-Dixmier, \({\rm{im}}(\sigma)=B\otimes \Bbbk\). Restricted to \(A\), we have \({\rm{im}}(\phi)\subseteq A\otimes \Bbbk\). Since \(A\) is Dixmier by part (1), we obtain \({\rm{im}}(\phi)= A\otimes \Bbbk\) as required. ◻

Proof. (1) Let \(\phi\colon A\longrightarrow A\) be an injective Poisson \(\Bbbk\)-algebra endomorphism and let \(\sigma=\phi\otimes K\). Then \(\sigma\) is an injective Poisson \(K\)-algebra endomorphism of \(B\). Since \(B\) is Dixmier as a Poisson \(K\)-algebra, \(\sigma\) is surjective. This forces \(\phi\) to be surjective.

(2) Let \(\phi\colon A\longrightarrow A\otimes R\) be an injective Poisson \(\Bbbk\)-algebra morphism and let \(\sigma=\phi\otimes K\). We consider \(\sigma\) as a Poisson \(K\)-algebra morphism from \(B\) to \((A\otimes R)\otimes K=(A\otimes K) \otimes_{K} (R\otimes K)=B\otimes_{K} S\), where \(S=R\otimes K\). Since \(B\) is \(S\)-Dixmier as a Poisson \(K\)-algebra, we have that \({\rm{im}}(\sigma)=B=A\otimes K\). This forces that \({\rm{im}}(\phi)=A\otimes \Bbbk\) inside \(A\otimes R\). Thus \(A\) is \(R\)-Dixmier as a Poisson \(\Bbbk\)-algebra.

(3) By part (1), \(A\) is Dixmier as a Poisson \(\Bbbk\)-algebra. Let \(\phi \colon A \longrightarrow A \otimes K\) be an injective Poisson \(\Bbbk\)-algebra morphism. Define \(\sigma\) to be the composition \[B := A \otimes K \xrightarrow{\phi \otimes \mathrm{Id}_K} (A \otimes K) \otimes K \cong A \otimes (K \otimes K) \xrightarrow{\mathrm{Id}_A \otimes m_K} A \otimes K =: B.\] Let us show that \(\sigma\) is injective. Since the canonical associativity isomorphism \[\iota\colon (A \otimes K)\otimes K \xrightarrow{\;\cong\;} A \otimes (K \otimes K)\] is bijective, so the composition \(\iota \circ (\phi \otimes \mathrm{Id}_K)\) is injective.

Next, consider the multiplication map \(m_K \colon K \otimes K \to K\). It admits a \(\Bbbk\)-linear section \[s\colon K \longrightarrow K \otimes_{\Bbbk} K,\qquad s(\lambda)=\lambda \otimes 1,\] since \(m_K(\lambda \otimes 1)=\lambda\). Tensoring with \(A\) yields a \(\Bbbk\)-linear map \[\mathrm{Id}_A \otimes s \colon A \otimes K \longrightarrow A \otimes (K \otimes K),\] which, for every \(m\) in the image of \(\iota \circ (\phi \otimes \mathrm{Id}_K)\), satisfies \[(\mathrm{Id}_A \otimes s)\circ (\mathrm{Id}_A \otimes m_K)(m)=m.\] Thus \(\sigma\) is injective. Since \(B\) is Dixmier as a Poisson \(K\)-algebra, \(\sigma\) is an automorphism. By the hypothesis, \(\mathop{\mathrm{Aut}}_{\Bbbk}(A)=\mathop{\mathrm{Aut}}_{K}(B)\), there is an \(\pi\in \mathop{\mathrm{Aut}}_{\Bbbk}(A)\) such that \(\sigma=\pi\otimes K\). For every \(a\in A\), \[\pi(a)\otimes 1=\sigma(a\otimes 1)= (Id_A\otimes m_K)\circ (\phi\otimes Id_K)(a\otimes 1)= (Id_A\otimes m_K)(\phi(a)\otimes 1)=\phi(a)\] which implies that \({\rm{im}}(\phi)\subseteq A\otimes \Bbbk\). Since \(A\) is Dixmier as a Poisson \(\Bbbk\)-algebra, \({\rm{im}}(\phi)= A\otimes \Bbbk\) as required. ◻

Proof. Let \(\sigma\) be an automorphism of \(A(d,\zeta)\) that is not the identity. By Lemma 14(3), \(\sigma\) extends to an automorphism of \(A\). Since \(\sigma\) is graded and \(\sigma(\zeta)\not\in \Bbbk \zeta\), it cannot be an automorphism of \(A(d,\zeta)\), contradiction. The assertion follows. ◻