Absolute Constants of Koras-Russell-like threefolds

Let $\eta\in F[X]$ be such that $\eta(X)=\sum_{i=0}^{n}c_{i}X^{i}$ for $c_{i}\in F$. Consider a ring homomorphism $\phi_{1}$ as in [1], then $\phi_{1}(x^{2}y+z^{2}+t^{3}+\eta(x))=\phi_{1}\left(x^{2}y+z^{2}+t^{3}+\sum_{i=0}^{n}c_{i}x^{i}\right)=\phi_{1}\left(\sum_{i=0}^{n}c_{i}x^{i}\right)+\phi_{1}^{2}(x)\phi_{1}(y)+\phi_{1}^{2}(z)+\phi_{1}^{3}(t)=\sum_{i=0}^{n}c_{i}\phi_{1}^{i}(x)+x^{2}(y+2zU-x^{2}U^{2})+(z-x^{2}U)^{2}+t^{3}=\sum_{i=0}^{n}c_{i}x^{i}+x^{2}y+t^{3}+2x^{2}zU-x^{4}U^{2}+z^{2}-2x^{2}zU+x^{4}U^{2}=\eta(x)+x^{2}y+z^{2}+t^{3}\cong 0$; hence, $\phi_{1}$ is a well-defined ring homomorphism, so it has an associated locally finite higher derivation $D$. Just as in [2], $D$ is iterative over $R_{k,\eta}$; and hence, $\phi_{1}$ is an exponential function of $R_{F,\eta}/F$. Only $x$ and $t$ are fixed by $\phi_{1}$ and $R_{F\eta}$ is a domain, so the ring of $\phi_{1}$-invariants is $F[x,t]$. Then $\operatorname{AK}(R_{F,\eta}/F)\subseteq F[x,t]$.
Again consider, $\phi_{2}\colon R_{F}\rightarrow R_{F}[U]$ as in [1], then $\phi_{2}(\eta(x)+x^{2}y+z^{2}+t^{3})=\phi_{2}(\eta(x))+\phi_{2}^{2}(x)\phi_{2}(y)+\phi_{2}^{2}(z)+\phi_{2}^{3}(t)+\phi_{2}\left(\sum_{i=0}^{n}c_{i}x^{i}\right)=\sum_{i=0}^{n}c_{i}\phi_{2}^{i}(x)+x^{2}(y+3t^{2}U-3x^{2}tU^{2}+x^{4}U^{3})+z^{2}+(t-x^{2}U)^{3}=\sum_{i=0}^{n}c_{i}x^{i}+x^{2}y+z^{2}+3x^{2}t^{2}U-3x^{4}tU^{2}+x^{6}U^{2}+t^{3}-3x^{2}t^{2}U+3x^{4}tU^{2}-x^{6}U^{3}=\eta(x)+x^{2}y+z^{2}+t^{3}\cong 0$; hence, $\phi_{2}$ is a well-defined ring homomorphism, so it has an associated locally finite higher derivation $D$. $D$ is iterative over $R_{k,\eta}$; and hence, $\phi_{2}$ is an exponential function. The ring of $\phi_{2}$-invariants is $F[x,z]$. Therefore, $\operatorname{AK}(R_{F,\eta}/F)\subseteq F[x,z]\cap F[x,t]=F[x]$. ∎

1. 1.

   Under $\omega_{1}$, the leading form of $f_{\eta_{c}}$ is $f_{c}$ a grade $0$ homogeneous polynomial over the grading $\omega_{1}$. Since $f_{c}$ is a prime element of $B$ for all $c\in k$, $\mathcal{F}_{1}$ is a degree filtration on $R_{\eta_{c}}$. Identifying the $\mathcal{F}_{1}$-leading forms of $x_{\eta_{c}},y_{\eta_{c}},z_{\eta_{c}},t_{\eta_{c}}\in R_{\eta_{c}}$ with $\Xi(x_{c}),\Xi(y_{c}),\Xi(z_{c}),\Xi(t_{c})\in S_{c}$ respectively, yields a natural $k$-isomorphism of the associated graded ring of $\mathcal{F}_{1}$ onto the integral domain $S_{c}$.

2. 2.

   Under $\omega_{2}$, the leading form of $f_{c}$ is $f_{0}$ a grade $6$ homogeneous polynomial over the grading $\omega_{2}$. Since $f_{0}$ is a prime element of $B$ for all $c\in k$, $\mathcal{F}_{2}$ is a degree filtration on $S_{c}$. Identifying the $\mathcal{F}_{2}$-leading forms of $\Xi(x_{c}),\Xi(y_{c}),\Xi(z_{c}),\Xi(t_{c})\in S_{c}$ with $\Xi(x_{0}),\Xi(y_{0}),\Xi(z_{0}),\Xi(t_{0})\in S_{0}$ respectively, yields a natural $k$-isomorphism of the associated graded ring of $\mathcal{F}_{2}$ onto the integral domain $S_{0}$.

3. 3.

   Finally, since $f_{0}$ is homogeneous under $\omega_{1}$, and $\omega_{2}$; it induces a $\mathbb{Z}$-grading on $S_{0}$ and clearly, induces an $\mathbb{N}$-grading on the subring $k[\Xi(z_{0}),\Xi(t_{0})]$.

∎

Clearly $A=k[x,y,z,t]$. Consider a product $\alpha y^{i}\in A$ where $1\leq i\in\mathbb{N}$ and $\alpha\in k[x,z,t]$. Then $\alpha=\beta x^{2}+ux+v$ for some $\beta\in k[x,z,t]$, $u,v\in k[z,t]$. Since $x^{2}y=-g\in k[x,z,t]$, so $\alpha y^{i}=(ux+v)y^{i}+\gamma y^{i-1}$ where $\gamma:=-g\beta\in k[x,z,t]$. Now the assertion readily follows by linearity. ∎

Fix $c\in k$ and $\eta_{c}\in k[X]$. Let $S:=S_{c},x:=x_{c},y:=y_{c},z:=z_{c},t:=t_{c}$ By (i) of 0.1. $\mathcal{F}_{1}$ is a degree filtration on $R$. Let $\delta$ denote the corresponding degree function on $R$. Suppose, if possible, $R^{\phi}\setminus{k[x,z,t]}\neq\emptyset$. Let $p\in R^{\phi}\setminus{k[x,z,t]}$. Applying 0.2, there are $\varepsilon(p)\in\mathbb{N}$, $u_{j},v_{j},\in k[z,t]$ for $1\leq j\leq\varepsilon(p)$ and $h\in k[x,z,t]$ such that $p=h+\sum_{j=1}^{\varepsilon(p)}(u_{j}x+v_{j})y^{j}$ where $(u_{\varepsilon(p)},v_{\varepsilon(p)})\neq(0,0)$. Our choice of $p$ forces $\varepsilon(p)\geq 1$. Let $m:=\varepsilon(p)$, $u:=u_{m}$ and $v:=v_{m}$. Note that $\delta((ux+v)y^{m})=2m$ if $v\neq 0$ and $\delta((ux+v)y^{m})=2m-1$ otherwise. Since $\delta((u_{j}x+v_{j})y^{j})\leq 2j\leq 2m-2$ for $0\leq j<m$, then $\delta((ux+v)y^{m})=\delta(p)\geq 1$. So, the $\mathcal{F}_{1}$-leading form $\Xi(p)\in S$ of $p$ is either $\Xi(u)\bar{x}\bar{y}^{m}$ or $\Xi(v)\bar{y}^{m}$ where $\Xi(u),\Xi(v)$ denote the leading forms of $u,v$ respectively. $\Xi(u),\Xi(v)\in k[\bar{z},\bar{t}]$. It is claimed there is $p\in R^{\phi}$ such that $\Xi(p)$ is either $\Xi(u)\bar{x}\bar{y}^{m}$ or $\Xi(v)\bar{y}^{m}$ with $m\in\mathbb{N}$ and $\Xi(v)\in k[\bar{z},\bar{t}]\setminus{k}$. Supposing otherwise, i.e. supposing $\Xi(p)=\Xi(v)\bar{y}^{m}$ with $0\neq\Xi(v)\in k$ for all $0\neq p\in R^{\phi}$, derives to a contradiction.

First suppose there is $v\in R^{\phi}\cap k[x,z,t]\setminus{k}$. Say $v=v_{0}+\sum_{i\geq 1}v_{i}x^{i}$ where $v_{i}\in k[z,t]$ for $i\geq 0$. If $v_{0}\in k[z,t]\setminus{k}$, then clearly $\Xi(v)=\Xi(v_{0})\in k[\bar{z},\bar{t}]\setminus{k}$ contrary to our supposition. If $v_{0}\in k$, then $0\neq v-v_{0}\in R^{\phi}$ and since $R^{\phi}$ is factorially closed, $x\in R^{\phi}$. But then, $\Xi(x)=\Xi(1)\bar{x}\bar{y}^{0}$ contrary to our supposition above. Hence $R^{\phi}\cap k[x,z,t]=k$. Then, for $\varepsilon(p)$ and $v$ from the expression of $p$ above, $\varepsilon(p)\geq 1$ and $0\neq v\in k$, for all $0\neq p\in R^{\phi}$. In particular, for each $p\in R^{\phi}\setminus{k}$, $\delta(p)=2\varepsilon(p)\geq 2$. Since the transcendence degree of $R$ over $R^{\phi}$ is $1$, the elements $z,t\in R$ must be algebraically dependent over $R^{\phi}$. Let $U,V$ be indeterminates and $0\neq P\in R^{\phi}[U,V]$ be such that $P(z,t)=0$. The polynomial $P$ is all not in $k[U,V]$ since $z,t$ are algebraically independent over $k$. Let $P(U,V)=\sum p(i,j)U^{i}V^{j}$ with $p(i,j)\in R^{\phi}$ for all $i,j$. Then, there is at least one $i,j$ such that $p(i,j)$ is not in $k$ and hence $\delta(p(i,j))\geq 2$. Partition $\textit{suppt}(P):=\{(i,j)\,|\,p(i,j)\neq 0\}$ into sets $M,M^{{}^{\prime}}$ such that $\delta(p(i,j))=2n:=\max\{\delta(p(i,j))\in\textit{suppt}(P)\}$ for all $(i,j)\in M$ and $\delta(p(i,j))<2n$ for all $(i,j)\in M^{{}^{\prime}}$. A straightforward verification shows that $P^{{}^{\prime}}(U,V)+\sum_{(i,j)\in M\setminus{M^{{}^{\prime}}}}p(i,j)U^{i}V^{j}=P(U,V)=\mu(U,V)y^{n}+\eta(U,V)xy^{n}+\sum_{r=1}^{n}\nu_{r}(U,V)y^{n-r}$ where $0\neq\mu(U,V)\in k[U,V]$, $\eta(U,V)\in k[z,t][U,V]$, $P^{{}^{\prime}}(U,V):=\sum_{(i,j)\in M^{{}^{\prime}}}p(i,j)U^{i}V^{j}$, and for $1\leq r\leq n$, $\nu_{r}(U,V)\in k[x,z,t][U,V]$.

Now $P(z,t)=0$ and since $z,t$ are algebraically independent over $k$, $\mu(z,t)\neq 0$. So,

$0\neq\mu(z,t)y^{n}=-\eta(z,t)xy^{n}-\sum_{r=1}^{n}\nu_{r}(z,t)y^{n-r}-P^{{}^{\prime}}(z,t)$.

This is absurd since $\delta(RHS)<2n$ whereas $\delta(\mu(z,t)y^{n})=2n$. In conclusion, there is $p\in R^{\phi}$ such that $\Xi(p)$ is either $\Xi(u)\bar{x}\bar{y}^{m}$ or $\Xi(v)\bar{y}^{m}$ with $m\in\mathbb{N}$ and $\Xi(v)\in k[\bar{z},\bar{t}]\setminus{k}$.

Next, let $\bar{\phi}\in\operatorname{Exp}(S/k)$ be the nontrivial grade-preserving exponential map induced by $\phi$. The, for $p\in R^{\phi}\setminus{k[x,z,t]}$, so $\Xi(p)\in S^{\bar{\phi}}$ and either $\Xi(p)=\Xi(u)\bar{x}\bar{y}^{m}$ or $\Xi(v)\bar{y}^{m}$ for some positive integer $m$. Since $S^{\bar{\phi}}$ is factorially closed in $S$, surely $\bar{y}\in S^{\bar{\phi}}$. Furthermore, let $p\in R^{\phi}$ such that either $\Xi(p)=\Xi(u)\bar{x}\bar{y}^{m}$ or $\Xi(p)=\Xi(v)\bar{y}^{m}$ with $m\geq 0$ and $\Xi(v)\in k[\bar{z},\bar{t}]\setminus{k}$. It readily follows that either $\bar{x}\in S^{\bar{\phi}}$ or $\Xi(v)\in S^{\bar{\phi}}\setminus{k}$. Therefore either $k[\bar{x},\bar{y}]\subseteq S^{\bar{\phi}}$ or $k[\bar{z},\bar{t}]\subseteq S^{\bar{\phi}}\neq k$. Neither of these is a possibility in view of Lemma 4.0.4 and Theorem 4.1 of [2]. Thus $R^{\phi}\subseteq k[x,z,t]$.

Lastly, suppose $x$ is not in $R^{\phi}$. Let $q,h\in R^{\phi}\subseteq k[x,z,t]$ be algebraically independent over $k$. Then, $q=q_{1}x+q_{2}$, $h=h_{1}x+h_{2}$ for $q_{1},h_{1}\in k[x,z,t]$ and $q_{2},h_{2}\in k[z,t]$. If there is $0\neq\rho(X,Y)\in k[X,Y]$ with $\rho(q_{2},h_{2})=0$, then $\rho(q,h)=x\sigma$ for some $\sigma\in k[x,z,t]$ and that forces $x,\sigma\in R^{\phi}$ contradicting our supposition. So, $q_{2},h_{2}$ are algebraically independent over $k$. Furthermore, letting $\bar{\phi}\in\operatorname{Exp}(S/k)$ denote the nontrivial grade-preserving exponential map induced by $\phi$, then $k[\Xi(q),\Xi(h)]\subseteq S^{\bar{\phi}}$. Now the elements $\bar{z},\bar{t}$ of $S$ both are algebraic over $k[\Xi(q),\Xi(h)]$ and also algebraic over $S^{\bar{\phi}}$. Since $S^{\bar{\phi}}$ is algebraically closed in $S$, so $k[\bar{z},\bar{t}]\subseteq S^{\bar{\phi}}$. Now the equation $\bar{x}^{2}\bar{y}=-\bar{z}^{2}-\bar{t}^{3}-c$ allows us to conclude that each of $\bar{x},\bar{y}$ is also in $S^{\bar{\phi}}$. Consequently, $S=k[\bar{x},\bar{y},\bar{z},\bar{t}]\subseteq S^{\bar{\phi}}$, i.e., $S^{\bar{\phi}}=S$. This is absurd since $\bar{\phi}$ is nontrivial. Hence $k[x]\subseteq R^{\phi}$. ∎