Question 3.3. Let $G$ be a finite local $k$ -group scheme. If there exists an injective homomorphism $\mathbb{X}(G){\hookrightarrow}(\mathbb{Q}_{p}/\mathbb{Z}_{p})^{\oplus \unicode[STIX]{x1D6FE}+n-1}$ , then does the group scheme $G$ belong to the set $\unicode[STIX]{x1D70B}_{A}^{\operatorname{loc}}(U)$ ?

Proof. First we remark on $\unicode[STIX]{x1D6FC}_{p}$ -torsors over $U$ . Since $U$ is affine,

$$\begin{eqnarray}H_{\operatorname{fpqc}}^{1}(U,\mathbb{G}_{a})=H^{1}(U,{\mathcal{O}}_{U})=0,\end{eqnarray}$$

we have

$$\begin{eqnarray}H_{\operatorname{fpqc}}^{1}(U,\unicode[STIX]{x1D6FC}_{p})\simeq \unicode[STIX]{x1D6E4}(U,{\mathcal{O}}_{U})/\unicode[STIX]{x1D6E4}(U,{\mathcal{O}}_{U})^{p}.\end{eqnarray}$$

On the other hand, since $U$ is an affine smooth integral scheme, there exists a dominant morphism $U\rightarrow \mathbb{A}_{k}^{1}$ , whence

(3.2) $$\begin{eqnarray}H_{\operatorname{fpqc}}^{1}(U,\unicode[STIX]{x1D6FC}_{p}){\hookleftarrow}H_{\operatorname{fpqc}}^{1}(\mathbb{A}_{k}^{1},\unicode[STIX]{x1D6FC}_{p})=\bigoplus _{p\nmid n}k\cdot t^{n},\end{eqnarray}$$

where $t$ is the coordinate of $\mathbb{A}^{1}$ . Furthermore, since $\unicode[STIX]{x1D6FC}_{p}$ is simple, any nonzero element of $H_{\operatorname{fpqc}}^{1}(U,\unicode[STIX]{x1D6FC}_{p})$ corresponds to a surjective homomorphism $\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U){\twoheadrightarrow}\unicode[STIX]{x1D6FC}_{p}$ (cf. (2.1)).

Now let us prove the proposition. It suffices to show the ‘if’ part. We prove this by induction on the order $\operatorname{dim}_{k}k[G]=p^{r}~(r>0)$ . From the assumption, $G$ is obtained by central extensions of $\unicode[STIX]{x1D6FC}_{p}$ or $\unicode[STIX]{x1D707}_{p}$ . If $\operatorname{dim}_{k}k[G]=p$ , then $G=\unicode[STIX]{x1D6FC}_{p}$ , or $=\unicode[STIX]{x1D707}_{p}$ and the statement is immediate from (3.2), or from the assumption. Since $G$ is a nontrivial nilpotent group scheme, the center $Z(G)$ is nontrivial. Let $H\subset Z(G)$ be a subgroup scheme of order $p$ . Then we get a central extension of finite $k$ -group schemes:

(3.3) $$\begin{eqnarray}1\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 1.\end{eqnarray}$$

Since $\operatorname{dim}_{k}k[G/H]<\operatorname{dim}_{k}k[G]$ and $\mathbb{X}(G/H)\subseteq \mathbb{X}(G)$ , by induction hypothesis, there exists a surjective homomorphism $\overline{\unicode[STIX]{x1D719}}:\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U)\rightarrow G/H$ . Since $U$ is affine, $H_{\operatorname{fpqc}}^{q}(U,\mathbb{G}_{a})=0$ if $q\neq 0$ , we have $H_{\operatorname{fpqc}}^{2}(U,\unicode[STIX]{x1D6FC}_{p})=0$ . On the other hand, $H_{\operatorname{fpqc}}^{1}(U,\mathbb{G}_{m})=\text{Pic}(U)$ is divisible (cf. [Sta, Tag 03RN, Proof of Lemma 53.68.3]) and $H_{\operatorname{fpqc}}^{2}(U,\mathbb{G}_{m})=\text{Br}(U)=0$ , we then also have $H_{\operatorname{fpqc}}^{2}(U,\unicode[STIX]{x1D707}_{p})=0$ . Therefore, we find that $H_{\operatorname{fpqc}}^{2}(U,H)=0$ and the exactness of (3.3), noticing that $H_{\operatorname{fpqc}}^{0}(U,G/H)=0$ , implies that the resulting sequence

$$\begin{eqnarray}0\rightarrow H_{\operatorname{fpqc}}^{1}(U,H)\rightarrow H_{\operatorname{ fpqc}}^{1}(U,G)\rightarrow H_{\operatorname{ fpqc}}^{1}(U,G/H)\rightarrow 0\end{eqnarray}$$

is an exact sequence of pointed sets (cf. [Reference GiraudGir71, p. 284, Remarque 4.2.10]). Therefore, the isomorphism (2.1) implies that there exists a lift $\unicode[STIX]{x1D719}:\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U)\rightarrow G$ of $\overline{\unicode[STIX]{x1D719}}$ and we obtain the following commutative diagram:

where $K\overset{\operatorname{def}}{=}\operatorname{Ker}(\overline{\unicode[STIX]{x1D719}})$ . If $f$ is nontrivial, then it is surjective, whence so is $\unicode[STIX]{x1D719}$ . Thus, from now on assume $f=0$ . In this case, the homomorphism $\unicode[STIX]{x1D719}$ factors through $G/H$ . Thus, the central extension (3.3) is trivial, i.e., $G=H\times (G/H)$ . We claim that

(3.4) $$\begin{eqnarray}H_{\operatorname{fpqc}}^{1}(U,H)\supsetneq \operatorname{Ker}(\operatorname{Hom}_{k}(\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U),H)\rightarrow \operatorname{Hom}_{k}(K,H)).\end{eqnarray}$$

If the claim (3.4) is true, then one takes a $k$ -homomorphism $g:\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U)\rightarrow H$ so that $g|_{K}\neq 0$ and the one $(g,\overline{\unicode[STIX]{x1D719}}):\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U)\rightarrow G=H\times (G/H)$ is surjective. Thus, it remains to show the claim (3.4). Notice that

$$\begin{eqnarray}\operatorname{Hom}_{k}(G/H,H)=\operatorname{Ker}(\operatorname{Hom}_{k}(\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U),H)\rightarrow \operatorname{Hom}_{k}(K,H)).\end{eqnarray}$$

Thus, if $H\simeq \unicode[STIX]{x1D6FC}_{p}$ , then the claim (3.4) follows from $\operatorname{dim}_{k}H_{\operatorname{fpqc}}^{1}(U,\unicode[STIX]{x1D6FC}_{p})=\infty$ . If $H\simeq \unicode[STIX]{x1D707}_{p}$ , then the claim (3.4) follows from the following inequality:

$$\begin{eqnarray}\operatorname{dim}_{\mathbb{F}_{p}}\operatorname{Hom}_{k}(G/H,\unicode[STIX]{x1D707}_{p})<\operatorname{dim}_{\mathbb{F}_{p}}\operatorname{Hom}_{k}(G,\unicode[STIX]{x1D707}_{p})\leqslant \unicode[STIX]{x1D6FE}+n-1=\operatorname{dim}_{\mathbb{F}_{p}}H_{\operatorname{fpqc}}^{1}(U,\unicode[STIX]{x1D707}_{p}).\end{eqnarray}$$

Here for the first inequality, we use $G\simeq \unicode[STIX]{x1D707}_{p}\times (G/H)$ . This completes the proof.◻

Question 4.1. Fix an integer $r>0$ .

1. (1) Does there exist any $k$ -morphism $f:U\rightarrow \unicode[STIX]{x1D6F4}$ so that the $\unicode[STIX]{x1D6F4}_{(r)}$ -torsor $f^{\ast }\unicode[STIX]{x1D6F4}^{(-r)}\rightarrow U$ defined by the following cartesian diagram is saturated?

   

2. (2) Furthermore, if it exists, for which $k$ -morphism $f:U\rightarrow \unicode[STIX]{x1D6F4}$ , is the resulting $\unicode[STIX]{x1D6F4}_{(r)}$ -torsor $f^{\ast }\unicode[STIX]{x1D6F4}^{(-r)}\rightarrow U$ saturated?

Proof. We will show this by induction on $r\geqslant 1$ . We will denote by $\unicode[STIX]{x1D719}^{(-r)}:\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U)\rightarrow \unicode[STIX]{x1D6F4}_{(r)}$ the homomorphism corresponding to the torsor $f^{\ast }\unicode[STIX]{x1D6F4}^{(-r)}$ . Assume $\unicode[STIX]{x1D719}^{(-r)}$ is surjective. Let us show that $\unicode[STIX]{x1D719}^{(-r-1)}$ is also surjective. Since $F^{(r)\ast }f^{\ast }\unicode[STIX]{x1D6F4}^{(-r)}$ is a trivial torsor, the composition

$$\begin{eqnarray}\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U^{(-r)})\xrightarrow[{}]{F_{\ast }^{(r)}}\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U)\overset{\unicode[STIX]{x1D719}^{(-r)}}{{\twoheadrightarrow}}\unicode[STIX]{x1D6F4}_{(r)}\end{eqnarray}$$

is trivial. We then obtain the following commutative diagram:

(4.1)

and the map $\unicode[STIX]{x1D719}^{(-r-1)}\circ F_{\ast }^{(r)}$ factors through $\unicode[STIX]{x1D6F4}_{(1)}^{(-r)}$ . We denote by $\unicode[STIX]{x1D713}$ the resulting homomorphism $\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U^{(-r)})\rightarrow \unicode[STIX]{x1D6F4}_{(1)}^{(-r)}$ . We are then reduced to showing the surjectivity of $\unicode[STIX]{x1D713}$ . The commutativity of the diagram (4.1) implies that $\text{Ind}_{\unicode[STIX]{x1D6F4}_{(1)}^{(-r)}}^{\unicode[STIX]{x1D6F4}_{(r+1)}}(Q)\simeq F^{(r)\ast }f^{\ast }\unicode[STIX]{x1D6F4}^{(-r-1)}$ , where $Q$ is the torsor over $U^{(-r)}$ corresponding to the morphism $\unicode[STIX]{x1D713}$ . On the other hand, by considering the tautological commutative diagram

we can find that

$$\begin{eqnarray}\text{Ind}_{\unicode[STIX]{x1D6F4}_{(1)}^{(-r)}}^{\unicode[STIX]{x1D6F4}_{(r+1)}}(\unicode[STIX]{x1D6F4}^{(-r-1)}\xrightarrow[{}]{F^{(1)}}\unicode[STIX]{x1D6F4}^{(-r)})=F^{(r)\ast }(\unicode[STIX]{x1D6F4}^{(-r-1)}\xrightarrow[{}]{F^{(r+1)}}\unicode[STIX]{x1D6F4}).\end{eqnarray}$$

Therefore, by the construction, $Q$ is nothing but the torsor defined by the cartesian diagram

where $f^{(-r)}$ is just the $r$ th Frobenius twist of $f$ .

Then, the saturatedness of the torsor $f^{\ast }\unicode[STIX]{x1D6F4}^{(1)}\rightarrow U$ indicates the saturatedness of the torsor $Q\rightarrow U^{(-r)}$ , or equivalently, the surjectivity of $\unicode[STIX]{x1D713}$ . This completes the proof.◻

Proof. We will show this by induction on $n>0$ . In the case where $n=1$ , then since $\unicode[STIX]{x1D6FC}_{p}$ is simple, the assertion is obvious. From now on assume that $n>1$ and that $\operatorname{dim}_{k}\langle \overline{f_{i}(t)}\,|\,1\leqslant i\leqslant n-1\rangle _{k}=n-1$ . Put $\text{}\underline{f^{\prime }}:=(f_{1}(t),\ldots ,f_{n-1}(t))$ . Then $P_{\text{}\underline{f^{\prime }}}$ is an $\unicode[STIX]{x1D6FC}_{p}^{\oplus n-1}$ -torsor over $\mathbb{A}_{k}^{1}$ . We denote by $\unicode[STIX]{x1D719}$ and $\unicode[STIX]{x1D719}^{\prime }$ the homomorphism $\unicode[STIX]{x1D70B}^{\operatorname{loc}}(\mathbb{A}_{k}^{1})\rightarrow \unicode[STIX]{x1D6FC}_{p}^{\oplus n}$ corresponding to $P_{\text{}\underline{f}}$ and the one $\unicode[STIX]{x1D70B}^{\operatorname{loc}}(\mathbb{A}_{k}^{1})\rightarrow \unicode[STIX]{x1D6FC}_{p}^{\oplus n-1}$ to  $P_{\text{}\underline{f^{\prime }}}$ , respectively. From the assumption, $\unicode[STIX]{x1D719}^{\prime }$ is surjection. Let $K\overset{\operatorname{def}}{=}\operatorname{Ker}(\unicode[STIX]{x1D719}^{\prime })$ . We then obtain the following commutative diagram.

Then we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}:\text{surjective} & \Longleftrightarrow & \displaystyle \unicode[STIX]{x1D713}~\neq ~1\nonumber\\ \displaystyle & \Longleftrightarrow & \displaystyle \text{pr}_{\text{n}}\circ \unicode[STIX]{x1D719}\in \operatorname{Hom}(\unicode[STIX]{x1D70B}^{\operatorname{loc}}(\mathbb{A}_{k}^{1}),\unicode[STIX]{x1D6FC}_{p})\setminus {\unicode[STIX]{x1D719}^{\prime }}^{\ast }\operatorname{Hom}(\unicode[STIX]{x1D6FC}_{p}^{\oplus n-1},\unicode[STIX]{x1D6FC}_{p})\nonumber\\ \displaystyle & \Longleftrightarrow & \displaystyle \overline{f_{n}(t)}\in (k[t]/k[t^{p}])\setminus \langle \overline{f_{1}(t)},\ldots ,\overline{f_{n-1}(t)}\rangle _{k}\nonumber\\ \displaystyle & \Longleftrightarrow & \displaystyle \operatorname{dim}_{k}\langle \overline{f_{i}(t)}\,|\,1\leqslant i\leqslant n\rangle _{k}=n.\nonumber\end{eqnarray}$$

This completes the proof. ◻

Proof. Let $P^{1\text{-}\text{strat}}=\{E^{(i)},\unicode[STIX]{x1D70E}^{(i)}\}$ . Let $\unicode[STIX]{x1D70B}:P\rightarrow \mathbb{G}_{m}$ be the structure morphism. We then have

$$\begin{eqnarray}\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}=\frac{k[x^{\pm 1},y_{ij}~(i,j=1,2)]}{(y_{11}^{2}-f_{22}(x)x^{-m},y_{12}^{2}-f_{12}(x)x^{-m},y_{21}^{2}-f_{21}(x)x^{-m},y_{22}^{2}-f_{11}(x)x^{-m})}.\end{eqnarray}$$

Let $\unicode[STIX]{x1D70C}_{P}$ (respectively $\unicode[STIX]{x1D70C}_{0}$ ) be the coaction $\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\rightarrow \unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\otimes k[\operatorname{GL}_{2(1)}]$ induced by the action $P\times \operatorname{GL}_{2(1)}\rightarrow P$ (respectively the one by the trivial action). Let $\unicode[STIX]{x1D704}$ be the antipode of $k[\operatorname{GL}_{2(1)}]$ . Then the composition

(4.5) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\otimes k[\operatorname{GL}_{2(1)}] & & \displaystyle \xrightarrow[{}]{\unicode[STIX]{x1D70C}\otimes \text{id}}\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\otimes k[\operatorname{GL}_{2(1)}]\otimes k[\operatorname{GL}_{2(1)}]\nonumber\\ \displaystyle & & \displaystyle \xrightarrow[{}]{\text{id}\otimes \unicode[STIX]{x1D704}\otimes \text{id}}\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\otimes k[\operatorname{GL}_{2(1)}]\otimes k[\operatorname{GL}_{2(1)}]\xrightarrow[{}]{\text{id}\otimes m}\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\otimes k[\operatorname{GL}_{2(1)}]\end{eqnarray}$$

gives a $k[\operatorname{GL}_{2(1)}]$ -comodule isomorphism

(4.6) $$\begin{eqnarray}(\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P},\unicode[STIX]{x1D70C}_{P})\otimes (k[\operatorname{GL}_{2(1)}],\unicode[STIX]{x1D70C}_{\text{reg}})\xrightarrow[{}]{\simeq }(\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P},\unicode[STIX]{x1D70C}_{0})\otimes (k[\operatorname{GL}_{2(1)}],\unicode[STIX]{x1D70C}_{\text{reg}}).\end{eqnarray}$$

If one writes $k[\operatorname{GL}_{2(1)}]=k[z_{ij}]/(z_{ij}^{2}-\unicode[STIX]{x1D6FF}_{ij})$ , then

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}\unicode[STIX]{x1D704}(z_{11}) & \unicode[STIX]{x1D704}(z_{12})\\ \unicode[STIX]{x1D704}(z_{21}) & \unicode[STIX]{x1D704}(z_{22})\end{array}\right)=\left(\begin{array}{@{}cc@{}}z_{11} & z_{12}\\ z_{21} & z_{22}\end{array}\right)^{-1}.\end{eqnarray}$$

Therefore, via the map (4.5), the element

$$\begin{eqnarray}e_{ij}\overset{\operatorname{def}}{=}\mathop{\sum }_{q=1}^{2}y_{iq}\otimes z_{qj}\in \unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P}\otimes k[\operatorname{GL}_{2(1)}]\end{eqnarray}$$

is mapped to $y_{ij}\otimes 1$ , which belongs to

$$\begin{eqnarray}((\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{P},\unicode[STIX]{x1D70C}_{0})\otimes (k[\operatorname{GL}_{2(1)}],\unicode[STIX]{x1D70C}_{\text{reg}}))^{\operatorname{GL}_{2(1)}}=k[x^{\pm 1}][y_{ij}\otimes 1\,|\,1\leqslant i,j\leqslant 2].\end{eqnarray}$$

Hence, by (4.6), we find that

$$\begin{eqnarray}E^{(0)}=k[x^{\pm 1}][e_{ij}\,|\,1\leqslant i,j\leqslant 2].\end{eqnarray}$$

Notice that $E^{(i)}\simeq k[x^{\pm 1}]^{\oplus 2^{4}}$ with free basis $\{e_{11}^{m_{11}}e_{12}^{m_{12}}e_{21}^{m_{21}}e_{22}^{m_{22}}\,|\,m_{ij}=0,1\}$ .

Let $A\in \operatorname{GL}_{2^{4}}(k[x^{(-1)\pm 1}])$ be the representation matrix of the inverse isomorphism $\unicode[STIX]{x1D70E}^{(0)-1}$ with respect to the basis

$$\begin{eqnarray}\displaystyle & & \displaystyle 1,\,\unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{21},\,\unicode[STIX]{x1D708}_{12}\unicode[STIX]{x1D708}_{22},\,\unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{12}\unicode[STIX]{x1D708}_{21}\unicode[STIX]{x1D708}_{22},;\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D708}_{11},\,\unicode[STIX]{x1D708}_{12},\,\unicode[STIX]{x1D708}_{21},\,\unicode[STIX]{x1D708}_{22},\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{12},\,\unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{22},\,\unicode[STIX]{x1D708}_{12}\unicode[STIX]{x1D708}_{21},\unicode[STIX]{x1D708}_{21}\unicode[STIX]{x1D708}_{22};\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{12}\unicode[STIX]{x1D708}_{21},\,\unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{12}\unicode[STIX]{x1D708}_{22},\,\unicode[STIX]{x1D708}_{11}\unicode[STIX]{x1D708}_{21}\unicode[STIX]{x1D708}_{22},\,\unicode[STIX]{x1D708}_{12}\unicode[STIX]{x1D708}_{21}\unicode[STIX]{x1D708}_{22}\nonumber\end{eqnarray}$$

for $\unicode[STIX]{x1D708}\in \{e,z\}$ . One then obtains

$$\begin{eqnarray}A=\left(\begin{array}{@{}ccccc@{}}C & O & O & O & \\ & B\otimes E & O & B\otimes D & \\ & & B\otimes B & O & \!\\ & O & & B\otimes x^{(-1)m}E\end{array}\right),\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle & \displaystyle B=\left(\begin{array}{@{}cc@{}}f_{11}^{\prime }(x^{(-1)}) & f_{21}^{\prime }(x^{(-1)})\\ f_{12}^{\prime }(x^{(-1)}) & f_{22}^{\prime }(x^{(-1)})\end{array}\right);\quad E=\left(\begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right); & \displaystyle \\ \displaystyle & \displaystyle C=\left(\begin{array}{@{}cccc@{}}1 & f_{11}^{\prime }(x^{(-1)})f_{21}^{\prime }(x^{(-1)}) & f_{12}^{\prime }(x^{(-1)})f_{22}^{\prime }(x^{(-1)}) & \ast \\ 0 & x^{(-1)m} & 0 & f_{12}^{\prime }(x^{(-1)})f_{22}^{\prime }(x^{(-1)})x^{(-1)m}\\ 0 & 0 & x^{(-1)m} & f_{11}^{\prime }(x^{(-1)})f_{21}^{\prime }(x^{(-1)})x^{(-1)m}\\ 0 & 0 & 0 & x^{m}\end{array}\right);\\ \displaystyle & \displaystyle D=\left(\begin{array}{@{}cc@{}}0 & f_{12}^{\prime }(x^{(-1)})f_{22}^{\prime }(x^{(-1)})\\ f_{11}^{\prime }(x^{(-1)})f_{21}^{\prime }(x^{(-1)}) & 0\end{array}\right),\end{eqnarray}$$

where $\{f_{ij}^{\prime }(x^{(-1)})\}$ is defined by

$$\begin{eqnarray}f_{ij}^{\prime }(x^{(-1)})^{2}=f_{ij}(x).\end{eqnarray}$$

Here, notice that $x^{(-1)2}=x$ and that $\det B=x^{(-1)m}$ . Furthermore, the assumption on $\{f_{ij}(x)\}$ implies the same condition on $\{f_{ij}^{\prime }(x^{(-1)})\}$ .

Now one can reduce the problem to calculating the right-hand side of the following equation:

$$\begin{eqnarray}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{G}_{m},1)}(\mathbb{I},P^{1\text{-}\text{strat}})\simeq \{(\text{}\underline{a},\text{}\underline{b})\in k^{\oplus 2^{4}}\times k[x^{\pm 1}]^{\oplus 2^{4}}\,|\,A\cdot \text{}\underline{a}=\text{}\underline{b}\}.\end{eqnarray}$$

First we consider the case where $2\mid m$ . In this case, without loss of generality, we may assume that $m=0$ . Note that the condition that $\det (f_{ij}(x))=1$ implies that the set

$$\begin{eqnarray}\{\overline{f_{11}f_{12}},\,\overline{f_{21}f_{22}}\}\end{eqnarray}$$

is also linearly independent over $k$ . Indeed, this follows from the equation:

$$\begin{eqnarray}\left(\begin{array}{@{}c@{}}f_{11}f_{12}\\ f_{21}f_{22}\end{array}\right)=\left(\begin{array}{@{}cc@{}}f_{12}^{2} & f_{11}^{2}\\ f_{22}^{2} & f_{21}^{2}\end{array}\right)\left(\begin{array}{@{}c@{}}f_{11}f_{21}\\ f_{12}f_{22}\end{array}\right).\end{eqnarray}$$

We will solve the simultaneous equations $A\cdot \text{}\underline{a}=\text{}\underline{b}$ from the bottom. Then the condition that the set $\{\overline{f_{11}^{\prime }},\,\overline{f_{21}^{\prime }}\}$ , or $\{\overline{f_{12}^{\prime }},\,\overline{f_{22}^{\prime }}\}$ is linearly independent implies that

$$\begin{eqnarray}a_{i}=0\quad (13\leqslant i\leqslant 16).\end{eqnarray}$$

Moreover, the conditions that $\{\overline{f_{11}^{\prime }f_{12}^{\prime }},\,\overline{f_{21}^{\prime }f_{22}^{\prime }}\}$ is linearly independent and that $\det B=f_{11}^{\prime }f_{22}^{\prime }-f_{12}^{\prime }f_{21}^{\prime }=1$ imply that

$$\begin{eqnarray}\displaystyle & a_{i}=0\quad (i=9,12), & \displaystyle \nonumber\\ \displaystyle & a_{10}=a_{11}. & \displaystyle \nonumber\end{eqnarray}$$

By solving the equations

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}C & O\\ O & B\otimes E\end{array}\right),\end{eqnarray}$$

again combined with the assumption on $\overline{f_{ij}^{\prime }}$ , we have

$$\begin{eqnarray}a_{i}=0\quad (2\leqslant i\leqslant 8).\end{eqnarray}$$

All the above computations then imply

$$\begin{eqnarray}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{G}_{m},1)}(\mathbb{I},P^{1\text{-}\text{strat}})\simeq k\oplus k\cdot \det (z_{ij})\simeq k[\unicode[STIX]{x1D707}_{2}].\end{eqnarray}$$

Finally, let us consider the case where $2\nmid m$ . In this case, the equations in the part $B\otimes x^{(-1)m}E$ cannot imply $a_{i}=0~(13\leqslant i\leqslant 16)$ . However, by solving the simultaneous equations

$$\begin{eqnarray}\left(\begin{array}{@{}cccc@{}}O & B\otimes E & O & B\otimes D\\ \end{array}\right)\text{}\underline{a}=\text{}\underline{b},\end{eqnarray}$$

we can conclude that

$$\begin{eqnarray}a_{i}=0\quad (5\leqslant i\leqslant 8~\text{or }13\leqslant i\leqslant 16).\end{eqnarray}$$

Next let us solve the part $C$ . Notice that $\{\overline{x^{(-1)m}},\,\overline{f_{11}^{\prime }f_{21}^{\prime }x^{(-1)m}}\}$ is linearly independent over $k$ . Thus, we find that $a_{4}=0$ . On the other hand, by the assumption that $\{\overline{f_{11}^{\prime }f_{21}^{\prime }},\,\overline{f_{12}^{\prime }f_{22}^{\prime }}\}$ is linearly independent, we also have $a_{2}=a_{3}=0$ . Finally let us solve the part $B\otimes B$ . Then the condition that $\overline{f_{11}^{\prime }f_{21}^{\prime }}\neq 0$ in $k[x^{(-1)\pm 1}]/k[x^{(-1)\pm 2}]$ implies that $a_{10}=a_{11}$ . Then the condition that $\{\overline{f_{11}f_{12}},\,\overline{f_{21}f_{22}},\,\overline{x^{m}}\}$ is linearly independent over $k$ implies that $a_{9}=a_{10}=a_{11}=a_{12}=0$ . Therefore, we can conclude that

$$\begin{eqnarray}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{G}_{m},1)}(\mathbb{I},P^{1\text{-}\text{strat}})=k.\Box\end{eqnarray}$$

Proof. By considering the composition

$$\begin{eqnarray}\mathbb{G}_{m}{\hookrightarrow}\mathbb{A}_{k}^{1}\xrightarrow[{}]{\text{}\underline{f}}\operatorname{SL}_{2}{\hookrightarrow}\operatorname{GL}_{2},\end{eqnarray}$$

we are reduced to the case where $\text{}\underline{f}:\mathbb{G}_{m}\rightarrow \operatorname{GL}_{2}$ with $\det (\text{}\underline{f})=1$ . Thus, the statement follows from the argument in the proof of Theorem 4.12, noticing that the condition

$$\begin{eqnarray}\operatorname{dim}_{k}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{G}_{m},1)}(\mathbb{I},P^{1\text{-}\text{strat}})=2\end{eqnarray}$$

implies the image of the homomorphism $\unicode[STIX]{x1D70B}^{\operatorname{loc}}(\mathbb{G}_{m})\rightarrow \operatorname{GL}_{2(1)}$ corresponding to the torsor $P=\text{}\underline{f}^{\ast }\operatorname{GL}_{2}^{(-1)}\rightarrow \mathbb{G}_{m}$ coincides with $\operatorname{SL}_{2(1)}$ .◻

Proof. It suffices to find a $k$ -morphism $\text{}\underline{f}\in \operatorname{GL}_{2}(k[x^{\pm 1}])=\text{Mor}_{k}(\mathbb{G}_{m},\operatorname{GL}_{2})$ satisfying the condition of Theorem 4.12. The morphism

$$\begin{eqnarray}\mathbb{G}_{m}\ni a\mapsto \left(\begin{array}{@{}cc@{}}{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{a^{4}}} & {\displaystyle \frac{1}{a^{2}}}\\ 1 & {\displaystyle \frac{1}{a}}\end{array}\right)\in \operatorname{GL}_{2}\end{eqnarray}$$

gives such a one. ◻

Proof. By considering $V:=V_{1}+\cdots +V_{m}$ , we are reduced to the case where $m=1$ . Let $V$ be a finite dimensional subspace of $k[x_{1},\ldots ,x_{n}]/k[x_{1}^{p},\ldots ,x_{n}^{p}]$ . Without loss of generality, we may assume that $V$ is of the form

$$\begin{eqnarray}V=\langle x_{1}^{m_{1}}\cdots x_{n}^{m_{n}}\,|\,0\leqslant m_{i}\leqslant d;\,p\nmid m_{i}~\text{for some }i\rangle\end{eqnarray}$$

for a sufficiently large integer $d>0$ . Take an integer $M>d$ with $p\nmid M$ and an integer $N>0$ with $p^{N}>d(M+1)$ . Let us define

$$\begin{eqnarray}g\overset{\operatorname{def}}{=}\mathop{\prod }_{i=1}^{n-1}(x_{i}^{p^{N}}+x_{i}^{M}).\end{eqnarray}$$

Note that the following subset of $k[x_{1},\ldots ,x_{n-1}]/k[x_{1}^{p},\ldots ,x_{n-1}^{p}]$ is linearly independent over $k$ :

$$\begin{eqnarray}\displaystyle x_{1}^{m_{1}}\cdots x_{n-1}^{m_{n-1}} & & \displaystyle \quad (0\leqslant m_{i}\leqslant d,m_{n}=0;~p\nmid m_{i}~\text{for some }1\leqslant i\leqslant n-1),\nonumber\\ \displaystyle \mathop{\prod }_{i=1}^{n-1}x_{i}^{m_{i}+m_{n}M} & & \displaystyle \quad (p\mid m_{i}~\text{for any }1\leqslant i\leqslant n-1,\text{whence }p\nmid m_{n}),\nonumber\\ \displaystyle \mathop{\prod }_{i=1}^{n-1}x_{i}^{m_{i}+m_{n}p^{N}} & & \displaystyle \quad (0\leqslant m_{i}\leqslant d;~m_{n}\neq 0;~p\nmid m_{i}~\text{for some }1\leqslant i\leqslant n-1).\nonumber\end{eqnarray}$$

One can then conclude that the polynomial $g$ gives a desired polynomial.◻

Proof of Theorem 4.17.

We will show that there exists a closed immersion $\unicode[STIX]{x1D704}:\mathbb{A}_{k}^{n-1}{\hookrightarrow}\mathbb{A}_{k}^{n}$ such that

$$\begin{eqnarray}\operatorname{dim}_{k}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{A}_{k}^{n-1},1)}(\mathbb{I},(\unicode[STIX]{x1D704}^{\ast }P)^{1\text{-}\text{strat}})=1.\end{eqnarray}$$

Note that $(\unicode[STIX]{x1D704}^{\ast }P)^{1\text{-}\text{strat}}=\unicode[STIX]{x1D704}^{\ast }(P^{1\text{-}\text{strat}})$ and that, from the assumption, we have

(4.7) $$\begin{eqnarray}\operatorname{dim}_{k}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{A}_{k}^{n},1)}(\mathbb{I},P^{1\text{-}\text{strat}})=1.\end{eqnarray}$$

Let $P^{1\text{-}\text{strat}}=\{E^{(i)},\unicode[STIX]{x1D70E}^{(i)}\}$ . By Serre’s conjecture on vector bundles on an affine space (cf. [Reference LamLam78]), the vector bundle $E^{(0)}$ is a free ${\mathcal{O}}_{\mathbb{A}_{k}^{n}}$ -module. Let

$$\begin{eqnarray}A\in \operatorname{GL}_{q}(k[x_{1}^{(-1)},\ldots ,x_{n}^{(-1)}])\end{eqnarray}$$

with $q=\operatorname{dim}_{k}k[G]$ , which is some power of $p$ , be the representation matrix of $\unicode[STIX]{x1D70E}^{(0)-1}$ with respect to some basis. The assumption (4.7) then amounts to saying that the vector space

$$\begin{eqnarray}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{A}_{k}^{n},1)}(\mathbb{I},P^{1\text{-}\text{strat}})\simeq \{\text{}\underline{a}\in k^{\oplus q}\,|\,A\cdot \text{}\underline{a}\in k[x_{1},\ldots ,x_{n}]^{\oplus q}\}\end{eqnarray}$$

is of dimension one. Then, by permuting basis if necessary, we may assume that

$$\begin{eqnarray}\{\text{}\underline{a}\in k^{\oplus q}\,|\,A\cdot \text{}\underline{a}\in k[x_{1},\ldots ,x_{n}]^{\oplus q}\}=\left\{\left.\left(\begin{array}{@{}c@{}}a\\ 0\\ \vdots \\ 0\end{array}\right)\right|a\in k\right\}.\end{eqnarray}$$

Then, the matrix $A$ has the following form:

$$\begin{eqnarray}A=\left(\begin{array}{@{}cccc@{}}a_{11} & \ast & \cdots \, & \ast \\ a_{21} & \ast & \cdots \, & \ast \\ \vdots & & \ddots & \\ a_{q1} & \ast & \cdots \, & \ast \end{array}\right)\quad \text{with }a_{i1}\in k[x_{1},\ldots ,x_{n}]~(1\leqslant i\leqslant q).\end{eqnarray}$$

Let us write

$$\begin{eqnarray}B\overset{\operatorname{def}}{=}\left(\begin{array}{@{}ccc@{}}a_{12} & \cdots \, & a_{1q}\\ & \cdots \, & \\ a_{q2} & \cdots \, & a_{qq}\end{array}\right).\end{eqnarray}$$

Then the condition that $\operatorname{dim}_{k}\operatorname{Hom}_{\operatorname{Strat}(\mathbb{A}_{k}^{1},1)}(\mathbb{I},P^{1\text{-}\text{strat}})=1$ is equivalent to the condition that

$$\begin{eqnarray}B\cdot \text{}\underline{b}\not \in k[x_{1},\ldots ,x_{n}]^{\oplus q}\quad (0\neq \text{}\underline{b}\in k^{\oplus q-1}).\end{eqnarray}$$

Therefore, we are reduced to showing the following statement: if a matrix

$$\begin{eqnarray}B=(b_{ij})\in \text{Matrix}_{s,t}(k[x_{1}^{(-1)},\ldots ,x_{n}^{(-1)}])\end{eqnarray}$$

satisfies the condition

$$\begin{eqnarray}B\cdot \text{}\underline{b}\not \in k[x_{1},\ldots ,x_{n}]^{\oplus t}\quad (0\neq \text{}\underline{b}\in k^{\oplus s}),\end{eqnarray}$$

then there exists a polynomial $g=g(x_{1}^{(-1)},\ldots ,x_{n-1}^{(-1)})$ such that the condition

$$\begin{eqnarray}B^{\prime }\cdot \text{}\underline{b}\not \in k[x_{1},\ldots ,x_{n-1}]^{\oplus t}\quad (0\neq \text{}\underline{b}\in k^{\oplus s}),\end{eqnarray}$$

is fulfilled. Here $B^{\prime }=(b_{ij}^{\prime })$ is a matrix with $b_{ij}^{\prime }=b_{ij}(x_{1}^{(-1)},\ldots ,x_{n-1}^{(-1)},g)$ . Indeed, by applying Lemma 4.18 to the subspaces

$$\begin{eqnarray}V_{i}\overset{\operatorname{def}}{=}\langle \overline{b_{i1}},\ldots ,\overline{b_{is}}\rangle \subset k[x_{1}^{(-1)},\ldots ,x_{n}^{(-1)}]/k[x_{1},\ldots ,x_{n}]\quad (1\leqslant i\leqslant t),\end{eqnarray}$$

we can find a polynomial $g=g(x_{1}^{(-1)},\ldots ,x_{n-1}^{(-1)})$ so that the homomorphism

$$\begin{eqnarray}k[x_{1}^{(-1)},\ldots ,x_{n}^{(-1)}]/k[x_{1},\ldots ,x_{n}]\rightarrow k[x_{1}^{(-1)},\ldots ,x_{n-1}^{(-1)}]/k[x_{1},\ldots ,x_{n-1}]\end{eqnarray}$$

induced by $x_{n}^{(-1)}\mapsto g(x_{1}^{(-1)},\ldots ,x_{n-1}^{(-1)})$ maps all the subspaces $V_{i}$ injectively into

$$\begin{eqnarray}k[x_{1}^{(-1)},\ldots ,x_{n-1}^{(-1)}]/k[x_{1},\ldots ,x_{n-1}].\end{eqnarray}$$

Then for any $\text{}\underline{b}\in k^{\oplus s}$ , the condition that $B^{\prime }\cdot \text{}\underline{b}\in k[x_{1},\ldots ,x_{n-1}]^{\oplus t}$ implies the condition that $B\cdot \text{}\underline{b}\in k[x_{1},\ldots ,x_{n}]^{\oplus t}$ . This completes the proof.◻

Proof. We will adopt the argument of [Reference SerreSer92, § 3.2]. Let us find a $k$ -morphism $\mathbb{A}_{k}^{1}\rightarrow \unicode[STIX]{x1D6F4}$ so that the resulting tower of torsors

$$\begin{eqnarray}\cdots \rightarrow \unicode[STIX]{x1D6F4}^{(-r)}|_{\mathbb{ A}_{k}^{1}}\rightarrow \cdots \rightarrow \unicode[STIX]{x1D6F4}^{(-1)}|_{\mathbb{ A}_{k}^{1}}\rightarrow \mathbb{A}_{k}^{1}\end{eqnarray}$$

is saturated. By virtue of Lemma 4.2, it suffices to find a $k$ -morphism $\mathbb{A}_{k}^{1}\rightarrow \unicode[STIX]{x1D6F4}$ so that $\unicode[STIX]{x1D6F4}^{(-1)}|_{\mathbb{A}_{k}^{1}}$ is saturated. Fix a maximal torus $T<\unicode[STIX]{x1D6F4}$ and a set of positive roots. Let $U^{+}<\unicode[STIX]{x1D6F4}$ (respectively $U^{-}<\unicode[STIX]{x1D6F4}$ ) be the unipotent radical corresponding to the positive roots (respectively the negative roots). Let us define a $k$ -morphism $f:U^{+}\times U^{-}\rightarrow \unicode[STIX]{x1D6F4}$ by the composition

$$\begin{eqnarray}U^{+}\times U^{-}{\hookrightarrow}\unicode[STIX]{x1D6F4}\times \unicode[STIX]{x1D6F4}\xrightarrow[{}]{m}\unicode[STIX]{x1D6F4}.\end{eqnarray}$$

Here the first map is the natural inclusion and the second one the multiplication of $\unicode[STIX]{x1D6F4}$ . Then the pulling-back $P\overset{\operatorname{def}}{=}f^{\ast }\unicode[STIX]{x1D6F4}$ defines a saturated $\unicode[STIX]{x1D6F4}_{(1)}$ -torsor over $U^{+}\times U^{-}$ . Indeed, notice that the diagram

commutes. Then the $\unicode[STIX]{x1D6F4}_{(1)}$ -torsor $i_{\pm }^{\ast }P\rightarrow U^{\pm }$ is reduced to the saturated $U_{(1)}^{\pm }$ -torsor $U^{\pm (-1)}\rightarrow U^{\pm }$ , i.e.,

$$\begin{eqnarray}i_{\pm }^{\ast }P\simeq \text{Ind}_{U_{(1)}^{\pm }}^{\unicode[STIX]{x1D6F4}_{(1)}}(U^{\pm (-1)})\overset{\operatorname{def}}{=}U^{\pm (-1)}\times ^{U_{(1)}^{\pm }}\unicode[STIX]{x1D6F4}_{(1)}.\end{eqnarray}$$

If we denote by $\unicode[STIX]{x1D719}:\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U^{+}\times U^{-})\rightarrow \unicode[STIX]{x1D6F4}_{(1)}$ (respectively $\unicode[STIX]{x1D713}^{\pm }:\unicode[STIX]{x1D70B}^{\operatorname{loc}}(U^{\pm })\rightarrow U_{(1)}^{\pm }$ ) the homomorphism corresponding to the torsor $P$ (respectively $i_{\pm }^{\ast }P$ ), this amounts to saying that the diagram homomorphism

commutes. Therefore, $\text{Im}(\unicode[STIX]{x1D719})\supset U_{(1)}^{\pm }$ , whence $\text{Im}(\unicode[STIX]{x1D719})=\langle U_{(1)}^{\pm }\rangle =\unicode[STIX]{x1D6F4}_{(1)}$ , where, for the last equality, see Remark 3.7(4). Therefore, $P$ is a saturated $\unicode[STIX]{x1D6F4}_{(1)}$ -torsor over $U^{+}\times U^{-}\simeq \mathbb{A}_{k}^{N}$ for some $N>1$ . Then by applying Theorem 4.17, we can conclude that there exists a closed immersion $\unicode[STIX]{x1D704}:\mathbb{A}_{k}^{1}{\hookrightarrow}U^{+}\times U^{-}$ so that $\unicode[STIX]{x1D704}^{\ast }P\rightarrow \mathbb{A}_{k}^{1}$ is saturated as well. Therefore, the $k$ -morphism $f\circ \unicode[STIX]{x1D704}:\mathbb{A}_{k}^{1}\rightarrow \unicode[STIX]{x1D6F4}$ gives a desired one. This completes the proof.◻