1 Introduction
For any commutative ring 
$R$, we will use the notation 
$A=R^{[n]}$ to mean that 
$A$ is a polynomial ring in 
$n$ variables over 
$R$. Now let 
$k$ be a field with algebraic closure 
$\bar{k}$ and 
$A$ be a 
$k$-algebra. We say that 
$A$ is an 
$\mathbb{A}^{n}$-form over 
$k$ if 
$A\otimes _{k}\bar{k}=\bar{k}^{[n]}$. It is well known that separable 
$\mathbb{A}^{1}$-forms are trivial (i.e., 
$k^{[1]}$) and that there exist nontrivial purely inseparable 
$\mathbb{A}^{1}$-forms over fields of positive characteristic. An extensive study of such algebras was made by Asanuma in [Reference Asanuma3]. Kambayashi established [Reference Kambayashi18] that separable 
$\mathbb{A}^{2}$-forms over a field 
$k$ are trivial. Over any field of positive characteristic, the nontrivial purely inseparable 
$\mathbb{A}^{1}$-forms can be used to give examples of nontrivial 
$\mathbb{A}^{n}$-forms for any integer 
$n>1$. However, the problem of the existence of nontrivial separable 
$\mathbb{A}^{3}$-forms over a field is still open in general. A few recent partial results on the triviality of separable 
$\mathbb{A}^{3}$-forms are mentioned in Remark 3.3.
Now let 
$R$ be a ring containing a field 
$k$. An 
$R$-algebra 
$A$ is said to be an 
$\mathbb{A}^{n}$-form over 
$R$ with respect to 
$k$ if 
$A\otimes _{k}\bar{k}=(R\otimes _{k}\bar{k})^{[n]}$, where 
$\bar{k}$ denotes the algebraic closure of 
$k$. In [Reference Dutta10], Dutta investigated separable 
$\mathbb{A}^{1}$-forms over any ring 
$R$ containing a field 
$k$ and obtained Theorem 2.9. He also observed Theorem 2.10 for 
$\mathbb{A}^{2}$-forms over any Principal Ideal Domain (PID) containing 
$\mathbb{Q}$.
In this paper, we prove a partial result on separable 
$\mathbb{A}^{3}$-forms over a field 
$k$ (Theorem 3.2) and extend the results on 
$\mathbb{A}^{2}$-forms (Theorems 2.8 and 2.10) to any one-dimensional Noetherian 
$\mathbb{Q}$-algebra (Theorem 3.7) and to any 
$\mathbb{Q}$-algebra having a fixed point free locally nilpotent derivation (Theorem 3.8). After receiving a preprint of our paper, Prof. M. Miyanishi informed us that a part of Theorem 3.2 has also been obtained recently in [Reference Gurjar, Masuda and Miyanishi13] by a different approach (see Remark 3.3 (4) for a precise statement).
2 Preliminaries
In this section we recall a few definitions and well-known results. All rings will be assumed to be commutative containing unity.
Definition 2.1. An 
$R$-algebra 
$A$ is said to be an 
$\mathbb{A}^{r}$-fibration over 
$R$ if the following hold:
(i)
$A$ is finitely generated over $R$.(ii)
$A$ is flat over $R$.(iii)
for every prime ideal of $R$.
for every prime ideal 
of 
Definition 2.2. Let 
$k$ be a field of characteristic 
$p$
$({\geqslant}0)$ with algebraic closure 
$\bar{k}$ and 
$R$ a 
$k$-algebra. An 
$R$-algebra 
$A$ is said to be an 
$\mathbb{A}^{n}$-form over 
$R$ (with respect to 
$k$) if 
$A\otimes _{k}\bar{k}=(R\otimes _{k}\bar{k})^{[n]}$.
Definition 2.3. Let 
$A=R^{[n]}$ and 
$F\in A$. 
$F$ is said to be a residual coordinate in 
$A$ if, for every prime ideal 
of 
$R$, 
, where 
$\bar{F}$ denotes the image of 
$F$ in 
.
Definition 2.4. A derivation 
$D$ on a ring 
$A$ is said to be a locally nilpotent derivation if, for each 
$a\in A$, there exists an integer 
$n\geqslant 0$ (depending on 
$a$), such that 
$D^{n}(a)=0$.
Definition 2.5. We say that a locally nilpotent derivation 
$D$ on a ring 
$A$ admits a slice if there exists 
$s$ in 
$A$ for which 
$D(s)=1$.
Definition 2.6. A locally nilpotent derivation 
$D$ on a ring 
$A$ is said to be fixed point free if 
$(DA)=A$, where 
$(DA)$ is the ideal of 
$A$ generated by 
$D(A)$.
Definition 2.7. Let 
$R$ be a ring and 
$D$ a locally nilpotent 
$R$-derivation on the polynomial ring 
$A=R^{[n]}$. Then the rank of the derivation 
$D$, denoted by 
$\text{rk}~(D)$, is defined to be the least integer 
$i$ such that there exist 
$X_{1},\ldots ,X_{n-i}\in \text{ker}\,D$ satisfying 
$A=R[X_{1},\ldots ,X_{n-i}]^{[i]}$.
We first state Kambayashi’s theorem [Reference Kambayashi18, Theorem 3] on the triviality of separable 
$\mathbb{A}^{2}$-forms over fields.
Theorem 2.8. Let 
$k$ and 
$L$ be fields such that 
$L$ is separably algebraic over 
$k$. Suppose 
$A$ is a 
$k$-algebra such that 
$A\otimes _{k}L=L^{[2]}$. Then 
$A=k^{[2]}$.
We now state a theorem on separable 
$\mathbb{A}^{1}$-forms over rings and a theorem on 
$\mathbb{A}^{2}$-forms over a PID due to Dutta [Reference Dutta10, Theorem 7 and Remark 8].
Theorem 2.9. Let 
$k$ be a field, 
$L$ a separable field extension of 
$k$, 
$R$ a 
$k$-algebra and 
$A$ an 
$R$-algebra such that 
$A\otimes _{k}L$ is isomorphic to the symmetric algebra of a finitely generated rank one projective module over 
$R\otimes _{k}L$. Then 
$A$ is isomorphic to the symmetric algebra of a finitely generated rank one projective module over 
$R$.
Theorem 2.10. Let 
$k$ be a field of characteristic zero, 
$R$ a PID containing 
$k$ and 
$A$ an 
$R$-algebra such that 
$A$ is an 
$\mathbb{A}^{2}$-form over 
$R$ with respect to 
$k$. Then 
$A=R^{[2]}$.
Next we quote a result on 
$\mathbb{A}^{2}$-fibrations due to Asanuma and Bhatwadekar [Reference Asanuma and Bhatwadekar2, Theorem 3.8 and Remark 3.13].
Theorem 2.11. Let 
$R$ be a one-dimensional Noetherian 
$\mathbb{Q}$-algebra. Let 
$A$ be an 
$\mathbb{A}^{2}$-fibration over 
$R$. Then there exists 
$H\in A$ such that 
$A$ is an 
$\mathbb{A}^{1}$-fibration over 
$R[H]$.
The following result on residual coordinates was proved by Bhatwadekar and Dutta for Noetherian rings containing 
$\mathbb{Q}$ [Reference Bhatwadekar and Dutta5, Theorem 3.2] and later generalized by van den Essen and van Rossum for general 
$\mathbb{Q}$-algebras [Reference van den Essen and van Rossum11, Theorem 3.4].
Theorem 2.12. Let 
$R$ be a 
$\mathbb{Q}$-algebra, 
$A=R^{[2]}$ and 
$F\in A$. If 
$F$ is a residual coordinate in 
$A$ then 
$A=R[F]^{[1]}$.
Next we state a theorem which follows from a fundamental result in the theory of locally nilpotent derivations [Reference Freudenburg12, Corollary 1.26].
Theorem 2.13. Let 
$k$ be a field of characteristic zero, 
$A$ a 
$k$-algebra, 
$D$ a locally nilpotent derivation on 
$A$ and 
$B:=\text{ker}\,D$. Then the following are equivalent:
(1)
$D$ admits a slice $s$.(2)
$A=B[s]=B^{[1]}$ and $D=d/ds$ on $A$.(3)
$D(A)=A$.
The following rigidity theorem is due to Daigle [Reference Daigle8, Theorem 
$2.5$].
Theorem 2.14. Let 
$k$ be a field of characteristic zero and 
$D$ be a locally nilpotent derivation on the polynomial ring 
$A=k^{[3]}$ with 
$\text{rk}\,(D)=2$. Let 
$X,W\in \text{ker}\,D$ be such that 
$A=k[X]^{[2]}=k[W]^{[2]}$. Then 
$k[X]=k[W]$.
The following result on fixed point free locally nilpotent derivations was obtained by Bhatwadekar and Dutta [Reference Bhatwadekar and Dutta6, Theorem 4.7] for any Noetherian 
$\mathbb{Q}$-algebra and later generalized to any 
$\mathbb{Q}$-algebra by Berson et al. [Reference Berson, van den Essen and Maubach4, Theorem 3.5]; [Reference Freudenburg12, Theorem 4.15].
Theorem 2.15. Let 
$R$ be a 
$\mathbb{Q}$-algebra, 
$A=R[X,Y]=R^{[2]}$, 
$D$ a fixed point free locally nilpotent 
$R$-derivation of 
$A$ and 
$B=\text{ker}~D$. Then 
$D$ admits a slice, 
$B=R^{[1]}$ and 
$A=B^{[1]}$.
Remark 2.16. A fixed point free locally nilpotent derivation on 
$k[X,Y,Z]$ has a slice [Reference Kaliman17]. But a fixed point free locally nilpotent 
$R$-derivation on 
$R[X,Y,Z]$ need not have a slice even if 
$R$ is a PID [Reference Bhatwadekar, Gupta and Lokhande7, Example 5.6].
3 Main results
In this section we will prove our main results. Note that if 
$k$ is a field of characteristic zero, 
$A$ a 
$k$-algebra and 
$L$ a field extension of 
$k$, then any 
$k$-linear locally nilpotent derivation 
$D$ on 
$A$ can be extended to a locally nilpotent derivation 
$D\otimes 1_{L}$ on 
$A\otimes _{k}L$ such that 
$(D\otimes 1_{L})(a\otimes \unicode[STIX]{x1D706})=D(a)\otimes \unicode[STIX]{x1D706}$ for all 
$a\in A$ and 
$\unicode[STIX]{x1D706}\in L$. We will first establish our main theorem on 
$\mathbb{A}^{3}$-forms over 
$k$ (Theorem 3.2). We begin with a special case of this result which holds for 
$\mathbb{A}^{3}$-forms over a PID 
$R$ with respect to 
$k$.
Proposition 3.1. Let 
$k$ be a field of characteristic zero with algebraic closure 
$\bar{k}$, 
$R$ a PID containing 
$k$ and 
$A$ be an 
$\mathbb{A}^{3}$-form over 
$R$ with respect to 
$k$. Suppose that there exists an 
$R$-linear locally nilpotent derivation 
$D$ on 
$A$ such that 
$\text{rk}\,(D\otimes 1_{\bar{k}})=1$. Then 
$A=R^{[3]}$.
Proof. Since 
$A$ is an 
$\mathbb{A}^{3}$-form over 
$R$ with respect to 
$k$, there exists a finite extension 
$L$ over 
$k$ such that 
$A\otimes _{k}L=(R\otimes _{k}L)^{[3]}$ and 
$\text{rk}\,(D\otimes _{k}1_{L})=1$. Let 
$B=\text{ker}\,D$. Set 
$\bar{R}:=R\otimes _{k}L$, 
$\bar{A}:=A\otimes _{k}L$, 
$\bar{B}:=B\otimes _{k}L$ and 
$\bar{D}:=D\otimes 1_{L}$. Then 
$\bar{A}=\bar{R}^{[3]}$ and 
$\text{ker}\,\bar{D}=\bar{B}$. Since 
$\text{rk}(\bar{D})=1$, we have 
$\bar{A}=\bar{B}^{[1]}$ and 
$\bar{B}=\bar{R}^{[2]}$. Hence, 
$B=R^{[2]}$ by Theorem 2.10. As 
$\text{Pic}(B)$ is trivial, 
$A=B^{[1]}$ by Theorem 2.9. Thus, 
$A=R^{[3]}$.◻
We now prove our main result on 
$\mathbb{A}^{3}$-forms.
Theorem 3.2. Let 
$k$ be a field of characteristic zero with algebraic closure 
$\bar{k}$ and 
$A$ be an 
$\mathbb{A}^{3}$-form over 
$k$. Suppose that there exists a 
$k$-linear locally nilpotent derivation 
$D$ on 
$A$ such that 
$\text{rk}\,(D\otimes 1_{\bar{k}})\leqslant 2$. Then 
$A=k^{[3]}$.
Proof. Since 
$A$ is an 
$\mathbb{A}^{3}$-form over 
$k$, there exists a finite Galois extension 
$L$ over 
$k$ with Galois group 
$G$ such that 
$A\otimes _{k}L=L^{[3]}$ and 
$\text{rk}\,(D\otimes _{k}1_{L})\leqslant 2$. Let 
$B=\text{ker}\,D$. Set 
$\bar{A}:=A\otimes _{k}L$, 
$\bar{B}:=B\otimes _{k}L$ and 
$\bar{D}:=D\otimes 1_{L}$. Then 
$\bar{A}=L^{[3]}$ and 
$\text{ker}\,\bar{D}=\bar{B}$.
If 
$\text{rk}(\bar{D})=1$, then 
$A=k^{[3]}$ by Proposition 3.1 (taking 
$R=k$).
We now consider the case 
$\text{rk}\,(\bar{D})=2$. We then have 
$X\in \bar{B}$ such that 
$\bar{A}=L[X]^{[2]}$. We show that there exists 
$W\in L[X]\cap A$ such that 
$L[X]=L[W]$.
We identify 
$A$ with its image in 
$\bar{A}$ under the map 
$a\rightarrow a\otimes 1$. Any 
$\unicode[STIX]{x1D70E}\in G$ can be extended to an 
$A$-automorphism of 
$\bar{A}$ by defining 
$\unicode[STIX]{x1D70E}(a\otimes l)=a\otimes \unicode[STIX]{x1D70E}(l)$, for all 
$a\in A$ and 
$l\in L$. Let
$$\begin{eqnarray}X=1\otimes l_{0}+e_{1}\otimes l_{1}+\cdots +e_{r}\otimes l_{r}\end{eqnarray}$$ where 
$1,e_{1},\ldots ,e_{r}$ form a part of a 
$k$-basis of 
$A$ and 
$l_{i}$’s are in 
$L$. Since the bilinear map 
$L\times L\longrightarrow k$ given by 
$(x,y)\mapsto \text{Tr}(xy)$ is nondegenerate (where 
$\text{Tr}(a):=\text{Trace}(a)$ for all 
$a$ in 
$L$), replacing 
$X$ by 
$lX$ (for some 
$l\in L$) if necessary we can assume that 
$\text{Tr}(l_{i})\neq 0$ for some 
$i\geqslant 1$. Thus
$$\begin{eqnarray}W:=\underset{\unicode[STIX]{x1D70E}\in G}{\sum }\unicode[STIX]{x1D70E}(X)=1\otimes \text{Tr}(l_{0})+e_{1}\otimes \text{Tr}(l_{1})+\cdots +e_{r}\otimes \text{Tr}(l_{r})\end{eqnarray}$$ is an element of 
$A\setminus k$. Note that 
$\unicode[STIX]{x1D70E}\bar{D}=\bar{D}\unicode[STIX]{x1D70E}$ and hence 
$\unicode[STIX]{x1D70E}(X)\in \bar{B}$. Since 
$\unicode[STIX]{x1D70E}$ is an automorphism of 
$\bar{A}$, by Theorem 2.14, 
$L[X]=L[\unicode[STIX]{x1D70E}(X)]$. Hence 
$\unicode[STIX]{x1D70E}(X)$ is linear in 
$X$ for each 
$\unicode[STIX]{x1D70E}$ and hence 
$\text{deg}_{X}W\leqslant 1$. But as 
$B\cap L=k$, 
$W\notin L$, so that 
$\text{deg}_{X}W=1$ which implies 
$L[X]=L[W]$.
So 
$\bar{A}=L[W]^{[2]}=(k[W]\otimes _{k}L)^{[2]}$. By Theorem 2.10, we get 
$A=k[W]^{[2]}$.◻
Remark 3.3. Let 
$k$ be a field of characteristic zero with algebraic closure 
$\bar{k}$ and 
$A$ an 
$\mathbb{A}^{3}$-form over 
$k$. We record below a few other results on the triviality of 
$A$.
(1) Daigle and Kaliman have proved [Reference Daigle and Kaliman9, Corollary 3.3] that if 
$A$ admits a fixed point free locally nilpotent derivation 
$D$, then 
$A=k^{[3]}$.
(2) Daigle and Kaliman have also proved [Reference Daigle and Kaliman9, Proposition 4.9] that if 
$A$ contains an element 
$f$ which is a coordinate of 
$A\otimes _{k}\bar{k}$, then 
$A=k^{[3]}$ and 
$f$ is a coordinate of 
$A$.
(3) Koras and Russell have proved [Reference Koras and Russell19, Theorem C] that if 
$A$ admits an effective action of a reductive algebraic 
$k$-group of positive dimension, then 
$A=k^{[3]}$.
(4) Recently, Gurjar, Masuda and Miyanishi have shown [Reference Gurjar, Masuda and Miyanishi13] that 
$A=k^{[3]}$ if 
$A$ admits either a fixed point free locally nilpotent derivation or a nonconfluent action of a unipotent group of dimension two. Their results give an alternative approach to Theorem 3.2 for the case 
$\text{rk}\,(\bar{D})=1$.
We now extend Theorems 2.8 and 2.10 to more general rings. For convenience, we first record a few easy lemmas.
Lemma 3.4. Let 
$R$ be a ring containing 
$\mathbb{Q}$ and 
$A=R^{[2]}$. If 
$H\in A$ is such that 
$A$ is an 
$\mathbb{A}^{1}$-fibration over 
$R[H]$, then 
$A=R[H]^{[1]}$.
Proof. Let 
be a prime ideal of 
$R$ and let 
$\bar{H}$ denote the image of 
$H$ in 
. Then 
is an 
$\mathbb{A}^{1}$-fibration over the PID 
and hence 
. Thus, 
$H$ is a residual coordinate of 
$A$. Hence, by Theorem 2.12, 
$A=R[H]^{[1]}$.◻
We now observe that Theorem 2.8 extends to separable 
$\mathbb{A}^{2}$-forms over a field 
$K$ with respect to a subfield 
$k$.
Lemma 3.5. Let 
$k$ be a field and 
$K$ a field extension of 
$k$. Let 
$A$ be a 
$K$-algebra such that 
$A\otimes _{k}L=(K\otimes _{k}L)^{[2]}$, for some finite separable field extension 
$L$ of 
$k$. Then 
$A=K^{[2]}$.
Proof. By hypothesis, we have 
$A\otimes _{K}(K\otimes _{k}L)=(K\otimes _{k}L)^{[2]}$. Since 
$L$ over 
$k$ is a finite separable extension, 
$K\otimes _{k}L$ is a finite direct product of separable extensions 
$L_{i}$ over 
$K$. Hence, we have 
$A\otimes _{K}L_{i}={L_{i}}^{[2]}$ (for each 
$i$), which implies 
$A=K^{[2]}$ by Theorem 2.8.◻
We now show that 
$\mathbb{A}^{2}$-forms are 
$\mathbb{A}^{2}$-fibrations.
Lemma 3.6. Let 
$k$ be a field of characteristic zero, 
$R$ be a 
$k$-algebra and 
$A$ be an 
$R$-algebra. Let 
$A$ be an 
$\mathbb{A}^{2}$-form over 
$R$ with respect to 
$k$. Then 
$A$ is an 
$\mathbb{A}^{2}$-fibration over 
$R$.
Proof. Let 
$A\otimes _{k}\bar{k}=(R\otimes _{k}\bar{k})[X,Y]$, where 
$\bar{k}$ is an algebraic closure of 
$k$. Let 
$X=\sum _{i=0}^{n}a_{i}\otimes \unicode[STIX]{x1D706}_{i}$ and 
$Y=\sum _{i=0}^{m}b_{i}\otimes \unicode[STIX]{x1D707}_{i}$, where 
$a_{i},b_{i}\in A$ and 
$\unicode[STIX]{x1D706}_{i},\unicode[STIX]{x1D707}_{i}\in \bar{k}$. Then 
$R[a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{m}]\subseteq A$ and the induced map 
$R[a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{m}]\otimes _{k}\bar{k}\longrightarrow A\otimes _{k}\bar{k}$ is an isomorphism. Hence 
$\bar{k}$ being faithfully flat over 
$k$, we have 
$A=R[a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{m}]$. Thus 
$A$ is a finitely generated 
$R$-algebra. Again, as 
$A\otimes _{k}\bar{k}$ is faithfully flat over 
$R\otimes _{k}\bar{k}$ and 
$\bar{k}$ is faithfully flat over 
$k$, 
$A$ is flat over 
$R$. Now it suffices to show 
, for each prime ideal 
of 
$R$.
Let 
be an arbitrary prime ideal of 
$R$. By hypothesis there exists a finite separable extension 
$L$ of 
$k$ such that 
$A\otimes _{k}L=(R\otimes _{k}L)^{[2]}$. Hence, 
. Hence by Lemma 3.5, 
.
Thus, 
$A$ is an 
$\mathbb{A}^{2}$-fibration over 
$R$.◻
We now extend Theorems 2.8 and 2.10 to any one-dimensional Noetherian ring containing a field of characteristic zero.
Theorem 3.7. Let 
$k$ be a field of characteristic zero and 
$R$ a one-dimensional Noetherian 
$k$-algebra. If 
$A$ is an 
$\mathbb{A}^{2}$-form over 
$R$ with respect to 
$k$, then there exists a finitely generated rank one projective 
$R$-module 
$Q$ such that 
$A\cong (\text{Sym}_{R}(Q))^{[1]}$.
Proof. By Lemma 3.6, 
$A$ is an 
$\mathbb{A}^{2}$-fibration over 
$R$ and hence by Theorem 2.11, there exists 
$H\in A$ such that 
$A$ is an 
$\mathbb{A}^{1}$-fibration over 
$R[H]$. Let 
$\bar{k}$ be an algebraic closure of 
$k$, 
$\bar{A}:=A\otimes \bar{k}$ and 
$\bar{R}:=R\otimes \bar{k}$. Since 
$\bar{A}=\bar{R}^{[2]}$ and 
$\bar{A}$ is an 
$\mathbb{A}^{1}$-fibration over 
$\bar{R}[H]$, we have 
$\bar{A}=\bar{R}[H]^{[1]}$ by Lemma 3.4. Thus by Theorem 2.9, 
$A\cong \text{Sym}_{R[H]}(Q_{1})$, for some finitely generated rank one projective 
$R[H]$-module 
$Q_{1}$. Set 
$R_{\text{red}}:=R/nil(R)$. Now
$$\begin{eqnarray}A/nil(R)A\cong \text{Sym}_{R[H]}(Q_{1})\otimes _{R}R_{\text{red}}=\text{Sym}_{R_{\text{red}}[H]}(Q_{1}\otimes _{R}R_{\text{red}}).\end{eqnarray}$$ Now by [Reference Itoh14, Section 2, Lemma 1], there exists a finitely generated rank one projective 
$R_{\text{red}}$-module 
$Q^{\prime }$ such that 
$Q_{1}\otimes _{R}(R_{\text{red}})=Q^{\prime }\otimes _{R_{\text{red}}}R_{\text{red}}[H]$. Thus,
$$\begin{eqnarray}A/nil(R)A\cong \text{Sym}_{R_{\text{red}}}(Q^{\prime })\otimes _{R_{\text{red}}}R_{\text{red}}[H]=(\text{Sym}_{R_{\text{red}}}(Q^{\prime }))^{[1]}.\end{eqnarray}$$ Now by [Reference Ischebeck and Rao15, Proposition 2.3.5], there exists a finitely generated rank one projective 
$R$-module 
$Q$ such that 
$Q\otimes _{R}R_{\text{red}}=Q^{\prime }$ and hence 
$A=(\text{Sym}_{R}(Q))^{[1]}$.◻
The following result shows that under the additional hypothesis that 
$A$ has a fixed point free locally nilpotent 
$R$-derivation, Theorem 3.7 can be extended to any ring containing a field of characteristic zero.
Theorem 3.8. Let 
$k$ be a field of characteristic zero, 
$R$ a ring containing 
$k$ and 
$A$ be an 
$\mathbb{A}^{2}$-form over 
$R$ with respect to 
$k$. Suppose 
$A$ has a fixed point free locally nilpotent 
$R$-derivation. Then there exists a finitely generated rank one projective 
$R$-module 
$Q$ such that 
$A\cong (\text{Sym}_{R}(Q))^{[1]}$.
Proof. Let 
$L$ be a finite extension of 
$k$ such that 
$A\otimes _{k}L=(R\otimes _{k}L)^{[2]}$. Let 
$D$ be a fixed point free locally nilpotent 
$R$-derivation of 
$A$ and 
$B=\text{ker}~D$. Set 
$\bar{R}:=R\otimes _{k}L$, 
$\bar{A}:=A\otimes _{k}L$, 
$\bar{B}:=B\otimes _{k}L$ and 
$\bar{D}:=D\otimes 1_{L}$. Then 
$\bar{A}=\bar{R}^{[2]}$, 
$\text{ker}\,\bar{D}=\bar{B}$ and 
$\bar{D}$ is a fixed point free locally nilpotent derivation of 
$\bar{A}$. Hence, by Theorem 2.15, 
$\bar{D}$ has a slice and 
$\bar{B}=\bar{R}^{[1]}$. Now, by Theorem 2.13, 
$D(\bar{A})=\bar{A}$. Thus, 
$D(A)\otimes _{k}L=D(\bar{A})=\bar{A}=A\otimes _{k}L$. Hence, by faithful flatness of 
$L$ over 
$k$, 
$D(A)=A$. So 
$A=B^{[1]}$ by Theorem 2.13. Since 
$\bar{B}=\bar{R}^{[1]}$, by Theorem 2.9, 
$B=\text{Sym}_{R}(Q)$ and hence 
$A=(\text{Sym}_{R}(Q))^{[1]}$, for some finitely generated rank one projective 
$R$-module 
$Q$.◻
Remark 3.9. Kahoui and Ouali have shown [Reference Kahoui and Ouali16, Corollary 3.2] that when 
$R$ is regular the above result holds for any 
$\mathbb{A}^{2}$-fibration over 
$R$.
Acknowledgments
The authors thank S. M. Bhatwadekar for the current version of Theorem 3.8 and also for his useful comments and suggestions. The second author acknowledges the Department of Science and Technology for their SwarnaJayanti Fellowship. The third author also acknowledges Council of Scientific and Industrial Research (CSIR) for their grant.
