On cancellation of variables of the form $bT^{n}-a$
over affine normal domains

Let $\sigma$ be the $k$-automorphism of $B$ induced by the $k$-automorphism $\tilde{\sigma}$ of $A[T]$ defined by $\tilde{\sigma}(T)=\omega T$ where $\omega$ is a primitive $n^{th}$ root of unit. Obviously, $\sigma$ has order $n$.

Since $B=k^{[2]}$, by Lemma 2.10 there exist variables $U^{\prime},V^{\prime}\in B$ and $\alpha,\ \beta\in k^{*}$ such that $\sigma(U^{\prime})=\alpha U^{\prime}$ and $\sigma(V^{\prime})=\beta V^{\prime}$ where $\alpha^{n}=\beta^{n}=1$. Let $t$ be the image of $T$ in $B$ and $C=A[a/b]$. Then $t^{n}=a/b$ and $B=A[t]=C[t]=C\oplus tC\oplus t^{2}C\oplus\cdots\oplus t^{n-1}C$. Observe that for all $x\in B,\ t\mid(x-\sigma(x))$. Thus $t\mid(1-\alpha)U^{\prime}$ and $t\mid(1-\beta)V^{\prime}$. Without loss of generality we may assume that $\alpha=1$ and hence we get that $V^{\prime}$ is a unit multiple of $t$ and the ring of invariant of $\sigma$ is $C=A[a/b]=k[U^{\prime},a/b]=k^{[2]}$.

Set $Y:=t$. We shall show that there exists $X\in A$ such that $B=k[X,Y]$, $b\in k[X]$ and $A=k[X,a]=k^{[2]}$.

If $b\in k^{*}$, then $A=A[a/b]=k[U^{\prime},a/b]$. Then setting $X:=U^{\prime}$ we have $A=k[X,a]$, $B=k[X,Y]$ and $C=k[X,a/b]$. Now, let $b\notin k^{*}$. Suppose $p_{1},p_{2},\cdots,p_{m}$ be distinct irreducible factors of $b$ in $C=k^{[2]}$. We shall show that $p_{i}$’s are pair wise comaximal.

Let $\mathfrak{p}_{i}=A\cap p_{i}C$. Since $a,b\in A\cap bC\subseteq\mathfrak{p}_{i}$, ht$(\mathfrak{p}_{i})>1$ for all $i=1,2,\cdots,m$. Since dim$(A)=2$, each $\mathfrak{p}_{i}$ is a maximal ideal of $A$. Let $\bar{k}$ denote an algebraic closure of $k$, $L_{i}$ be a subfield of $\bar{k}$ isomorphic to $A/\mathfrak{p}_{i}$ and let $L$ be the subfield of $\bar{k}$ generated by the fields $L_{1},L_{2},\dots,L_{m}$. Then $L_{i}$ is an algebraic extension of $k$ and $C/\mathfrak{p}_{i}C=(A/\mathfrak{p}_{i})[\zeta_{i}]=L_{i}[\zeta_{i}]$ where $\zeta_{i}$ is the image of $a/b$ in $C/\mathfrak{p}_{i}C$. Since $\mathfrak{p}_{i}C\subseteq p_{i}C$, it follows that $\zeta_{i}$ is transcendental over $L_{i}$ (otherwise $p_{i}\mid a$ and hence $p_{i}$ is a non-zero divisor in $C$ which is an integral domain, a contradiction) and $\mathfrak{p}_{i}C$ is a prime ideal of $C$. As ht$\ p_{i}C=1$ and $\mathfrak{p}_{i}C\neq 0$, we have $p_{i}C=\mathfrak{p}_{i}C$. This shows that $p_{i}$ are pairwise comaximal in $C$ and hence in $B$.

Let $g(\zeta_{i})$ be the image of $U^{\prime}$ in $C/p_{i}C=L_{i}[\zeta_{i}]$. Then $U^{\prime}-g(a/b)$ is divisible by $p_{i}$ in $C\otimes_{k}L_{i}$. But $U^{\prime}-g(a/b)=U^{\prime}-g(Y^{n})$ is a variable in both $C\otimes_{k}L_{i}$ and $B\otimes_{k}L_{i}$. Hence $U^{\prime}-g(a/b)$ is a constant multiple of $p_{i}$. Thus $C\otimes_{k}L_{i}=L_{i}[p_{i},a/b]$, $B\otimes_{k}L_{i}=L_{i}[p_{i},Y]$, and for $i\neq j$, $(p_{i},p_{j})B\otimes_{k}L=B\otimes_{k}L$. Set $X:=p_{1}$. Using Lemma 2.9, we have $p_{i}=\lambda_{i}X+\pi_{i}$ for $\lambda_{i}\in L^{*}$ and $\pi_{i}\in L$. So, we have $b\in L[X]$. This shows that $X$ is integral over $A\otimes_{k}L$ and hence over $A$. As $X\in A[a/b]$ and $A$ is a normal domain, we have $X\in A$. Since $L|_{k}$ is faithfully flat, it follows that $B=k[X,Y]$ with $X\in A,Y=t$ and $b\in k[X]$; and $C=k[X,a/b]$.

Now, to complete the proof, we are only left to show that $A=k[X,a]$. First we claim that $bA\cap k[X,a]=bk[X,a]$. We repeat the argument in ([25], pg. 98) to prove this claim. Let $h\in bA\cap k[X,a]$. Then

Since $a\in bC$, it follows that $h_{0}(X)\in bC$. But since $C=k[X,a/b](=k^{[2]})$, we get $h_{0}(X)\in bk[X]$ and hence $h_{0}(X)\in bk[X,a]$. So, we may replace $h$ by $h-h_{0}(X)=h_{1}(X)a+h_{2}(X)a^{2}+\cdots+h_{d}(X)a^{d}$. Let $h^{\prime}=h_{1}(X)+h_{2}(X)a+\cdots+h_{d}(X)a^{d-1}$. Then $h=h^{\prime}a$. Since there is no height one prime ideal of $A$ which contains both $a$ and $b$, and since $h^{\prime}a\in bA$, it follows from the normality of $A$ that (the associative prime ideals of $a$ are of height one) $\displaystyle h^{\prime}/b\in\underset{\mathfrak{p}\in\ \text{Spec}(A);\ ht(\mathfrak{p})=1}{\overset{}{\bigcap}}A_{\mathfrak{p}}=A$. Therefore, $h^{\prime}\in bA$. Now we argue as before that $h_{1}(X)\in bk[X,a]$. We continue this process to conclude that $h_{i}(X)\in bk[X,a]$ for $0\leq i\leq d$, which proves the claim. Now, since $b\in k[X,a]$ is a non-zero element, applying Lemma 2.8 we have $A=k[X,a]=k^{[2]}$.

Thus, if $k$ contains all the $n^{th}$ roots of unity, then there exist variables $X,Y$ in $B$ such that $Y$ is the image of $T$ in $B$, $b\in k[X]$, $A=k[X,a]=k^{[2]}$ and $A[T]=k[X,bT^{n}-a,T]$.

Now we take the other case.

Let $E$ be the field obtained by adjoining all the $n^{th}$ roots of unity to $k$ and let $g=bT^{n}-a$. Since $p\nmid n$, $E$ is a Galois extension over $k$. By Case - I, we get variables $X^{\prime}$ and $Y^{\prime}$ of $B\otimes_{k}E$ ($=(A\otimes_{k}E)[T]/(g)=E^{[2]}$) such that $Y^{\prime}$ is the image of $T$, $b\in E[X^{\prime}]$ and $A\otimes_{k}E=E[X^{\prime},a]$. As $E|_{k}$ is separable, we have $A={k[a]}^{[1]}$ by Theorem 2.7. If $b\in A\backslash k$, then, by Lemma 2.11, we get $X\in A={k[a]}^{[1]}=k^{[2]}$ such that $A=k[X]^{[1]}$, $b\in k[X]$ and $E[X]=E[X^{\prime}]$. Since $E|_{k}$ is faithfully flat, $E[X^{\prime},a]=E[X,a]$ and $k[X,a]\subseteq A$, we have $A=k[X,a]$. If $b\in k$, then we choose $X$ to be any complementary variable of $a$ in $A$. Now $A[T]=k[X,a,T]=k[X,bT^{n}-a,T]$, as $b\in k[X]$. This completes the proof. ∎

Fix $P\in$ Spec($R$). Letting $\overline{a}$ and $\overline{b}$ respectively denote the images of $a$ and $b$ in $A\otimes_{R}k(P)$, we have $\displaystyle{B\otimes_{R}k(P)=\frac{(A\otimes_{R}k(P))[T]}{(\overline{b}T^{n}-\overline{a})}=k(P)^{[2]}}$. Then since $A\otimes_{R}k(P)$ is a normal affine $k(P)$-domain, by Theorem 3.13, there exist variables $X_{P},Y_{P}$ in $B\otimes_{R}k(P)$ such that $Y_{P}$ is the image of $T$ in $B\otimes_{R}k(P)$, $X_{P}\in A\otimes_{R}k(P)$, $A\otimes_{R}k(P)=k(P)[X_{P},\overline{a}]=k(P)^{[2]}$ and $A[T]\otimes_{R}k(P)=k(P)[T,\overline{b}T^{n}-\overline{a}]^{[1]}=k(P)^{[3]}$.

This shows that $a$ is a residual variable of $A$ over $R$ and $(bT^{n}-a,T)$ is a pair of residual variables of $A[T]$ over $R$. Since $R$ is a Noetherian domain, and since $A$ is finitely generated flat $R$-algebra, by Theorem 2.12 we have $A=R[a]^{[1]}=R^{[2]}$ and $A[T]=R[bT^{n}-a,T]^{[1]}$, if either $\Omega_{R}(A)$ is a stably free module or $R$ is a factorial domain. This completes the proof. ∎

Note that since $A[T]=R^{[3]}$, $A$ is a finitely generated flat residually factorial $R$-domain; and by ([10], Lemma 2.1), $\Omega_{R}(A)$ is a stably free $A$-module.

Fix a prime ideal $P$ of $R$. If $b\notin PA_{P}$, then applying Theorem 3.13, we have $A\otimes_{R}k(P)=(R[a]\otimes_{R}k(P))^{[1]}$ and $A[T]\otimes_{R}k(P)=(R[bT^{n}-a,T]\otimes_{R}k(P))^{[1]}$. If $b\in PA_{P}$, then $(A\otimes_{R}k(P))^{[1]}=(R[bT^{n}-a]\otimes_{R}k(P))^{[2]}=(R[a]\otimes_{R}k(P))^{[2]}$. By a result of Abhyankar-Eakin-Heinzer ([1], 4.5) we have $(A\otimes_{R}k(P))=(R[a]\otimes_{R}k(P))^{[1]}$, and hence $A[T]\otimes_{R}k(P)=k(P)[\bar{a},T]^{[1]}=k(P)[\bar{b}T^{n}-\bar{a},T]^{[1]}=(R[bT^{n}-a,T]\otimes_{R}k(P))^{[1]}$. Since $P\in$ Spec($R$) is arbitrary, using Theorem 2.12, we get $A=R[a]^{[1]}=R^{[2]}$ and $A[T]=R[bT^{n}-a,T]^{[1]}$. This completes the proof. ∎