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Throughout, let $k$ be an algebraically closed field. For two $k$-algebras $A,B$ let us write $A \cong_k B$ to mean that $A,B$ are isomorphic as $k$-algebras.

It is known that if $A$ is an integral domain which is finitely generated $k$-algebra and such that $A[T] \cong_k k[X_1,...,X_n] $ , then $A \cong_k k[X_1,...,X_{n-1}]$ as $k$-algebras, when $2\le n \le 3$.

A proof for $n=2$ case can be fond in On the Uniqueness of the Coefficient Ring in a Polynomial Ring

and a proof of $n=3$ case can be found in An algebraic proof of a cancellation theorem for surfaces.

My questions are :

Are similar cancellation theorems known for Laurent polynomial or power series rings ? i.e. let $A$ be an integral domain which is a finitely generated $k$-algebra , such that $A[T,T^{-1}] \cong_k k[X_i, X_i^{-1} : 1 \le i \le n]$ (resp. $A[[T]] \cong_k k[[X_1,...,X_n]]$ ) ; then when can we say that $A\cong_k k[X_i, X_i^{-1} : 1 \le i \le n-1]$ (resp. $A \cong_k k[[X_1,...,X_{n-1}]] $) ?

Any reference towards these questions will also be very appreciated.

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  • $\begingroup$ Possibly related for power series : math.stackexchange.com/questions/2747943 $\endgroup$ – Watson May 12 '18 at 9:32
  • $\begingroup$ @Watson: Thanks ... I will take a look ... at first glance the content is slightly different since I consider $k$-algebra isomorphisms, still thanks for the reference $\endgroup$ – user111524 May 12 '18 at 15:40
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The Laurent polynomial ring is fairly trivial. If $n=1$, clearly $A=k$ and if $n=2$, $A$ is rational and smooth, so $A=k[u, f(u)^{-1}]$ and the rest should be clear. So, let us assume $n\geq 3$.

Under the isomorphism, $T=\alpha X_1^{a_1}\cdots X_n^{a_n}$ for some $\alpha\in k^*$ and $a_i\in\mathbb{Z}$. One first checks that the gcd of the $a_i$'s is one. If not, $T=\alpha z^d, z\in A[T,T^{-1}]^*$, $d>1$. But, the unit group of $A[T,T^{-1}]$ is $A^*\times\mathbb{Z} T$. Easy to see that $d$ has to be one.

Since $n\geq 3$, we can find integers $b_i, 2\leq i\leq n$ so that $a_1+\sum_{i=2}^n a_ib_i$=1. Consider the automorphism $k[X_i, X_i^{-1}1\leq i\leq n]$ given by $X_1\mapsto X_1$, $X_i\mapsto X_iX_1^{b_i}$ for $i>1$. Composing, we see that $T\mapsto \alpha X_1^{a_1}X_2^{b_2}X_1^{a_2b_2}\cdots X_n^{b_n}X_1^{a_nb_n}= \alpha X_1X_2^{b_2}\cdots X_n^{b_n}$. Now, another easy automorphism will arrange it so that $T\mapsto X_1$. Then it is clear that $A\cong k[X_i,X_i^{-1}, 1<i\leq n]$.

Edit: This is to address OP's question below about the case $n=2$. If $n=2$, then $\dim A=1$ and since its fraction field is contained in a purely transcendental extension, by (variations of) Luroth, the fraction field must be purely transcendental. Since $A$ is an affine algebra over an algebraically closed field which is integrally closed, this implies $A\cong k[u, f(u)^{-1}]$ for some $f(u)$. But, $A^*=k^*\times\mathbb{Z}$ and the units of the right side is isomorphic to $k^*\times\mathbb{Z}^r$ where $r$ is the number of distinct roots of $f(u)$. Since $r=1$, this says $f(u)$ must have exactly one root and the rest is clear, I hope.

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  • $\begingroup$ Could you please elaborate on the $n=2$ case ? How did you arrive at $A=k[u,f(u)^{-1}]$ (what is $f$ ) ? What do you mean by $A$ is rational and smooth ? $\endgroup$ – user111524 May 11 '18 at 15:32
  • $\begingroup$ @users Have added an edit. $\endgroup$ – Mohan May 11 '18 at 16:20

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