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Let $k$ be a field. Given two sets of elements of $k$, $\{a_1,\ldots,a_n\}$ and $\{b_1,\ldots,b_n\}$ are the $k$-algebras $k[x,(x-a_1)^{-1},\ldots,(x-a_n)^{-1}]$ and $k[x,(x-b_1)^{-1},\ldots,(x-b_n)^{-1}]$ always isomorphic?

For $n=1$ the statement is true as we have an isomorphism $k[x,(x-a)^{-1}]\cong k[x,x^{-1}]$ for all $a\in k$ given by $x\mapsto x+a$.

For $n=2$, by the above argument any such ring is isomorphic to $k[x,x^{-1},(x-a)^{-1}]$ for some $a\in k$. Any two such rings are isomorphic by scalar multiplication, i.e, $x\mapsto ab^{-1}x$ gives an isomorphism $k[x,x^{-1},(x-a)^{-1}]\cong k[x,x^{-1},(x-b)^{-1}]$.

I think this should generalise to arbitrary $n$ by induction but I can't seem to find the right map $k[x,x^{-1},(x-a_1)^{-1},\ldots,(x-a_n)^{-1}]\cong k[x,x^{-1},(x-a_1)^{-1},\ldots,(x-b_n)^{-1}]$ where only the last term is different.

I was initially interested in the version of this question for polynomials in many variables but then realized I couldn't completely prove the single variable case.

hm2020
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Arun Kumar
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    Welcome back, returning contributor - I can see it's been a while since you asked a question here on MSE. At present, this question is only the statement of the problem, which does not meet our standards here. Please see here on Meta for some ways to improve your question. (I think there is some fun math to be done here, but you'll need to improve your question before you can get an answer.) – KReiser Apr 04 '25 at 18:50
  • Yeah, please add what you know about this question, and how did you run into it. Such as, which values of $n$ have you seriously looked at. What KReiser said! – Jyrki Lahtonen Apr 04 '25 at 19:40
  • Apologies, I forgot the rules (it's been a while). I have added details on how much I have tried. – Arun Kumar Apr 05 '25 at 04:46
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    Thank you for adding that, I think it helps your question quite a bit. – KReiser Apr 05 '25 at 04:53

1 Answers1

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No, these rings are not uniquely determined by $n$ unless $n\leq 2$.

Viewing these rings as the coordinate algebras of $\Bbb P^1\setminus \{a_1,\cdots,a_n,\infty\}$ and $\Bbb P^1\setminus \{b_1,\cdots,b_n,\infty\}$, an isomorphism of coordinate algebras induces an isomorphism of these varieties, and such an isomorphism extends uniquely to an automorphism of $\Bbb P^1$ by the curve-to-projective extension theorem. Such an automorphism of $\Bbb P^1$ must interchange the sets $\{a_1,\cdots,a_n,\infty\}$ and $\{b_1,\cdots,b_n,\infty\}$. But the automorphism group of $\Bbb P^1$ is $PGL(2)$, which is only three-dimensional, whereas a choice of $n+1$ distinct points in $\Bbb P^1$ lives in an open set $U$ of $(\Bbb P^1)^{n+1}$, which is of dimension $n+1$. As soon as $n+1>3$, the orbit of one group of $n+1$ distinct points under the $PGL(2)$ action is too small to be the whole of $U$. (On the other hand, the natural action of $PGL(2)$ on $\Bbb P^1$ is three-transitive, confirming your work that the ring is uniquely determined for $n\leq 2$.)

Here is one explicit example in the case when $n=3$. Choose $a_1=0$, $a_2=1$, $a_3=2$, and $b_1=0$, $b_2=1$, $b_3=3$. If there was an element of $PGL(2)$ which moved one of these to the other, then as elements of $PGL(2)$ preserve the cross ratio, we must have that for some ordering of each of our points, their cross ratios are equal. Depending on the order we take the cross-ratio in, there are six possible values: if the value in one specific order is $\lambda$, the other potential cross-ratios are $\frac{1}{\lambda}$, $1-\lambda$, $\frac{1}{1-\lambda}$, $\frac{\lambda-1}{\lambda}$, and $\frac{\lambda}{\lambda-1}$. For the $a$ in the order $\infty,0,1,2$ we get $\lambda = 2$. For the $b$ in the order $\infty,0,1,3$ we get $\lambda = 3$. But none of $\frac{1}{\lambda}=\frac12$, $1-\lambda=-1$, $\frac{1}{1-\lambda}=-1$, or $\frac{\lambda-1}{\lambda}=\frac12$ are equal to 3. (Alternatively, if you know a little bit about elliptic curves, this is the verification that $j(2)\neq j(3)$ where $j$ is the $j$-invariant.)

KReiser
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