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Let $k$ be an algebraically closed field and let$c,d$ be distinct elements of $k$.

Why there is no ring isomorphism between $k[x,\frac{1}{x}]$ and $k[x,(x-c)^{-1},(x-d)^{-1}]$?

I guess one approach is to look at the units, however (just for curiosity) I would like to see if there are other ways of answering this (i.e using tools from commutative algebra or any sort)

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2 Answers

up vote 4 down vote accepted

First note that $ k[x,(x-c)^{-1}, (x-d)^{-1}] =k[x-c, (x-c)^{-1}, (x-d)^{-1}]$ so that we can reduce to showing that $A=k[x, x^{-1}]$ is not isomorphic to $B=k[x, x^{-1},(x-d)^{-1}] $ if $d\neq 0$.

Translating into geometry, this is equivalent to showing that $X=Spec(A)=\mathbb P^1_k \setminus \lbrace P,Q \rbrace $ is not isomorphic to $Y=Spec(B)=\mathbb P^1_k \setminus \lbrace P,Q,R \rbrace $ .
But if there were such an isomorphism $f:X\to Y$ it would extend to a morphism $\hat f: \mathbb P^1_k \to \mathbb P^1_k$ because $X$ is nonsingular: Hartshorne, Chapter I, Proposition 6.8.
This is absurd because any non-constant morphism $\mathbb P^1_k \to \mathbb P^1_k$ must be surjective and $\hat f$ cannot be surjective: the two points $\hat f(P)$ and $\hat f(Q)$ cannot fill the three holes $P,Q,R$ that $Y$ has with respect to $\mathbb P^1_k$ !

[The little argument above is actually just a rigorous proof of the visually obvious fact that $\mathbb P^1_k$ minus two points is not isomorphic to $\mathbb P^1_k$ minus three points!]

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If $k$ is an integral domain, then $k[x,x^{-1}]^*/k^*$ is isomorphic to $\mathbb{Z}$, generated by $x$, and $k[x,(x-c)^{-1},(x-d)^{-1}]^* / k^*$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$, generated by $x-c$ and $x-d$. These groups are not isomorphic.

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