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Let $K$ be a field and $K[x,x^{-1}] \subset K(x)$ the ring of Laurent polynomials.

How to show that $K[x,x^{-1}] \cong K[x,y]/(xy-1)$?

My idea was:

Firstly, I noticed that:

Let $\varphi: K \to K[x,x^{-1}]$ be a ring homomorphism and $\sigma: \mathbb{Z} \to K[x,x^{-1}], k \mapsto x^k$. So $K[\mathbb{Z}] \cong K[x,x^{-1}]$. Since it's an equivalence relation it has to be shown that

$\Phi:K[x,y]\to K[x,x^{-1}], \sum \limits_{(m,n)\in \mathbb{N^2}}^{< \infty} a_{m,n}x^my^n \mapsto \sum \limits_{(m,n)\in \mathbb{N^2}}^{< \infty} a_{(m,n)}x^my^{-n}$ is a ring epimorphism with $\mathrm{ker}(\Phi)=(xy-1) \vartriangleleft K[x,y]$.

Now I don't know what to do next.

Is this way correct?

Or is there another possibility to show the isomorphism?

KReiser
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Tartulop
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1 Answers1

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I think doing the map $\Phi:k[x,y]\to k[t,t^{-1}]$ via $x\mapsto t$ and $y\mapsto t^{-1}$ is a good idea. Note that $\Phi$ is automatically a $k$-algebra homomorphism (in particular, this is a ring homomorphism) because $x,y$ are algebraically independent and they generate the domain ring. That is, once we specify where the generators of the domain (i.e. $x$ and $y$) are sent, then it extends to a $k$-algebra homomorphism (or just a ring homomorphism, if you like).

Then you just show that $\ker(\Phi)=\langle xy-1\rangle$ and that the map is surjective, and then you're done by first isomorphism theorem. Surjectivity is immediate, and the kernel computation isn't so bad.

Dave
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    Alternatively, you could define the inverse map similarly (depending on the exact definition of $k[x,x^{-1}]$ of course), via $x^n \mapsto x^n$ if $n \ge 0$ or $x^n \mapsto y^{-n}$ if $n < 0$. (Though showing this is a homomorphism probably involves several cases to show it works on products of monomials, depending on the signs of the exponents and the sign of the sum.) And then, you get the proof of the kernel identification pretty much automatically. – Daniel Schepler Apr 13 '22 at 18:08
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    Or, I guess if the definition of $k[x, x^{-1}]$ is fairly canonically isomorphic to $k[\mathbb{Z}]$, I guess you could also use the description of $\mathbb{Z}$ as the Grothendieck group of $\mathbb{N}$ (i.e. equivalence classes of $\mathbb{N} \times \mathbb{N}$ under $(a,b) \sim (c,d) \overset{def}{\leftrightarrow} a+d=b+c$). – Daniel Schepler Apr 13 '22 at 18:10
  • It's clear that $k[x,x^{-1}]=k[x]_x$, so one could use the universal property of localization to explicitly define the inverse map. – Dave Apr 13 '22 at 18:21