I mean clearly one uses the isomorphism $\phi$ that sends to $x$ to $x$ and $y$ to $\frac{1}{x}$. And also clearly is $(xy-1)\subseteq\ker(\phi)$. I just struggle to prove the other inclusion.
Can you guys maybe help me?
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1What is your definition/construction of $k[x, 1/x]$? – Jendrik Stelzner Apr 30 '16 at 13:37
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What do you mean? – Chris B. Apr 30 '16 at 13:40
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1What Jendrik means is that the most common way to define the relation between the elements $x$ and $\frac1x$ is exactly by this quotient, i.e. that $x\cdot\frac1x-1=0$. – Arthur Apr 30 '16 at 13:44
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Ohh ok. Didn't know that, well i mean it more or less as a polynomial ring in the variables $x$ and $x^{-1}$ (If that makes sense) – Chris B. Apr 30 '16 at 13:46
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So we have a polynomial $$p(x,y)=\sum_{j,k\ge0}a_{j,k}x^jy^k$$such that $$p(x,1/x)=0,$$and we want to show that there exists a polynomial $r(x,y)$ with $$p(x,y)=(1-xy)r(x,y).$$
In general $$p(x,1/x)=\sum_{n\in\Bbb Z}b_nx^n,$$where $$b_n=\sum_{j-k=n}a_{j,k}.$$Since $p(x,1/x)=0$ we have $$\sum_{j-k=n}a_{j,k}=0$$for every $n\in\Bbb Z$. Let $$p_n(x,y)=\sum_{j-k=n}a_{j,k}x^jy^k.$$
Assume first that $n\ge0$. Then we have $$p_n(x,y)=x^n\sum_{j-k=n}a_{j,k}(xy)^k=x^nq_n(xy),$$where $$q_n(t)=\sum_{j-k=n}a_{j,k}t^k.$$Now $q_n(1)=0$, so there exists a polynomial $r_n(t)$with $$q_n(t)=(1-t)r_n(t).$$Hence $$p_n(x,y)=(1-xy)x^nr_n(xy).$$
We've shown that $1-xy$ divides $p_n(x,y)$ for $n\ge0$. The proof for $n<0$ is similar, except factoring out $y^{-n}$ from $p_n$ instead of $x^n$.
David C. Ullrich
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@ChrisB. Maybe not quite trivial, but not all that hard; after all I figured it out without knowing any algebra to speak of. Close to straightforward if you think of it this way: (i) figure out exactly what $p(x,1/x)=0$ says about the coefficients of $p$. (Did you do that? If not you could have and should have.) (ii) Then the answer to (i) makes splitting $p=\sum p_n$ fairly natural; now we just have to show that $1-xy$ divides each $p_n$. (iii) Again, look carefully at $p_n$; notice that it's really a monomial times some polynomial in $xy$ and you're done. – David C. Ullrich Apr 30 '16 at 14:10
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People should give some reason for downvotes. If there's an error above I'd like to know about it. Maybe it's just inappropriately elementary or detailed? – David C. Ullrich Apr 30 '16 at 22:06
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For any commutative ring $k$, the ring $k[x; 1/x]$ is characterized by the following "mapping property": for any other ring $R$ containing $k$ and a unit $u \in R$, there is a homomorphism $\psi: k[x; 1/x] \to R$ such that $\psi(a) = a$ for $a \in k$ and $\psi(x) = u$. We can apply this to $k[x,y]/(xy-1)$: since $k \cap (xy-1)$ is the trivial ideal, $k$ is isomorphic to its image in the quotient ring $k[x,y]/(xy-1)$. Further, $x$ is a unit there, with inverse $y$. So construct a homomorphism $\psi: k[x;1/x] \to k[x,y]/(xy-1)$ by mapping $\psi(a)=a$ for $a \in k$ and $\psi(x) = x$. This will be an inverse to your homomorphism.