Let $k$ be a field of characteristic different from $2$, and $A=k[x,y]/(y^2-x^2(x+1))$. Let $\hat A$ be the $(x,y)A$-adic completion. How can I show that $\hat A\simeq k[[u,v]]/(uv)$?
Qing Liu: Algebraic Geometry and Arithmetic Curves, ex. 1.3.10.
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Actually $\hat A=k[[x,y]]/(y^2-x^2(x+1))$. Now notice that there is $f(x)\in k[[x]]$ such that $x+1=f(x)^2$ (here you need the characteristic $\ne 2$). So, $y^2-x^2(x+1)=y^2-(xf(x))^2$, that is, $y^2-x^2(x+1)=(y-xf(x))(y+xf(x))$. Now just perform a change of variables $u=y-xf(x)$, $v=y+xf(x)$ (here you also need the characteristic $\ne 2$).
user26857
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I just believe the accepting measures what is the best answer in my opinion, and up/downvoting means how most mathematicians thinks the usefulness of the answer. I'm more interested about learning things than thinking whether some others thinks the answers are useful. Maybe I should try to learn to change my habit. – Jaakko Seppälä Feb 27 '15 at 22:01
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Considerer the map $\varphi : k[[u,v]]/(uv) \to \widehat{A}$ defined by sending (the image of) $u$ to $y-x\alpha$ and (the image of) $v$ to $y-x\alpha$ where $\alpha$ is an element of $\widehat{A}$ such that $\alpha^2$ is equal to the image of $x^2(1+x)$. To show that such an $\alpha$ exists, it suffices to show that the image of $1+x$ is a square in $\widehat{A}$, which amounts to show that $1+x$ is a square in $k[[x,y]]$. Show that $\varphi$ is an isomorphism.
Olórin
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