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Let $Y$ be the affine plane curve given by the equation $y^2=F(x)$, where $F$ is a polynomial in one variable of odd degree over a field of characteristic not equal to 2. Let $\xi\in Y$.

  1. Suppose $y(\xi)\neq0$. Then why is $x$ a local parameter at $\xi$?
  2. Suppose $y(\xi)=0$. Then why is $y$ a local parameter at $\xi$?
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Are you missing an assumption on distinct roots? I don't think the second claim is true when $F(x)=0$ has roots with nontrivial multiplicity. – Alon Amit May 5 '12 at 7:03
Yes, exactly as you pointed out, $F$ should not have any repeated roots. – Alan Lee May 5 '12 at 7:30
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Put $G(x,y)=y^2-F(x)$. Now if $y(\xi)\ne 0$, then $\partial G/\partial y\ne0$ at $\xi$, so $x$ is a local parameter at $\xi$ by the implicit function theorem ($y$ can be locally eliminated from $G(x,y)=0$). If $y(\xi)=0$, then $\partial G/\partial x=-dF/dx\ne0$ at $\xi$ since $x$ is not a multiple root of $F$. Hence, $x$ can be eliminated and $y$ is a local parameter.

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I agree with the elimination arguments using the Implicit Function Theorem, but won't this procedure produce functions that are not rational (maybe we have to take radicals somewhere)? – Alan Lee May 15 '12 at 17:42

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