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Consider the curve $C=Z(-x_1^5+x_0^{4}x_2-x_2^5) \subset \mathbb{P}^2$. Dehomogenize with respect to $x_2$ to obtain an irreducible affine curve $U_2=Z(-y^5+x^4-1)$. Notice that $C=U_2 \cup P$ where $P$ is the point at infinity $(1:0:0)$. Define a 1-form on $U_2$: \begin{equation} \omega= \frac{dx}{5y^4}=\frac{dy}{4x^3}. \end{equation} I want to study this differential form at the point $P$. With a change of coordinate $x=v^{-1}$, $y=uv^{-1}$, I need to express $\omega$ at $P$ in terms of $du$ $\textit{only}$ (multiplied by an appropriate non-vanishing rational function, as $u$ is a uniformising parameter in this context, if you are familiar with the notion) but when I try to calculate $du$ and $dv$ from $dx$ and $dy$, I inevitably obtain an expression that contains both $du$ and $dv$ and I don't know how to get rid of the $dv$. I would like to know whether I am miscalculating or there is another way to obtain that expression for $\omega$.

Edit: if I try to write the first form $\omega$ in terms of $u,v$ I get:

\begin{equation} \frac{-(v^2dv)}{5u^{4}} = \frac{-5u^4v^{2} du}{5u^4(1-5v^4)} = \frac{-v^{2} du}{1-5v^4} \end{equation} but this rational function vanishes at $P$ which should not happen.

My end goal is to write $\omega = f u^n \, du$ with $n \in \mathbb{Z}$ and $f$ a rational function with $f(P) \neq 0$.

Viktor Vaughn
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Karl
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  • What is the equation for $U_2$ in terms of $u$ and $v$? This should allow you to eliminate either $du$ or $dv$. – Viktor Vaughn Nov 26 '18 at 15:46
  • Thanks for your reply. $-u^5+v-v^5$ but how does this help me eliminate the $dv$? Should I differentiate it? – Karl Nov 26 '18 at 16:03
  • Yes, differentiate both sides (set it $=0$): this gives you a relation between $du$ and $dv$. – Viktor Vaughn Nov 26 '18 at 16:20
  • $-5u^4du+dv-5v^4dv$ which I would then substitute in the equation of my edit? – Karl Nov 26 '18 at 16:22
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    $5u^4 du = (1-5v^4)dv$, so you can solve for $du$ in terms of $dv$ or vice versa. – Viktor Vaughn Nov 26 '18 at 16:42
  • Ok I think I wrote it in the form you meant now (see edit) but the function I am looking for shouldn't vanish at $P$. Is there something wrong in my calculation? – Karl Nov 26 '18 at 17:17
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    Your calculation looks correct. I even did the computation on a computer using Magma and it says that $\omega$ has a zero of order $10$ at $P$. Why do you say this "should not happen"? – Viktor Vaughn Nov 26 '18 at 19:04
  • Because my aim is to write the last term as $u^n$ times a rational function that does not vanish at $P$. $n$ will then be the order of $\omega$ at $P$. But that expression involves no $u$ $\textit{and}$ it vanishes at $P$ so I don't know how to make it apparent that there is this zero of order 10 – Karl Nov 26 '18 at 19:43
  • I mean the last term except the du - the coefficient of the du basically – Karl Nov 26 '18 at 19:49
  • I don't see immediately how to rewrite $-\frac{v^2}{1 - 5v^4}$ in that form, but you can use the discrete valuation to see the order of vanishing nonetheless. Since $v = u^5 + v^5$ then $v$ has order of vanishing $\geq 5$ at $P$. Since $u^5$ has order of vanishing $5$ and $v^5$ has order of vanishing $\geq 25$, then $v$ also has order of vanishing $5$ by properties of discrete valuations. Thus $v^2$ has order of vanishing $10$. – Viktor Vaughn Nov 26 '18 at 21:09
  • Thank you, I understand. – Karl Nov 26 '18 at 21:28
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    Actually, here's a way to get it in the form you want. It suffices to write $v^2$ in the desired form since $1 - 5v^4$ doesn't vanish at $P$. Since $u^5 = v - v^5$, then $u^{10} = (v-v^5)^2 = v^2 - 2v^6 + v^{10} = v^2(1 - 2v^4 + v^6)$. Thus $v^2 = u^{10}/(1 - 2v^4 + v^6)$, as desired. – Viktor Vaughn Nov 26 '18 at 21:30

1 Answers1

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Here's a summary of the comments.

Rewriting the equations for $U_2$ in terms of $u$ and $v$, we find $$ \left(\frac{u}{v}\right)^5 = \frac{1}{v^4} - 1 \implies u^5 = v - v^5 $$ and thus obtain an affine curve $U: u^5 = v - v^5$. Since $x = X_0/X_2$ and $y = X_1/X_2$, then $$ v = \frac{1}{x} = \frac{X_2}{X_0} \quad \text{and} \quad u = yv = \frac{y}{x} = \frac{X_1}{X_0} $$ so $P = (1:0:0)$ corresponds to $(u,v) = (0,0)$. In $(u,v)$ coordinates, $\omega$ transforms into $$ \omega = \frac{dx}{5y^4} = \frac{d(1/v)}{5(u/v)^4} = \frac{-\frac{1}{v^2} dv}{5 \frac{u^4}{v^4}} = -\frac{v^2}{5 u^4} dv $$ Differentiating the equation for $U$, we find $5u^4 du = dv - 5v^4 dv = (1-5v^4)dv$, so $$ dv = \frac{5u^4}{1 - 5 v^4} du \, . $$ Thus $$ \omega = -\frac{v^2}{5 u^4} \frac{5u^4}{1 - 5 v^4} du = -\frac{v^2}{1 - 5 v^4} du \, . $$ Let $\DeclareMathOperator{\m}{\mathfrak{m}} \m = (u,v)$ be the maximal ideal of the local ring at $P$ (so $\m = (u)$ since $u$ is a uniformizer at $P$), and let $\newcommand{\ord}{\operatorname{ord}_P} \ord$ be the associated discrete valuation (the order of vanishing). Since $v = u^5 + v^5$, we see that $v \in \m^5$, so $\ord(v) \geq 5$. Then $\ord(v^5) \geq 25$ and since $\ord(u^5) = 5$, then $\ord(u^5) \neq \ord(v^5)$, and hence $$ \ord(v) = \min\{\ord(u^5), \ord(v^5)\} = 5 $$ by properties of discrete valuations (see here). Thus $v^2$ has valuation $10$, so $v^2$ and $u^{10}$ differ multiplicatively by a unit. To find this unit, we square the equation for $U$: \begin{align*} u^{10} = (v - v^5)^2 = v^2 - 2v^6 + v^{10} = v^2(1 - 2v^4 + v^8) \end{align*} and find that $v^2 = \frac{u^{10}}{1 - 2v^4 + v^8}$. Thus $$ \omega = -\frac{v^2}{1 - 5 v^4} du = -\frac{\frac{u^{10}}{1 - 2v^4 + v^8}}{1 - 5 v^4} du = -\frac{1}{(1 - 2v^4 + v^8)(1 - 5 v^4)} u^{10} \, du $$ as desired.

Viktor Vaughn
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