I'd like to ask remark 1.1 in chapter 2 of Silverman's "Arithmetic of elliptic curves".

Let $K$ be a field and $C$ be a curve with a smooth point $P$, after proving $\bar{K}[C]_P$ is a discrete valuation ring in proposition 1.1, Silverman mentions that if $P\in C(K)$, then $K(C)$ contains uniformizers for $P$. In another word, there are uniformizers that are defined over K at $P$.

I want to know how to prove this and if there are some references of algebraic curves in Silverman's style in chapter II.


  • $\begingroup$ I don't typically think about non-algebraically closed fields, but I think the idea is that if $P$ is defined over $K$, then evaluation is defined, so we have a surjective map $\mathcal O_{C,P} \to K$ given by $f \mapsto f(P)$. Its kernel is a maximal ideal, any generator of which is a uniformizer. $\endgroup$ Dec 2, 2020 at 2:00
  • 1
    $\begingroup$ @TabesBridges. Does $\mathcal O_{C,P}=\bar{K}[C]_p$? I have studied some algebraic geometry, but when I see this chapter of Silverman, I am somewhat confused. $\endgroup$
    – user832207
    Dec 2, 2020 at 2:11
  • $\begingroup$ $\mathcal O_{C,P}$ is the local ring of $C$ at $P$. Whether this cleanly fits into Silverman's exposition I'm not sure. $\endgroup$ Dec 2, 2020 at 2:44
  • $\begingroup$ This should basically be covered by this question. $\endgroup$
    – KReiser
    Dec 2, 2020 at 6:22

1 Answer 1


The short answer: Let $P = (x_P,y_P)$. Since $P$ is a smooth point on $C$, $\left.\displaystyle \frac{dy}{dx}\right|_P$ is well defined on $C$; note that the value may well be $\infty$. Then either $\left.\displaystyle \frac{dy}{dx}\right|_P = 0$ or $\left.\displaystyle \frac{dy}{dx}\right|_P \neq 0$ (the latter case includes the case $\left.\displaystyle \frac{dy}{dx}\right|_P = \infty$.)

If $\left.\displaystyle \frac{dy}{dx}\right|_P = 0$, then $x-x_P$ is a uniformizer. Otherwise $y-y_P$ is a uniformizer. I won't prove this for you because proving this on your own is a necessary step towards being able to understand the rest of the book, and so I suggest you do so.

To answer your reference request, I recommend Lorenzini, "An Invitation to Arithmetic Geometry."

  • $\begingroup$ It seems that you've assumed $C$ is a plane curve in your answer. Is this sufficient to prove it in general? I can imagine trying to reduce to this case using projection, but it seems like this could fail if $K = \mathbb{F}_p$. $\endgroup$ Dec 5, 2020 at 1:40
  • $\begingroup$ The same thing works in any number of variables. As long as one partial derivative is nonzero, that coordinate is a uniformizer. The partial derivatives can't all be zero since the point is a smooth point. $\endgroup$
    – djao
    Dec 5, 2020 at 9:44

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