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I am stumped trying to understand Silverman's definition of $\operatorname{ord}_P(f)$, the (normalized) valuation on $\bar K[C]_P$ (which denotes the localization of a curve $C$'s coordinate ring at $P$), namely $$\operatorname{ord}_P\colon\;\bar K[C]_P\to\{0,1,2,\dots\}\cup\{\infty\} \\\operatorname{ord}_P(f)=\sup\,\{\,d\in\mathbb Z\,:\,f\in M_P^d\,\}\text.$$ Here $M_P$ is an ideal of $\bar K[C]$, defined as $$M_P=\{\,f\in\bar K[C]\,:\,f(P)=0\,\}\text.$$

Of course I am well aware how the involved operations are defined, I just don't have any "feeling" for them. What does it mean for a polynomial to be in $M_P^d$?

HRpLMvKZW0geYs8h
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1 Answers1

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Actually, valuation is a very geometric concept in the case of varieties and curves. The valuation of a function $f$ at a point (divisor, in the case of varieties) $P$ is the multiplicity of $P$ as a root of $f$: it is $>0$ if $P$ is a zero of $f$ and $<0$ if $P$ is a pole of $f$.

Another useful definition is to use a local parametrization $g(t)$ (by formal series) around $P$ (called a “uniformizer”, but think of implicit functions if you want), such that $g(0) = P$, $g'(0) \neq 0$, and $g(t)$ locally parametrizes the curve $C$. Then for any function $f$ on $C$, you may write the composed series $f(g(t))$ in the form $$ f(g(t)) = \sum_{i \geq n} a_i t^i, \quad a_n \neq 0,$$ and the index $n$ of the first non-zero coefficient is exactly the valuation of $f$ at $P$.

Example: for the elliptic curve $y^2 = x^3 + x$ over the rationals and the point $P = (0,0)$, we may use the local parametrization $t \mapsto (x(t), t^2)$ where $x(t)$ is the inverse series to $x^3 + x$, given by $x(t) = t - t^3 + 3 t^5 - 12 t^7 + \dots$. (Check that this does give a point of the curve!). Since $x(t) = t + \dots$, the valuation of $x$ at this point is $1$; since $y(t) = t^2$, the valuation of $y$ at this point is $2$. The important fact is that we could have chosen any other uniformizer (such as $x$ itself), the valuations would have remained the same.

Circonflexe
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  • Following your exhortation, I tried to check that $(x(t), t^2)$ lies on the curve, but it doesn't seem to. Substituting $y = t^2$ and $x = t - t^3 + 3 t^5 - 12 t^7 + \dots$ into the equation $y^2 = x^3 + x$, on the lefthand side I get $t^4$, while on the right I get $t - 55 t^9 + 108 t^{11} - 279 t^{13} + \cdots$. If this map gives a parametrization, these should agree modulo $t^m$ for all $m$. – Viktor Vaughn Mar 22 '24 at 20:02
  • I think that $x$ vanishes to order $2$ at $P$ and it is instead $y$ that is a uniformizer at $P$. Writing $x = y^2 - x^3$, we see that $x$ has to vanish to at least order $2$ at $P$. See my answer here for more details. – Viktor Vaughn Mar 22 '24 at 20:05
  • Ah, I see now. The mistake is just where the $t^2$ goes. Letting $g$ be the compositional inverse of $x^3 + x$, applying $g$ to both sides of $t^2 = x^3 + x$ yields $g(t^2) = x$. So the correct parametrization is $t \mapsto (g(t^2), t)$. And from this we see that $y = t$ is a uniformizer and $x = g(t^2) = t^2 - t^6 + 3 t^{10} - 12 t^{14} + \cdots$, so $x$ vanishes to order $2$, as I claimed. – Viktor Vaughn Mar 25 '24 at 17:42