Essentially, I am looking for a definition, which makes this a tricky question.

Consider a rational map $\phi: X \dashrightarrow \Bbb P^m$ of complex irreducible projective varieties. I want to define the "order of approxomation" of points $y\in\Bbb P^m$ that lie in the closure of the image of $\phi$. This notion should express how hard it is to approximate $y$. Let me explain what I mean by this:

If we fix an embedding $X\subseteq \Bbb P^n$ and and affine open $\Bbb A^n \subseteq \Bbb P^n$, then $\phi$ is given by regular functions $\phi_0,\ldots,\phi_m\in\Bbb C[x_1,\ldots,x_n]$ where the $x_i$ are the coordinates on $\Bbb A^n$. We can find laurent series $p_1,\ldots,p_n\in\operatorname{Frac}(\Bbb C[\![t]\!])$ such that $p=(p_1,\ldots,p_n)$ satisfies the equations of $X$ (i.e. $p$ is a $\operatorname{Frac}(\Bbb C[\![t]\!])$-rational point on $X$) and $$ \phi_i(p_1,\ldots,p_n)=y_i+t\cdot(\cdots). $$ See for example this article by Lehmkuhl & Lickteig, the Corollary on page 11 (and Prop. 1).

Let $k$ be minimal with $t^k p_i \in \Bbb C[\![t]\!]$ for all $1\le i\le n$, i.e. the highest power of a negative exponent occuring in the $p_i$. The order of approximation of $y$ is the minimum $k$, taken over all choices of power series $p_i$ with the above property.

Let us look at an example:

Consider the rational map \begin{align*} \phi : \Bbb A^3 &\longrightarrow \Bbb P^2 \\ (a,b,c) & \longmapsto [a^3 : b^3 : b^2c]\end{align*} we can see this as a rational map $\Bbb P^3\dashrightarrow \Bbb P^2$. The point $y=[1:0:1]$ is not in the image of this map, because $b^3=0$ implies $b^2c=0$. However, $\phi(1, t, t^{-2}) = [1:t^3:1]$. In this case, $y$ has order of approximation at most $2$ and at least $1$. I am quite sure it is $2$.

My problem is: The above definition depends on fixing an embedding of $X$. It feels to me that this should be intrinsic, depending only on the indeterminacies of $\phi$. However, I may very well be wrong. So the question is:

Can the order of approximation be defined intrinsically on $X$?

By the way, if you have an affirmative answer with additional assumptions on $X$ (e.g. smoothness), that is very welcome.

  • $\begingroup$ (1) Your definition seems to depend not only on an embedding of $X$, but also on the choice of $\mathbb{A}^n\subset \mathbb{P}^n$. (2) What is $X$ in your example? $\mathbb{A}^3$ is not projective. Moreover, $\phi$ is not a morphism: it is undefined where $a=b=0$. $\endgroup$ Oct 1, 2016 at 16:48
  • $\begingroup$ @LaurentMoret-Bailly: Since the rational map $\phi$ is completely defined by its behaviour on some (any) affine open of $X$, I think i can pick any such affine open and then it depends on an embedding of that chart, right? $\endgroup$ Oct 1, 2016 at 17:04
  • $\begingroup$ @LaurentMoret-Bailly: concerning (2), $\phi$ is a rational map, not a morphism. $\endgroup$ Oct 1, 2016 at 17:05
  • $\begingroup$ Concerning (1) in my comment, $y$ may be in the image of one chart, say $U$, but not another, say $U'$. The order of approximation calculated with $U$ (resp. $U'$) will then be $>0$ (resp. $0$). $\endgroup$ Oct 2, 2016 at 6:18
  • $\begingroup$ Concerning (2), I was just objecting to the fact that in the example, you do call $\phi$ a morphism. $\endgroup$ Oct 2, 2016 at 6:21

1 Answer 1


Since there has been no answer yet, I will summarize my own thoughts on the matter in an answer. Let $\Bbb K:=\operatorname{Frac}(\Bbb C[\![t]\!])$ be the field of Laurent series. There is a natural morphism $$\iota:\operatorname{Spec}(\Bbb K)\to\operatorname{Spec}(\Bbb C[\![t]\!]).$$ I define an approximation of $y$ to be a morphism $$p:\operatorname{Spec}(\Bbb K)\to X$$ such that there exists a $$\bar p:\operatorname{Spec}(\Bbb C[\![t]\!])\to\Bbb P^m$$ with $q(\langle t\rangle)=y$ and $\phi\circ p=\bar p\circ\iota$, i.e., so that the following diagram commutes: $$ \begin{array}{cccccc} \operatorname{Spec}(\Bbb K) &\xrightarrow{\;i\;} &\operatorname{Spec}(\Bbb C[\![t]\!]) \\ \Big\downarrow p && \Big\downarrow\bar p \\ X & \xrightarrow{\;\phi\;} & \Bbb P^m \\ \end{array} $$

Assuming $X$ affine, we can define the order of $p$ with respect to an embedding of $X$ as follows: Choose generators $\Bbb C[X]=\Bbb C[x_1,\ldots,x_n]$, let $p_i\in\Bbb K$ be the image of $x_i$ under the map $p^\sharp:\Bbb C[X]\to\Bbb K$ and take the maximum pole order of any of the $p_i$ to be the order of $p$. Then, define the order of $y$ to be the minimum order of any approximation of it.

The question is the following: Fix an approximation $p$ of $y$ which has order $r$ with respect to the above choice of generators. For another set of generators $\Bbb C[X]=\Bbb C[y_1,\ldots,y_k]$, is there an approximation $q$ of $y$ which has order $r$ with respect to the $y_i$?

Unfortunately, I don't know if this is true.

  • $\begingroup$ Can you show the setup in a commutative diagram? That would clarify all the p's. $\endgroup$
    – Matt F.
    Oct 6, 2016 at 19:29
  • $\begingroup$ @MattF.: I don't really know how to properly draw a diagram with MathJax. $\endgroup$ Oct 8, 2016 at 12:09
  • 1
    $\begingroup$ @JeskoHüttenhain I added a commutative diagram, hope it reflects what you meant. $\endgroup$ Jan 29, 2018 at 16:56

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