# Puiseux series expansion for space curves?

This result is apparently well known and used by many people. I am, however, quite frustrated that I cannot seem to find a proof that I can understand. For plane algebraic curves, this is not too hard. For an irreducible polynomial $F(x,y) \in \mathbb{C}[x,y]$ nonconstant in both $x$ and $y$ with $F(0,0) = 0$, one can formally expand $y$ as a Puiseux series in $x$ $$y(x) = \sum_{i=0}^\infty y_i x^{\alpha_i}$$ The Puiseux theorem actually states that such a series converges (in some sense) near $x=0$. Alternatively, we can construct Riemann surface over the point $(0,0)\in \mathbb{C}^2$ of the algebraic function $y(x)$ which has the normal representation as holomorphic element $$\begin{eqnarray} x &=& t^m \cr y &=& \sum_{i=1}^\infty y_i t^i \end{eqnarray}$$ With some abuse of notation, I think we can even say the two are the same. I can find many sources, among which I like Walker's representation the best.

Now how about space algebraic curves (of one complex dimension) in $\mathbb{C}^n$? For a polynomial system $F(x_0,\ldots,x_n) = (f_1(x_0,\ldots,x_n),\ldots,f_n(x_0,\ldots,x_n))$ where each $f_i \in \mathbb{C}[x_0,\ldots,x_n]$ with $F(0,\ldots,0) = (0,\ldots,0)$, and lies only on a one (complex) dimensional irreducible component of $V(F)$, if we fix a place of the this algebraic curve centred at $(0,\ldots,0)$, we should be able to find a parametrization $$\begin{eqnarray} x_0 &=& t^m \cr x_k &=& \sum_{i=1}^\infty c_{k,i} t^i \end{eqnarray}$$ with convergent power series. Or equivalently, we could express $x_1,\ldots,x_n$ as Puiseux series in $x_0$ that converge in certain sense.

I could only find "proofs" that reference Hironaka's resolution of singularity, which I don't think I can understand any time soon. I'm hoping to find a proof using only complex geometry or basic complex algebraic geometry. In particular, I was thinking maybe I can repeatedly apply Weierstrass preparation theorem together with Puiseux theorem, however, I'm not quite sure how to continue after the first step.

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The algebraic closure of $\mathbb{C}((t))$ is $\cup_n \mathbb{C}((t^{1/n}))$. The proof of this obviously reduce to the case of plane curves and implies what you want. Serre's Corps Locaux, for instance, has an algebraic proof. – Felipe Voloch Feb 19 '12 at 3:52
I'm not sure if I understand this. I know the algebraic closure of the ring of formal Laurent series is the formal Puiseux series, and this form a basis for the formal local parametrization of plane algebraic curves. But I still don't see how this extends to the case of space algebraic curves, unless I'm missing something. I just did a quick check on Serre's Local Fields (my French is, uh..., terrible), and I couldn't find it. Do you happen to remember which chapter is it in? Either way, a purely algebraic proof is probably not enough, as it require the concept of convergences. – ssquidd Feb 19 '12 at 4:15
A space algebraic curve, by virtue of being an algebraic curve, has a function field of transcendence degree one over $\mathbb{C}$, so for any coordinate function $t$ that you pick, the other coordinate functions will be algebraic functions on $t$, thus will be inside the algebraic closure of $\mathbb{C}((t))$. This does not give convergence in the complex topology. I missed that part of your question and I don't know a good reference offhand. The reference in Serre is Ch. 4, Prop. 8. – Felipe Voloch Feb 19 '12 at 4:36

If $(C,0)$ is a germ of complex curve in $\mathbb{C}^n$, then you can find coordinates $(z_1, z')$ and a polydisc $V=V_1\times V'$ centered in $0$ such that the canonical projection $V\ni (z_1,z')\mapsto \pi(z_1,z')=z_1\in V_1$ is a ramified covering when restricted to $C$ with $p$ sheets. Let $S$ be the ramification locus, then $S$ is an analytic set in $V_1$, that is, a discrete set of points. Therefore, upon taking a smaller $V_1$, we can think that $S=\{0\}$ or it is empty. Let us suppose $S$ is not empty, otherwise the projection is a local biholomorphism with the unit disc and this gives the thesis with $z_1=t$, $z_k=f_k(t)$, $f_k\in\mathcal{O}(D )$. Now, $C^*=C\cap \pi^{-1}(V_1\setminus S)$ is a $p-$sheeted covering of $V_1\setminus S=D\setminus\{0\}$, therefore it is isomorphic to the standard $p-$sheeted covering: $$D\setminus\{0\}\ni t\mapsto t^p\in D\setminus\{0\}.$$ This means that there exists a map $g:D\setminus\{0\}\to C^*$ such that $\pi\circ g (t)=t^p$; this map extends clearly to a holomorphic bijection by setting $g(0)=0$ (it is continuous in $0$ and holomorphic outside). Therefore, we have $$z_1=t^p$$ $$z_k=g_k(t)=\sum_{j=0}^\infty a_{kj}t^j.$$