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8 votes
1 answer
418 views

Ramification and Puiseux series in Several Variables

In Algebraically Closed Fields Analogous to Fields of Puiseux Series, Rayner makes the following observation: given the equation $Z^2 = (X+Y)Y$, we can solve this equation for $Z$ using a ...
A. Thomas Yerger's user avatar
7 votes
2 answers
1k views

How to treat Puiseux series as functions?

I have been reading about Puiseux series in the context of the Newton–Puiseux algorithm for resolution of singularities of algebraic curves in $\mathbb{C}^2$. Given a curve $f(x,y)=0$ with $f$ a ...
Gutiérrez's user avatar
1 vote
0 answers
205 views

Puiseux's theorem's converse

Puiseux's theorem asserts that given a polynomial equation $P(x,y)=0$, its solutions in $y$, viewed as functions of $x$, may be expanded as Puiseux series that are convergent in some neighbourhood of ...
user1337's user avatar
  • 473
8 votes
2 answers
1k views

Singular points of algebraic varieties and parametrization by Puiseux series

Let $V\subset \mathbb{R}^n$ (or $\mathbb{C}^n$ if that makes anything easier) be an algebraic variety and $p\in V$ a possibly singular point. Let $U\subset V$ be a sufficiently small neighborhood of $...
Igor Khavkine's user avatar
2 votes
2 answers
438 views

Determine asymptotic behavior of algebraic curves

Take an example polynomial $f(x, y) = y^2 x + y^3 - x^2$. A solution to $f(x,y)=0$ exists with Puiseux series given by $y(x) = x^{2/3} - x/3 + x^{4/3}/9+\cdots$. I got this by having Mathematica ...
Josh Burkart's user avatar