Algebraic curve cannot suddenly end - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:31:48Z http://mathoverflow.net/feeds/question/80368 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80368/algebraic-curve-cannot-suddenly-end Algebraic curve cannot suddenly end alext87 2011-11-08T09:23:33Z 2011-11-08T13:36:54Z <p>This is a literature request for (hopefully) an English version to a rigorous proof that a complex algebraic curve cannot abruptly end. </p> <p>That is, if the algebraic curve enters a closed region it must also leave it. </p> <p>This has a historic significance because Gauss's proof in his Phd thesis assumed this property holds. From looking around it seems that A Ostrowski rigorously proved the result around the 1920's. Is this correct? I am unable to find the title of the paper. </p> <p>Is there also a proof that a real algebraic curve does not end abruptly?</p> <p>I don't regard this property as obvious, but it doesn't seem to be well commented in the literature. Maybe, I'm wrong. </p> <p>Thanks in advance.</p> <p>Abruptly end: Given an irreducible polynomial $p$, we define $V(p)$ to be the complex algebraic curve associated to $p$. I say that $V(p)$ does not abruptly end at $(x,y)\in\mathbb{C}^2$ with $p(x,y)=0$ if there is a disc small enough so that the boundary contains exactly two points in $V(p)$. </p> <p>(This is a first attempt and maybe needs some corrections.)</p> http://mathoverflow.net/questions/80368/algebraic-curve-cannot-suddenly-end/80384#80384 Answer by auniket for Algebraic curve cannot suddenly end auniket 2011-11-08T13:14:34Z 2011-11-08T13:36:54Z <p>Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to <i> both </i> sides of the origin and therefore $C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is <i> odd </i>, and consequently, $(1)$ and $(2)$ give <i> different </i> real curves, and it follows that $C_1$ does not end abruptly.</p> <p>PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in \lbrace q/2, q\rbrace$, $\zeta^j$ is non-real, so the corresponding parametrization does not give any real points.</p>