08A40-DecomposableCurve.tex

\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{DecomposableCurve}
\pmcreated{2013-03-22 19:19:38}
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\pmtitle{decomposable curve}
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\pmtype{Definition}
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\pmdefines{decomposable}
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\begin{document}
An algebraic curve
$$f(x,\,y) \;=\; 0$$
is \emph{decomposable}, if the polynomial \,$f(x,\,y)$\, is \PMlinkescapetext{reducible} in\, $\mathbb{R}[x,\,y]$; that is, if there are polynomials\, $g(x,\,y)$\, and \,$h(x,\,y)$\, with positive degree in\, $\mathbb{R}[x,\,y]$\, such that 
$$f(x,\,y) \;=\; g(x,\,y)\,h(x,\,y).$$

\textbf{Example.}\, The quadratic curve
\begin{align}
\frac{x^2}{a^2}\!-\!\frac{y^2}{b^2} \;=\; 0
\end{align}
is decomposable, since the equation may be written
$$\left(\frac{x}{a}\!+\!\frac{y}{b}\right)\!\left(\frac{x}{a}\!-\!\frac{y}{b}\right) \;=\; 0$$
or equivalently
$$\frac{x}{a}\!+\!\frac{y}{b} \;=\; 0 \quad \lor \quad \frac{x}{a}\!-\!\frac{y}{b} \;=\; 0.$$
Thus the curve (1) consists of two intersecting lines.\\

Analogically, one can say that an algebraic surface 
$$g(x,\,y,\,z) \;=\; 0$$
is \emph{decomposable}, e.g.\, $(x\!+\!y\!+\!z)^2\!-\!1 = 0$\, which consists of two parallel planes.

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