function fieldFunctionField2005-11-10 10:06:222008-05-19 11:55:10Definitionrational function fieldgeometric extensiongenus of a function fielddegree of a prime% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Cl}{\operatorname{Cl}}Let $F$ be a field.
\begin{defn}
The rational function field over $F$ in one variable ($x$), denoted by $F(x)$, is the field of all rational functions $p(x)/q(x)$ with polynomials $p,q\in F[x]$ and $q(x)$ not identically zero.
\end{defn}
\begin{defn}
A function field (in one variable) over $F$ is a field $K$, containing $F$ and at least one element $x$, transcendental over $F$, such that $K/F(x)$ is a \PMlinkid{finite}{FiniteExtension} algebraic extension.
\end{defn}
Let $\overline{F}$ be a fixed algebraic closure of $F$.
\begin{defn}
Let $K$ be a function field over $F$ and let $L$ be a finite extension of $K$. The extension $L/K$ of function fields is said to be geometric if $L\cap \overline{F}=F$.
\end{defn}
\begin{exa}
The extension $\Rats(\sqrt{x})/\Rats(x)$ is geometric, but $\Rats(\sqrt{2})(x)/\Rats(x)$ is not geometric.
\end{exa}
\begin{thm}[Thm. I.6.9 of \cite{hart}] Let $K$ be a function field over an algebraically closed field $F$. There exists a nonsingular projective curve $C_K$ such that the function field of $C_K$ is isomorphic to $K$.
\end{thm}
\begin{defn}
Let $K$ be a function field over a field $F$. Let $K'=K\overline{F}$ which is a function field over $\overline{F}$, a fixed algebraic closure of $F$, and let $C_{K'}$ be the curve given by the previous theorem. The genus of $K$ is, by definition, the genus of $C_{K'}$.
\end{defn}
\begin{defn}
Let $K$ be a function field over a field $F$. A prime in $K$ is by definition a discrete valuation ring $R$ with maximal $P$ such that $F\subset R$ and the quotient field of $R$ is equal to $K$. The prime is usually denoted $P$ after the maximal ideal of $R$. The degree of $P$, denoted by $\deg P$, is defined to be the dimension of $R/P$ over $F$.
\end{defn}
\begin{exa}
Let $K=F(x)$ be the rational function field over $F$ and let $\mathcal{O}=F[x]$. The prime ideals of $\mathcal{O}$ are generated by monic irreducible polynomials in $F[x]$. Let $P=(f(x))$ be such a prime. Then $R_P=\mathcal{O}_P$, the localization of $\mathcal{O}$ at the prime $P$ is a discrete valuation ring with $F\subset \mathcal{O}_P$ and the quotient field of $R_P$ is $K$. Thus $R_P=\mathcal{O}_P$ is a prime of $K$.
One can define an `extra' prime in the following way. Let $R_\infty=\mathcal{O}_\infty=F[\frac{1}{x}]$ and let $P_\infty=(\frac{1}{x})$ be the prime ideal of $R_\infty$ generated by $\frac{1}{x}$. The localization ring $(R_\infty)_{P_\infty}$ is a prime of $K$, called the prime at infinity.
\end{exa}
\begin{thebibliography}{00}
\bibitem{hart} R. Hartshorne, {\em Algebraic Geometry}, Springer-Verlag, New York.
\bibitem{rosen} M. Rosen, {\em Number Theory in Function Fields}, Springer-Verlag, New York.
\end{thebibliography}