function field
Let $F$ be a field.
Definition 1.
The rational function field^{} over $F$ in one variable ($x$), denoted by $F\mathit{}\mathrm{(}x\mathrm{)}$, is the field of all rational functions $p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{/}q\mathit{}\mathrm{(}x\mathrm{)}$ with polynomials^{} $p\mathrm{,}q\mathrm{\in}F\mathit{}\mathrm{[}x\mathrm{]}$ and $q\mathit{}\mathrm{(}x\mathrm{)}$ not identically zero.
Definition 2.
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 $K\mathrm{/}F\mathit{}\mathrm{(}x\mathrm{)}$ is a http://planetmath.org/node/FiniteExtensionfinite algebraic extension^{}.
Let $\overline{F}$ be a fixed algebraic closure^{} of $F$.
Definition 3.
Let $K$ be a function field over $F$ and let $L$ be a finite extension^{} of $K$. The extension $L\mathrm{/}K$ of function fields is said to be geometric if $L\mathrm{\cap}\overline{F}\mathrm{=}F$.
Example 1.
The extension $\mathbb{Q}(\sqrt{x})/\mathbb{Q}(x)$ is geometric, but $\mathbb{Q}(\sqrt{2})(x)/\mathbb{Q}(x)$ is not geometric.
Theorem 1 (Thm. I.6.9 of [1]).
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 $K$.
Definition 4.
Let $K$ be a function field over a field $F$. Let ${K}^{\mathrm{\prime}}\mathrm{=}K\mathit{}\overline{F}$ which is a function field over $\overline{F}$, a fixed algebraic closure of $F$, and let ${C}_{{K}^{\mathrm{\prime}}}$ be the curve given by the previous theorem. The genus of $K$ is, by definition, the genus of ${C}_{{K}^{\mathrm{\prime}}}$.
Definition 5.
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 $F\mathrm{\subset}R$ 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 $\mathrm{deg}\mathit{}P$, is defined to be the dimension^{} of $R\mathrm{/}P$ over $F$.
Example 2.
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 ${R}_{\mathrm{\infty}}={\mathcal{O}}_{\mathrm{\infty}}=F[\frac{1}{x}]$ and let ${P}_{\mathrm{\infty}}=(\frac{1}{x})$ be the prime ideal of ${R}_{\mathrm{\infty}}$ generated by $\frac{1}{x}$. The localization ring ${({R}_{\mathrm{\infty}})}_{{P}_{\mathrm{\infty}}}$ is a prime of $K$, called the prime at infinity.
References
- 1 R. Hartshorne, Algebraic Geometry^{}, Springer-Verlag, New York.
- 2 M. Rosen, Number Theory^{} in Function Fields, Springer-Verlag, New York.
Title | function field |
Canonical name | FunctionField |
Date of creation | 2013-03-22 15:34:35 |
Last modified on | 2013-03-22 15:34:35 |
Owner | alozano (2414) |
Last modified by | alozano (2414) |
Numerical id | 8 |
Author | alozano (2414) |
Entry type | Definition |
Classification | msc 11R58 |
Synonym | algebraic function field |
Related topic | SimpleTranscendentalFieldExtension |
Defines | rational function field |
Defines | geometric extension |
Defines | genus of a function field |
Defines | degree of a prime |