By Henning Stichtenoth

This is an improved version of a favored textbook that offers a merely algebraic, self-contained and in-depth exposition of the idea of functionality fields. It comprises various workouts, a few rather basic, a few relatively difficult.

**Read Online or Download Algebraic Function Fields and Codes (Graduate Texts in Mathematics) PDF**

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**Extra resources for Algebraic Function Fields and Codes (Graduate Texts in Mathematics)**

**Sample text**

36) and the Strict Triangle Inequality. Putting x := y + z we obtain vPi (x − xi ) = vPi (y + (z − xi )) = ni (i = 1, . . 38). For P ∈ S \ {P1 , . . 37). 34 1 Foundations of the Theory of Algebraic Function Fields Next we investigate elements in F which have only one pole. 6. Let P ∈ IPF . Then for each n ≥ 2g there exists an element x ∈ F with pole divisor (x)∞ = nP . Proof. 17 we know that ((n − 1)P ) = (n − 1)deg P + 1 − g and (nP ) = n · deg P + 1 − g, hence L ((n − 1)P ) L (nP ). Every element x ∈ L (nP ) \ L ((n − 1)P ) has pole divisor nP .

Proof. Suppose that deg B = 2g − 2 and (B) ≥ g. Choose a canonical divisor W . Then g ≤ (B) = deg B + 1 − g + (W − B) = g − 1 + (W − B) , therefore (W − B) ≥ 1. 12 that W ∼ B. Next we come to a characterization of the rational function ﬁeld. 3. , F = K(x) for some x which is transcendental over the ﬁeld K. (2) F/K has genus 0, and there is some divisor A ∈ Div(F ) with deg A = 1. Proof. 18. (2) ⇒ (1): Let g = 0 and deg A = 1. 17. 5 (b)). Since (A ) = 2, there exists an element x ∈ L (A ) \ K, so (x) = 0 and (x) + A ≥ 0.

Then dim ΩF (A) = i(A) = (A) − deg A + g − 1 ≥ 1 , hence ΩF (A) = 0. 8. For x ∈ F and ω ∈ ΩF we deﬁne xω : AF → K by (xω)(α) := ω(xα) . It is easily checked that xω is again a Weil diﬀerential of F/K. In fact, if ω vanishes on AF (A) + F then xω vanishes on AF (A + (x)) + F . Clearly our deﬁnition gives ΩF the structure of a vector space over F . 9. ΩF is a one-dimensional vector space over F. Proof. Choose 0 = ω1 ∈ ΩF (we already know that ΩF = 0). It has to be shown that for every ω2 ∈ ΩF there is some z ∈ F with ω2 = zω1 .