By Henning Stichtenoth
This booklet hyperlinks matters: algebraic geometry and coding thought. It makes use of a unique method in response to the idea of algebraic functionality fields. insurance contains the Riemann-Rock theorem, zeta services and Hasse-Weil's theorem in addition to Goppa' s algebraic-geometric codes and different conventional codes. it will likely be invaluable to researchers in algebraic geometry and coding conception and desktop scientists and engineers in details transmission.
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Extra info for Algebraic Function Fields and Codes (Graduate Texts in Mathematics, Volume 254)
Ig of P . We have i1 = 1 and ig ≤ 2g − 1 . Proof. 6, and 0 is a pole number. We have the following obvious characterization of gap numbers: i is a gap number of P ⇐⇒ L ((i − 1)P ) = L (iP ) . 17. 8. 39) exactly g − 1 numbers 1 ≤ i ≤ 2g − 1 with L ((i − 1)P ) L (iP ). The remaining g numbers are gaps of P . Finally we must show that 1 is a gap. Suppose the converse, so 1 is a pole number of P . Since the pole numbers form an additive semigroup, every n ∈ IN is a pole number, and there are no gaps at all.
CΩ (D, G) is an [n, k , d ] code with parameters k = i(G − D) − i(G) and d ≥ deg G − (2g − 2) . Under the additional hypothesis deg G > 2g − 2, we have k = i(G − D) ≥ n + g − 1 − deg G. If moreover 2g − 2 < deg G < n then k = n + g − 1 − deg G . Proof. Let P ∈ IPF be a place of degree one and let ω be a Weil diﬀerential with vP (ω) ≥ −1. We claim that ωP (1) = 0 ⇐⇒ vP (ω) ≥ 0 . 3 which states that for an integer r ∈ ZZ, 52 2 Algebraic Geometry Codes vP (ω) ≥ r ⇐⇒ ωP (x) = 0 for all x ∈ F with vP (x) ≥ −r .
One checks easily that ϕ is surjective and that the kernel of ϕ is AF (A1 ). Consequently deg A2 − deg A1 = deg P = [FP : K] = dim(AF (A2 )/AF (A1 )) . Step 2. Let A1 , A2 ∈ Div(F ) and A1 ≤ A2 as before. Then dim((AF (A2 ) + F )/(AF (A1 ) + F ) = (deg A2 − (A2 )) − (deg A1 − (A1 )) . 25) Proof of Step 2. 26) where σ1 and σ2 are deﬁned in the obvious manner. In fact, the only non-trivial assertion is that the kernel of σ2 is contained in the image of σ1 . In order to prove this, let α ∈ AF (A2 ) with σ2 (α+AF (A1 )) = 0.