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Then the complement Q = D − Q ∼ (n − m∗ )P∞ + E. As in the first proof, Q + ER ∼ ((a0 + deg E)(q + 1) − a1)P∞ and a1 ≤ a0 + deg E. 42. Let A = B = (a0 (q + 1) − q)P∞ , Z = (q − a0 − 1)P∞ , 1 ≤ a0 ≤ q − 1. Then the code CΩ (G, D) with G = (2a0 −1)(q +1)−a0 has d = d∗ +(q −a0 −1). This corresponds to the case G = ((q − 2)(q + 1) + (2a0 + 1 − q)(q + 1) − a0 )P∞ in the theorem above, with a0 ≥ 2a0 + 1 − q if and only if a0 ≤ q − 1. For Hermitian one-point codes of length q 3 , the q 3 finite rational points form a complete intersection with coordinate ring 2 F[x, y]/(y q + y − xq+1 , xq − x) = xi y j : 0 ≤ i ≤ q 2 − 1, 0 ≤ j ≤ q − 1 .

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