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By Ivan Soprunov

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To show K is algebraically closed consider any polynomial f ∈ K[x]. Again, each of the coefficients of f lies in some finite field in the above union, so we can choose k ≥ 1 such that all coefficients of f lie in Fpk! e. f ∈ Fpk! [x]. Now let α be a root of f . Then we obtain a finite extension Fpk! ⊂ Fpk! (α) of some degree d. d divides n!. Therefore Fpk! (α) ⊂ Fpn! ⊂ K. This shows that α ∈ K. Finally, to show that K is the smallest algebraically closed field containing Fp , ¯ p must contain Fpn! for any n since F ¯ p must contain roots of irreducible note that F ¯ p must contain and, hence, equal polynomials over Fp of degree n!.

63. Let C, E be projective curves with equations F = 0 and G = 0 as above. Define the (local) intersection number (C·E)p0 at p0 = (x0 : y0 : z0 ) to be the multiplicity of (y0 : z0 ) as a root of the resultant R(F, G) ∈ K[y, z]. In other words, (C ·E)p0 is the largest integer k such that (z0 y −y0 z)k divides R(F, G). Note that if p0 is not a common point of C and E then (C · E)p0 = 0. If p0 lies on a common component of C and E then (C · E)p0 is undefined as then the resultant R(F, G) is identically zero.

Xn ) and call it the homogeneous coordinate of the corresponding point in Pn . In particular, we have the projective line: Pn = P2 = {lines in A3 passing through (0, 0, 0)}. Two points (x1 , y1 , z1 ) ∼ (x2 , y2 , z2 ) if and only if (x2 , y2 , z2 ) = λ(x1 , y1 , z1 ) for some λ ∈ F∗ , in which case they define the same line in A3 through (0, 0, 0). The expression (x : y : z) denotes the equivalence class of (x, y, z) under the equivalence relation ∼. 3. Constructing the projective plane. Note that if z �= 0 then (x : y : z) = A2 �→ P2 , �x z � : yz , 1 .

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Algebraic Curves and Codes [Lecture notes] by Ivan Soprunov

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