By Ivan Soprunov

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**Example text**

0 0 0 0 0 0 0 0 0 It remains to notice that the first term is F (p0 ) and the expression in the square brackets is the dot product of ∇p0 F and p = (x, y, z). � Now suppose p0 lies on C. e. L intersects C at p0 . e. L is tangent to C at p0 . 48. Suppose a line L = {p0 + pt | t ∈ F} is tangent to C at p0 . Then F (p0 ) = 0 and ∇p0 F · p = 0. Consequently, ∇p0 F · p = 0 is an equation of the tangent line to C at p0 . Let us write the equation of the tangent line in coordinates: x ∂F ∂F ∂F (p0 ) + y (p0 ) + z (p0 ) = 0.

F g = c(x)h for some non-constant c(x) ∈ F[x] and h ∈ F[x][y]. Let p(x) be an irreducible factor of c(x). The identity f g = c(x)h in (F[x]/�p(x)�) [y] becomes f¯g¯ = 0. This implies that either f¯ = 0 or g¯ = 0 in (F[x]/�p(x)�) [y] (remember that this is a polynomial ring over a field, hence it has no zero divisors). e. either f or g is not primitive. e. � � f (x) F(x) = | f, g ∈ F[x], g �= 0 . 23. (Gauss’s Lemma) A polynomial f is irreducible in F[x, y] if and only if it is irreducible in F(x)[y].

Note that the three coordinate lines intersect at three special points with coordinates (0 : 0 : 1), (1 : 0 : 0), and (0 : 1 : 0). 4. 4. Three coordinate lines in the projective plane. 38 from the viewpoint of the new definition. 41. First, consider two parallel lines in A2 : L1 = {(u, v) ∈ A2 | v − 2u = 0}, L2 = {(u, v) ∈ A2 | v − 2u − 2 = 0}. Where (u, v) are the aﬃne coordinates. Recall that in Uz we have u = xz , v = yz , so the two equations become yz − 2 xz = 0 and yz − 2 xz − 2 = 0. If we clear the denominators we obtain y − 2x = 0 and y − 2x − 2z = 0.

### Algebraic Curves and Codes [Lecture notes] by Ivan Soprunov

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