By Ivan Soprunov

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We will illustrate both reasons using the following example. 38. Let C be the parabola defined by f (x, y) = x2 − y ∈ R[x, y]. Consider any line L with equation ay + bx + c = 0 and assume a �= 0. The xcoordinates of the intersection points are given by ax2 +bx+c = 0. This polynomial has two real roots if b2 − 4ac > 0, one multiple real root if b2 − 4ac = 0, and two complex roots if b2 − 4ac < 0. Thus, if we consider C over complex numbers then the number of intersections with non-vertical line is always 2, which is the degree of C.

Let N = n−1 and assume C has N +1 double 2 points which we denote by p1 , . . , pN +1 . Choose n − 3 more points on C, call them � n� q1 , . . , qn−3 . 69 there is a curve E of degree n − 2 containing them. Let us now count the intersection number of C and E. Each pi comes with intersection number 2 and each qi comes with intersection number at least 1, which gives the total of at least 2(N + 1) + n − 3 = n(n − 2) + 1. This contradicts Bezout’s theorem as deg C = n and deg E = n − 2. Also E cannot be a component of C because C is irreducible and has larger degree than E.

8. Let Fq be a finite field of characteristic p. The map σ : Fq → F q , is called the Frobenius automorphism. α �→ αp Here some of its properties. 9. Let Fq be a finite field of q = pn elements. Then (1) for any α, β ∈ Fq we have (α + β)p = αp + β p ; (2) the map α �→ αp is an automorphism of Fq which fixes Fp ; (3) the Galois group of all automorphisms of Fq which fix Fp , Gal(Fq ) = {φ : Fq → Fq | φ(a) = a, ∀a ∈ Fp }, is cyclic of order n, generated by σ. 26 2. ALGEBRAIC CURVES Proof. (1) By the binomial formula p � � � p p−i i p (α + β) = α β = αp + ���� .

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

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