By Ivan Soprunov

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**Additional info for Algebraic Curves and Codes [Lecture notes]**

**Example text**

53). Suppose C has two double points p1 , p2 . e. (C ·E)p1 = (C ·E)p2 = 2. But, by Bezout’s theorem, this is impossible as deg C · deg E = 3 and E cannot be a component of C since we assumed C to be irreducible. � � The same idea works in general. 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.

Divides n! for any n ≥ 1, so these are indeed subfields. 2. Algebraic Closure. Remember the Fundamental Theorem of Algebra which says that any polynomial of degree n over complex numbers has exactly n complex roots, counting with multiplicities. This is the property of the complex numbers begin algebraically closed. What are algebraically closed fields of positive characteristic? We will start with the definition. 13. A field F is called algebraically closed if every polynomial f ∈ F[x] has a root in F.

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 + ���� . . +β p = αp + β p , i i=0 =0 � � where the middle terms are all zero since p divides pi for 1 ≤ i ≤ p − 1 and p is the characteristic of the field.