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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!. Therefore, F to K. � 28 2. 3. Polynomial Rings. Now we will review what we know about polynomials in one variable and see what remains true for polynomials in several variables.

We obtain � 4 � �λ a0 λ3 a1 λ2 a2 λa3 0 � � � � 0 λ3 a0 λ2 a1 λa2 a3 �� � 0 0 �� = λ4+3+2+1+0 R(y, z). λ1+0 λ2+1+0 R(λy, λz) = �� λ4 b0 λ3 b1 λ2 b2 3 2 � 0 λ b0 λ b1 λb2 0 �� � � 0 0 λ2 b0 λb1 b2 � Comparing the powers of λ on both sides we see that R(λy, λz) = λ6 R(y, z). The general case goes along the same lines. The only thing we will check is that the powers of λ on both sides will give as the desired answer λnm : λ(m−1)+···+1+0 λ(n−1)+···+1+0 R(λy, λz) = λ(n+m−1)+···+1+0 R(y, z), so the power of λ on the right hand side is (n+m)(n+m−1) and the one on the left 2 n(n−1) hand side is m(m−1) + .

47. We have F (p0 + pt) = F (p0 ) + (∇p0 F · p)t + higher order terms in t, � � ∂F ∂F where ∇p0 F = ∂F (p ), (p ), (p ) is the gradient of F at p0 . 0 0 0 ∂x ∂y ∂z Proof. By linearity it is enough to prove the statement for any monomial F = xi y j z k . Write the expansions in t (x0 + xt)i = xi0 + ixi−1 0 xt + . . , (y0 + yt)j = y0j + jy0j−1 yt + . . , (z0 + zt)k = z0k + kz0k−1 zt + . . Taking the product of the three expansions we obtain � � j k j−1 k k−1 i i j xi0 y0j z0k + ixi−1 y z (x) + x jy z (y) + x y kz (z) t + ...

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Algebraic Curves and Codes [Lecture notes] by Ivan Soprunov


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