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By Ivan Soprunov

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I Here σ i is the composition of σ with itself i times: σ i = σ · · ◦ σ�, so σ i (α) = αp . � ◦ ·�� i i times j Indeed, if σ i = σ j then αp = αp for any α ∈ Fq , in particular, when α is a i j primitive element. Then αp −p = 1, which implies that ord(α) = pn − 1 divides pi − pj . Since both i, j are less than n this is only possible when i = j. � The next result describes all possible subfields of Fq . 10. Let K be a subfield of Fq , where q = pn . Then K is isomorphic to Fpk for some divisor k of n.

Intersection with a line: (a) two real points, (b) two complex points, (c) one real, one infinite points If a = 0 the line L is vertical and has equation bx + c = 0. It intersects C � 2� only once: the intersection point is − cb , cb2 . e. its y-coordinate is very large. So if we equip C with one extra point “at infinity” then any vertical line with intersect C at two points, one of which is this extra point. Notice that this way we obtain a compact curve, the limit of a sequence of points on C with the y-coordinate approaching infinity is the infinite point we added to C.

There exist u, v ∈ F(x)[y] such that uf + vg = 1. After clearing the denominators in u, v we get u1 f + v1 g = c(x) for some c(x) ∈ F[x] and u1 , v1 ∈ F[x, y]. If (α, β) is a common zero of f, g then c(α) = 0. Therefore there could be only finitely many such α. Furthermore, β must be a root of f (α, y), so there are only finitely many such β as well. � It is an interesting question how many common zeroes f and g can have. We will answer this question in the case of an algebraically closed field.

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