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I tried simple example $F_{2}$ and $y=x^3$. The there are only 2 points (0,0) and (1,1). Then how to prove that every line through the origin is a tangent line?

Metso
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  • Hint: write an equation of the line, find the intersections with the curve and check it is tangent at the point you found. – Wojowu Feb 05 '19 at 18:55
  • @Wojowu: this line is $y=x$ and it goes through (0,0) and (1,1). But it's only one line. This exercise states that every line. – Metso Feb 05 '19 at 19:01
  • @Wojowu: It looks like there is only one tangent line in this filed? – Metso Feb 05 '19 at 19:09
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    I see your point, there are actually two tangent lines - the one at $(0,0)$ ($y=0$) and the one at $(1,1)$ ($y=x$), but we still are missing one line - it turns out to be the tangent at the point at infinity. I don't know if whatever book you are reading has covered projectivizations, but if you only consider affine points you are right the statement is false. – Wojowu Feb 05 '19 at 19:15

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Suppose $L$ is a line through the origin, so $L$ is given by the equation $y=mx$ for some $m$. The intersection of $L$ with the curve $y=x^{p+1}$ is $mx=x^{p+1}$ which is equivalent to $x(x^p - m)=0$, or $x(x-\sqrt[p]{m})^p=0$ because we are in characteristic p. By looking at the zeros of this equation, the intersection points correspond to $x=0$ (the origin) or $x=\sqrt[p]{m}$ with multiplicity $p>1$, so that is why the line $L$ is tangent to the curve.

Ref:Tangent lines to a curve passing through a given point

Metso
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