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I am trying to show that if a line given by $y = mx + b$ intersects an Elliptic Curve given by $E(\mathbb{K}): y^2 = x^3 + Ax + B$ in three points then the line is not tangent to the curve.

Given that char$(\mathbb{K}) \neq 2,3$ and $\mathbb{K}$ is algebraically closed.

Also that if they intersect in two points, the line is tangent to the curve. And if they intersect in one point, the intersection is an inflection point.

I have tried to characterize the points of intersections and compare the slope of the line and curve at those points but I'm not getting anywhere.

any help is deeply appreciated.

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    $\begingroup$ It would help if you told us what definition of tangency you are using. $\endgroup$ – Mariano Suárez-Álvarez Feb 13 '12 at 6:49
  • $\begingroup$ So basically using this equality $(mx +b)^2 = x^3 + Ax +B$ I tried to characterize the three points of intersection i.e. $(mx +b)^2 - x^3 - Ax -B = (x - x_1)(x - x_2)(x - x_3)$. But I'm not sure where to go from here. $\endgroup$ – bonyankan Feb 13 '12 at 7:07
  • $\begingroup$ Have you read my comment? $\endgroup$ – Mariano Suárez-Álvarez Feb 13 '12 at 7:11
  • $\begingroup$ Sorry Mariano, I was trying to use the fact that $m$ is the slope of the line and try to somehow compare it with the derivative of the curve at those three points of intersections, i.e. $x_1, x_2, x_3$ I don't think this is a good approach as I have been thinking about it for some times. Do you have any other approach in mind? $\endgroup$ – bonyankan Feb 13 '12 at 7:16
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    $\begingroup$ Do you know what the characteristic of a field is? $\endgroup$ – Will Jagy Feb 13 '12 at 7:25
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I will work here over a field of characteristic zero, and in fact, to fix ideas, I will be working just over $\mathbb{Q}$ (you can work out what may happen if the characteristic of the field of definition is positive and $>3$... not much changes). I will also only work out here the following case: if a line and an elliptic curve intersect in two points, the line is tangent to the curve.

Let $y=f(x)$ be a function, differentiable at a point $x=a$. Let us define tangent as follows: we say that a line $L: y=g(x)=mx+n$ is tangent to the graph of $f(x)$ at $x=a$, if $$g(a)=f(a)\quad \text{ and }\quad g'(a)=f'(a).$$

With this definition in mind, let $E/\mathbb{Q}$ be an elliptic curve given by $y^2=x^3+Ax+B$, let $y=f(x)=\sqrt{x^3+Ax+B}$ and let $L: y=g(x)=mx+n$ be a line such that the intersection of $L$ and $E$ intersect exactly at 2 points, $(a_1,b_1)$ and $(a_2,b_2)$. This means that $$ g(x)^2 - f(x)^2$$ is a polynomial of degree $3$ with only $2$ roots, namely $a_1$ and $a_2$, so one of them is a repeated root. Let us say $a_1$ is repeated, then $$ g(x)^2 - f(x)^2=(x-a_1)^2(x-a_2).$$ Thus, if we differentiate at a point $x$ we obtain: $$ 2g(x)g'(x) - 2f(x)f'(x)=(x-a_1)(2(x-a_2)+(x-a_1)). \quad (\diamond)$$

  • If $b_1\neq 0$, evaluate ($\diamond$) at $x=a_1$. Note that $f(a_1)=g(a_1)=b_1$ by assumption, because $E$ and $L$ intersect at $(a_1,b_1)$. If $b_1\neq 0$, then $g'(a_1)=f'(a_1)$, and therefore $L$ is tangent to the graph of $f(x)$ at $x=a_1$ or, equivalently, $L$ is tangent to $E$ at $(a_1,b_1)$.

  • Otherwise, if $b_1=0$, then $f(x)$ is not differentiable at $x=a_1$, and in fact either (i) $\lim_{x\to a_1^+} f'(x)=\infty$ and $f(x)$ is not defined on $(a_1,\epsilon)$, or (ii) $\lim_{x\to a_1^-} f'(x)=\infty$ and $f(x)$ is not defined on $(\epsilon,a_1)$, for some $\epsilon>0$. Either way, by taking an appropriate limit in ($\diamond$) we find that $\lim_{x\to a_1} g'(x) = \infty$ which implies that $L$ must be a vertical line $x=a_1$. Since in this case ($b_1=0$) the tangent line to $E$ is also $x=a_1$, we conclude that $L$ is tangent to $E$, as claimed.

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