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Let $E:y^2-xy+y-x^3=0$ over a field $K$, $P=(0,0)$. If $\text{char}(K)\neq2$, $E$ is an elliptic curve for which I can easily get $n*P$. Now, something goes wrong with $\text{char}(K)=2$: Basically, I draw the tangent line in $P$ (a horizontal line) which should then intersect $E$ in $P'$ and I would get $2*P=-P'$.

Unfortunately, this horizontal line never intersects $E$ except in $P$ itself. Who can help? Probably, there is something that $\text{char}(K)=2$ hints at?

mikemike
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  • How are you calculating the tangent line at $P$, in characteristic $2$ anyways? – Álvaro Lozano-Robledo Apr 29 '13 at 19:09
  • @mikemike: in characteristic $0$, $P$ is a $3$-torsion point by your computation (because $P'=P$). In characteristic $2$, as pointed out by Álvaro, $E$ is a singular cubic, you can't add two arbitrary points. But still, as long as your point $P$ is a smooth point (this is your case), and the intersection of the tangent line with $E$ lies in the smooth locus, the same result holds: $2P=-P'$. So the answer for your $E$ and $P$ is $2P=-P$. –  Apr 30 '13 at 03:09
  • @ÁlvaroLozano-Robledo: the tangent line given exactly as in characteristic $0$. For simplicity, if $X$ is a hypersurface defined by a polynomial $f$ in some affine space $\mathbb A^d$, then the tangent hyperplane is at $P:=(a_1,\dots, a_d)$ is $$\frac{\partial f}{\partial x_1}(P)(x_1-a_1)+\cdots + \frac{\partial f}{\partial x_d}(P)(x_d-a_d)=0.$$ –  Apr 30 '13 at 03:13
  • Oops, I calculated the derivatives too quickly in my head and thought $(0,0)$ was the singular point. – Álvaro Lozano-Robledo Apr 30 '13 at 04:19

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