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I got the following question where I stuck at the moment. Given is the elliptic curve (EC) equation: $E: y^2+3xy+y=x^3+4x+4$ over the finite field ${\bf F}_5$ The first task is now to find out all possible points on the curve by just trying out all possible values for x.

For this I got the possible values: $E({\bf F}_5)=\{(2,0),(2,3),(4,1),P(inf)\}$ (don't know if its 100% correct) P(inf) is the infinite point of the curve.

The next task is to find the characteristic polynomial of the Frobenius endomorphism and to use this in order to calculate the number of points in $E({\bf F}_{25}).$

In order to get the characteristic polynomial I found for the trace of the Frobenius endomorphism is: $\alpha_5=5+1-t=4,$ so must be $t=2.$

To get now to the char. polyn. I have to find the roots for the equation of the char. polyn.: $X(T)=T^2-tT+q=T^2-2T+5.$

But my problem is now, that I cannot find any roots that fit to this equation $T^2+2T+5.$ Maybe you can help me? Thank you!

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  • $\begingroup$ the points are ok $\endgroup$
    – 111
    Commented Jan 11, 2015 at 12:38
  • $\begingroup$ Hello Ovomaltine! At which university do you study? I have the same exercise and I wonder if we are at the same uni. :) $\endgroup$
    – user175343
    Commented Jan 12, 2015 at 22:40
  • $\begingroup$ It might be :) RUB $\endgroup$
    – Ovomaltine
    Commented Jan 15, 2015 at 18:12

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The roots of $T^2-2T+5$ are $\rho_{1,2}=1\pm 2{\rm i},$ so $$\#E({\bf F}_{p^2})=p^2+1-(\rho_1^2+\rho_2^2)=26-(-6)=32.$$

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  • $\begingroup$ Thank you for this fast answer. Can you please give me a short hint how you got the solutions of 1+-2i as the complex roots of the char.polyn.? Thank you $\endgroup$
    – Ovomaltine
    Commented Jan 11, 2015 at 13:29
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    $\begingroup$ you try to solve a problem with elliptic curves and you never seen the quadratic formula? (en.wikipedia.org/wiki/Quadratic_formula) $\endgroup$
    – 111
    Commented Jan 11, 2015 at 13:55

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