I am an amateur and a beginner in the topic
A theorem states
Let K be a field of characteristic p
an elliptic curve is supersingular iff $card(E(K)) = 1$ mod $ p$
supersingular means: the subgroup E[p] of points of order p of E is trivial, equal to the infinite point only (while for ordinary curves, E[p] is the cyclical group of order p)
Then lets consider this example: the curve E defined by $y^2 = x^3+2x+1$ on $F_5$
i can easily check on SageMath that its points are
[(0 : 1 : 0), (0 : 1 : 1), (0 : 4 : 1), (1 : 2 : 1), (1 : 3 : 1), (3 : 2 : 1), (3 : 3 : 1)]
E has the infinite point and 6 points of order 7, the group E is the cyclical group of order 7.
There are thus no points of order 5 except from the infinity point, and E[5] is trivial, which means E is supersingular
But that exemple seems to contradict the theorem, because card(E) = 7 is NOT equal to 1 mod 5.
Please could you tell me what is wrong ? Is it my understanding of the definitions ? or some bad calculation in this example ?
Thank you
Best Answer
thanks, i realize i hadn't fully absorbed the fact that E[n] is a subgroup of $E(\bar{K})$ and not of E(K).
in my particular example, there are no elements of order 5 in $E(F_5)$ but i could see effectively on Sage that $E(F_{5^{2n}})$ has 4 points of order 5 (plus the point at infinity whose order == 1 divides 5)
The theorem states that effectively E[5] is the cyclic group of order 5.
Thank you very much