Ciro Santilli @cirosantilli 34

One major difference between the elliptic curve over a finite field or the elliptic curve over the rational numbers the elliptic curve over the real numbers is that not every possible $x$ generates a member of the curve.

This is because on the Equation "Definition of the elliptic curves" we see that given an $x$, we calculate $x^3+ax+b$, which always produces an element $y^2$.

But then we are not necessarily able to find an $y$ for the $y^2$, because not all fields are not quadratically closed fields.

For example: with $a=1$ and $b=1$, taking $x=1$ gives:and therefore there is no $y\in \mathbb{Q}$ that satisfies the equation. So $x=1$ is not on the curve if we consider this elliptic curve over the rational numbers.

That $x$ would also not belong to Elliptic curve over the finite field ${\mathbb{F}}_4$, because doing everything $\mathrm{m}\mathrm{o}\mathrm{d}4$ we have:Therefore, there is no element $y\in {\mathbb{F}}_4$ such that $y\times y=2$ or $y\times y=3$, i.e. $2$ and $3$ don't have a multiplicative inverse.

For the real numbers, it would work however, because the real numbers are a quadratically closed field, and $\sqrt{3}\in \mathbb{R}$.

For this reason, it is not necessarily trivial to determine the number of elements of an elliptic curve.

This construction takes as input:and it produces an elliptic curve over a finite field of order $p$ as output.

The constructions is used in the Birch and Swinnerton-Dyer conjecture.

To do it, we just convert the coefficients $a$ and $b$ from the Equation "Definition of the elliptic curves" from rational numbers to elements of the finite field.

For example, suppose we have $a=3/4$ and we are using $p=11$.

For the denominator $4$, we just use the multiplicative inverse, e.g. supposing we havewhere $4^{-1}=3\mathrm{m}\mathrm{o}\mathrm{d}11$ because $4\times 3=1\mathrm{m}\mathrm{o}\mathrm{d}11$, related: math.stackexchange.com/questions/1204034/elliptic-curve-reduction-modulo-p