The rank of $y^2=x^3\pm i$

Question

How Can I calculate the rank of curves $y^2=x^3\pm i$ over Q(i)? Is there any soft function to do it?

Answer 1

The two curves are isomorphic via: $(x,y) \mapsto (-x,i y)$, so we only need to do one of them. Let $$ E \; : \; y^2=x^3+i. $$ By inspection, this has a point $P=(-i,1+i)$. In fact $P$ has infinite order (for example, $3P$ has $5$ in the denominator of the $x$-coordinate). Now plugging the curve into MAGMA shows that an upper bound for the rank is $1$ (this is computed using $2$-descent). So the rank must be $1$. If you have MAGMA, here is what you type:

K:=QuadraticField(-1); E:=EllipticCurve([0,0,0,0,i]); RankBound(E);

However, this would be a nice exercise for you do with a descent via $2$-isogeny. Note that $x^3+i$ factors as $(x-i)(x^2+ix-1)$, so $E$ has a point of order $2$.