Consider the two elliptic curves $$E_3: y^2+y=x^3+x^2+x \enspace [Cremona:19A3]$$ and $$E_1: y^2+y=x^3+x^2−9x−15 \enspace [Cremona:19A1]$$ Let $\varphi$ be the $3$-isogeny from $E_3$ to $E_1$. I want to prove that kernel of $\varphi$ contains a point over the rationals. I have computed the isogeny in Magma.
Input:
E:=EllipticCurve("19a3"); E2:=IsogenousCurves(E)[2]; A,B:=IsIsogenous(E,E2); A; B; Kernel (B)
Output:
true Elliptic curve isogeny from: CrvEll: E to CrvEll: E2 taking (x : y : 1) to ((x^3 + 2*x + 1) / x^2 : (x^3*y - 2*x*y - x - 2*y - 1) / x^3 : 1) Subgroup scheme of E defined by x
Now $E_3$ contains the rational $3$-torsion points $(0,0), (0,1), (0,-1)$. Since the isogeny map contains 'x' in the denominator, my confusion is where will the above 3 points be mapped ? Also, if someone can help me to find out a rational point in Ker $\varphi$, I will be very grateful.
Since the isogeny is defined in projective coordinates, you can write it as well in this form: $$[x:y:1] \mapsto [x(x^3+2x+1):x^3y-2xy-x-2y-1:x^3].$$ Now it is clear that $[0:0:1]$, $[0:1:1]$, and $[0:-1:1]$, map to $[0:1:0]$, the point at infinity in $E_1$. Indeed, they map to $[0:-1:0]=[0:-3:0]=[0:1:0]$, respectively, which are all the same point in projective coordinates, equal to the point at infinity. In other words, all three points are in the kernel of the map.