Automorphism group of a cuspidal "elliptic" curve

Question

I have a cuspidal elliptic curve ($\Delta=0$ and $J$-invariant$=0$), $E$ in a field $K$ of characteristic $3$ and I'm trying to show that its automorphism group is $K^*$. These are my calculations: $E$ has the form: $$Y^2Z=X^3+a_6Z^3$$ The substitutions preserving this form are:$$X=u^2X+rZ$$ $$Y=u^3Y$$ $$Z=Z$$ Then automorphism of $E$ have: $$a_6(u^6-1)-r^3=0$$ How can I proceed from here?

Answer 1

$\Delta=0$ means that your cubic projective curve is not an elliptic curve. Did you mean the automorphisms of the singular projective curve ? $$E/K:\qquad ZY^2=X^3+aZ^3$$ (singular at the point $[-1:0:1/a]$)

It is the projective closure of the affine curve $$D:y^2=x^3+a=(x+a)^3$$

With $C:z^2=w$, $x=w-a,y=wz$ we get a birational map $$E\to D\to C \to \Bbb{A}^1, [x:y:1]\to (x,y)\to (x+a,y/(x+a))\to y/(x+a)$$ Which sends $[-1:0:1/a]\to (-1,0)\to 0$.

Any automorphism of $E$ induces a birational map $\Bbb{A^1\to A^1}$, and since $E$ has only one singular point, only the birational maps $\Bbb{A^1\to A^1}$ fixing the image of $[-1:0:1/a]$ induce an automorphism of $E$.

$Aut(\Bbb{A^1})=Aut(K(z)/K)=PGL_2(K)$, and the subgroup of $PGL_2(K)$ fixing $0$ is $K^*$.