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Consider the deformation family $y^2=x^3+t^2ax+t^3b$ of cuspidal curve $y^2=x^3$ over $k[t]/(t^4)$.

Show that the automorphism $x'=x(1+4t), y'=y(1+6t+6t^2)$ of the family over $k[t]/(t^3)$ does not lift to an automorphism of this family over $k[t]/(t^4)$.

By showing this, we can know that the local deformation functor of $y^2=x^3$ is not pro-representable.

I'm trying to solve this probloem.

1) To lift it to an automorphism of that family over $k[t]/(t^4)$, I need $x'=x(1+4t)+ft^2+gt^3, y'=y(1+6t+6t^2)+ht^3$ for some $f,g,h\in k[x,y]$ satisfying $u(y^2-x^3-t^2ax-t^3b)=y'^2-x'^3-t^2ax'-t^3b$, where $u=l+mt+nt^2+st^3$ is a unit in $(k[t]/(t^4))[x,y]$

2) After calculation of $u(y^2-x^3-t^2ax-t^3b)=y'^2-x'^3-t^2ax'-t^3b$, I can get

$l=1, m=12, n(y^2-x^3)-axl=48(y^2-x^3)-(2+f)x^2-ax,$

$s(y^2-x^3)-axm-bl=64(y^2-x^3)+8y^2+2yh-24x^2f-3x^2g-4ax-b$.

I can't improve anymore...help

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