Surface defined as an image of a morphism Let
$$\phi:\mathbb{C}^2\to\mathbb{C}^3\;,\;(s,t)\to(s+t,s^2+2st,s^3+3s^2t)$$
I have to prove that the image of this morphism is the surface $M=\{(x,y,z)\in\mathbb{C}^3: 4x^3z-3x^2y^2-6xyz+4y^3+z^2=0\}$.
My work:
Take $f(x,y,z)=4x^3z-3x^2y^2-6xyz+4y^3+z^2\in\mathbb{C}[x,y,z]$, and lets name $I=\phi(\mathbb{C}^2)$. If $(x,y,z)\in I$ then exists some $(s,t)\in\mathbb{C}^2$ such as $x=s+t$, $y=s^2+2st$ and $z=s^3+3s^2t$. Plugging those values into $f$ we can see that $f(x,y,z)=0$ and since $M=Z(f)$ we have that $I\subset M$. But now I'm stuck at proving the reciprocal, $M\subset I$. Any help is welcomed!
Thanks
The question appears in the book Divisibilidad en anillos conmutativos by J.M. Gamboa. This is a basic book about rings with a little introduction to polynomial rings, so I think that we should not need to use the material in this link to prove the claim.
 A: Let
$$
\phi_1(s,t)=s+t, \
\phi_2(s,t)=s^2+2st, \
\phi_3(s,t)=s^3+3s^2t \tag{1}
$$
You have already checked that
$$
f(\phi_1(s,t),\phi_2(s,t),\phi_3(s,t)) = 0 \tag{2}
$$
Conversely, let $(x,y,z)\in{\mathbb C}^3$ such that $f(x,y,z)=0$.
Now, consider $g(s)=\phi_2(s,x-s)$. Then
$$g(s)=s^2+2s(x-s)-y=2sx-s^2-y=-(s^2-2sx+y) \tag{3}$$
This is a polynomial of degree exactly $2$ in $\mathbb C$, so it has two complex (possibly equal) roots $s_1$ and $s_2$ . We have $s_1+s_2=2x$ and $s_1s_2=y$. For $k=1,2$ put $t_k=x-s_k$ ; we then have
$$
\phi_1(s_k,t_k)=x, \phi_2(s_k,t_k) =y (k=1,2) \tag{4}
$$
Next, consider the polynomial $h(Z)=f(x,y,Z)$. This is a polynomial of degree $2$ in $Z$, so it has two complex (possibly equal) roots. By (2), for $k=1,2$ we have that $u_k=\phi_3(s_k,t_k)$ is a root of $h$. It is not clear at this point if $u_1$ and $u_2$ correspond to the same root of $h$ or not. We have to make some more computations to check that out. Note that
$$\begin{array}{lcl}
u_k &=& \phi_3(s_k,t_k)=s_k^2(s_k+3t_k) \\ 
&=& (2s_kx-y)(s_k+3(x-s_k)) \\
&=& -(2s_kx-y)(2s_k-3x) \\
&=& -\big(4s_k^2x-(6x^2+2y)s_k+3xy\big) \\ 
&=& -\big(4(2s_kx-y)x-(6x^2+2y)s_k+3xy\big) \\
&=& -\big((2x^2-2y)s_k-xy\big) \\ 
&=& (2y-2x^2)s_k+xy \\ 
\end{array}\tag{5}
$$
It follows that
$$u_1+u_2=(2y-2x^2)(s_1+s_2)+2xy=2x(2y-2x^2)+2xy=-4x^3+6xy \tag{6}$$
But this is exactly the opposite of the degree $1$ coefficient of $h(Z)$. It follows that the roots of $h$ are exactly $u_1$ and $u_2$ (with possibly $u_1=u_2$), so that
$$
f(x,y,Z)=(Z-u_1)(Z-u_2) \tag{7}
$$
From $f(x,y,z)=0$ we then deduce that $z=u_k=\phi_3(s_k,t_k)$ for some $k\in\lbrace 1,2 \rbrace$, which finishes the proof.
