Finding the defining equations for a simple quotient variety First of all let me note that I have no experience at all with modern algebraic geometry so if at all possible I would appreciate an answer not involving the concept of a scheme. I have however some experience with commutative algebra at the level of Atiyah's book & Spec.
Here is my question:
Let $G$ be the finite subgroup of order 3 of GL(2,$\mathbb{C}$) generated by $M=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$. This acts on $\mathbb{C}[X_1,X_2]$ by substitution so that the ring of invariants is generated by $f=XY(X+Y), g=X^2+XY+Y^2, h=(X-Y)(2X^2+5XY+2Y^2)$. 
We observe that $h^2+27f^2=4g^3$. I'm told that $V=${$X^2+27Y^2-4Z^3=0$} is the equation of the image of $\phi:\mathbb{C}^2\rightarrow\mathbb{C}^3, (s,t)\mapsto(h,f,g)$ but I don't understand why $\phi$ surjects onto $V$ (other than by direct check). It also seems to me that a better way to deal with this is to say the image is the isomorphic to $\mathbb{C^2}/G$ and thus the question becomes why is the equation of $V$ the same as the equation of the quotient variety. Another perhaps harder question is suppose we are given $V$ as defined above and we want a $\phi'$ that maps to it, how similar would $\phi'$ be to $\phi$.  
Sorry for what must be a basic question. I would be happy for any references to any elementary material. 
 A: I suggest you to change coordinates. Since you work over $\mathbb{C}$, the matrix is diagonalisable and the action is up to conjugation of the form $(x,y)\mapsto (\omega x,\omega^2 y)$ where $\omega$ is a third root of unity. Then it is easy to see that the invariants are generated by $x^3,y^3,xy$ (just look at an invariant and check that it is a sum of monomials which are invariants, then check that all of these are generated).
The morphism $(x,y)\mapsto (x^3,y^3,xy)$ has image inside the variety given by $uv=w^3$ (choosing $u=x^3,v=y^3,w=xy$). To get the surjectivity, take a point $(u,v,w)$ in your image. Choose $x,y$ with $x^3=u,y^3=v$ and observe that $(xy)^3=uv=w^3$. Hence, multiplying $x$ with a third root of unity, you can change the value of $x$ and obtain that $xy=w$. This provides a point sent on $(u,v,w)$.
A: I'm sure this holds in greater generality, but here is a version of the result Mohan is alluding to that suffices for you:
Lemma. Let $A$ be a finitely generated $k$-algebra, for $k$ a field, and let $G$ be a finite group that acts by $k$-algebra automorphisms on $A$. Let $A^G$ be the ring of $G$-invariant elements of $A$. Then, $A^G \subseteq A$ is an integral extension.
Proof. Let $a_1,a_2,\ldots,a_s$ be a set of generators for $A$ over $k$. For $1 \le r \le s$ and $1 \le j \le \lvert G \rvert$, define $e_{jr}$ by the following formula:
$$\prod_{g \in G}(t-g(a_r)) = t^{\lvert G \rvert} - e_{1r}t^{\lvert G \rvert - 1} + e_{2r}t^{\lvert G \rvert - 2} - \cdots + (-1)^{\lvert G \rvert} e_{\lvert G \rvert r}$$
and let $E$ be the sub-$k$-algebra of $A$ generated by the $e_{jr}$.
We claim $E \subseteq A^G$ and that $A$ is integral over $E$. Let $S_r = G \cdot a_r$ be the orbit of $a_r$ under the action of $G$. For each $g$, there is a permutation $\sigma_g$ of $S_r$ such that $g(a_r) = \sigma_g(a_r)$, hence the elementary symmetric polynomials $e_{jr}$ in the $g(a_r)$ are in $A^G$. Since this holds for each $j,r$, we have that the sub-$k$-algebra $E$ generated by the $e_{jr}$ is contained in $A^G$.
Now since each $a_r$ is a root of the monic polynomial $\prod_{g \in G} \left( t-g(a_r) \right)$ with coefficients in $E$, we see each $a_r$ is integral over $E$, hence also over $A^G$, and so $A^G \subseteq A$ is an integral extension. $\blacksquare$
Finally, let $A = \mathbf{C}[x_1,x_2]$, and let $A^G$ be the ring of invariants under your group action. In your case, $k = \mathbf{C}$ is algebraically closed, and so by the Nullstellensatz, every point $y$ in the variety $Y$ associated to $A^G$ corresponds to a maximal ideal $\mathfrak{m} \subseteq A^G$. Since $A$ is integral over $A^G$, it satisfies Lying Over [Atiyah–Macdonald, Thm. 5.10], and so there exists a prime ideal $\mathfrak{p} \subseteq A$ such that $\mathfrak{p} \cap A^G = \mathfrak{m}$. Moreover, $\mathfrak{p}$ is maximal by [Atiyah–Macdonald, Cor. 5.8], and so corresponds to a point $x \in \mathbf{A}^2_\mathbf{C}$. The regular map $\mathbf{A}^2_\mathbf{C} \to Y$ is exactly the map associated to the $\mathbf{C}$-algebra homomorphism $A^G \hookrightarrow A$, so under this map, we have $x \mapsto y$. Thus, the map $\mathbf{A}^2_\mathbf{C} \to Y$ is surjective.

EDIT: I was asked to justify why a variety $Y$ associated to $A^G$ exists. This is because of the following
Lemma. $A^G$ is a finitely generated $k$-algebra.
Proof. We have a chain of $k$-algebra extensions $k \subseteq A^G \subseteq A$. Now $k$ is noetherian, $A$ is a finitely generated $k$-algebra by hypothesis, and $A$ is integral over $A^G$ as shown in the previous Lemma. Thus, by [Atiyah–Macdonald, Prop. 7.8], $A^G$ is a finitely generated $k$-algebra. $\blacksquare$
Now if $A^G$ is generated by, say, $b_1,\ldots,b_s$, then there is a surjection $k[x_1,\ldots,x_s] \twoheadrightarrow A^G$ mapping $x_i \mapsto b_i$. Letting $I$ be the kernel of this homomorphism, the affine variety $Z(I) \subseteq \mathbf{A}^s_k$ has the coordinate ring $A^G$, and this variety $Z(I)$ is the variety $Y$ we want.
