Here is a problem form my abstract algebra course :
Let $R = \frac{\mathbb{Q}[u,v,w]}{(u^2v^2 - w^3)}$. Find finitely many monomials $x^{a_j}y^{b_j}$ in $S = \mathbb{Q}[x,y]$ such that $R \simeq \mathbb{Q}[x^{a_1}y^{b_1}, ... ,x^{a_r}y^{b_r}].$
My idea was to defined a simple isomorphism $\phi : \mathbb{Q}[u,v,w] \to \mathbb{Q}[x,y]$ defined by $\phi(u) = x$ and $\phi(v) = y$.
Then apply it to the ideal : $$\phi(u^2v^2 -w^3) = \phi(u^2)\phi(v^2) - \phi(w^3) = x^2y^2 - w^3 $$
And finally try to express this result as a sum fo monomials in x and y and this monomials would be my answer.
But now I'm stuck and really not sure about my method, but I can't find a better way to approach this problem. Any help appreciated !
Your approach does not work, since $\phi$ does not extend on the quotient and also $\phi(w)$ is not defined.
Let $k$ be any commutative ring. For a homomorphism of $k$-algebras $$k[u,v,w]/\langle u^2 v^2 - w^3 \rangle \to k[x,y],$$you need to find three elements $a,b,c \in k[x,y]$ satisfying the relation $a^2 b^2 = c^3$. One first idea is to take $a = x^3$, $b = y^3$, $c = x^2 y^2$. Notice that $a^2 b^2 = c^3$ indeed holds, both sides are $x^6 y^6$.
So we get a homomorphism $k[u,v,w]/\langle u^2 v^2 - w^3 \rangle \to k[x,y]$ with $[u] \mapsto x^3$, $[v] \mapsto y^3 $, $[z] \mapsto x^2 y^2$. The image is therefore $k[x^3,y^3,x^2 y^2]$. The only thing that remains is to show that the homomorphism is injective. And this is where it gets a bit tricky.
Let $[f] \in k[u,v,w]/\langle u^2 v^2 - w^3 \rangle$ be in the kernel, meaning $f \in k[u,v,w]$ satisfies $f(x^3,y^3,x^2 y^2)=0$ in $k[x,y]$. We need to show that $f$ is divisible by $u^2 v^2 - w^3$.
In $k[u,v,w] \cong k[u,v]\bigl[w\bigr]$ we can write
$$f \equiv f_0 + f_1 w + f_2 w^2 \bmod w^3 - u^2 v^2$$
with $f_0,f_1,f_2 \in k[u,v]$, namely by polynomial divison with the monic polynomial $w^3 - u^2 v^2 \in k[u,v]\bigl[w\bigr]$. Hence, we may assume $f = f_0 + f_1 w + f_2 w^2$. Then,
$$0 = f(x^3,y^3,x^2 y^2) = f_0(x^3,y^3) + f_1(x^3,y^3) x^2 y^2 + f_2(x^3,y^3) x^4 y^4$$
in $k[x,y]$. Now look at the monomials of the form $x^{3i} y^{3j}$ in this polynomial. They only appear in $f_0(x^3,y^3)$ (prove this!). It follows that $f_0(x^3,y^3)=0$ and hence $f_0(x,y)=0$. By looking at the monomials of the form $x^{3i+3} y^{3j+2}$ we find $f_1=0$, and finally $f_2=0$. Thus, $f = 0$.