Generator of the kernel of $\phi : \mathbb{C}[x,y]\to\mathbb{C}[t]$ defined by $f(x,y)\mapsto f(t^2-t, t^3-t^2)$ I need to prove that $(x^3-y^2+xy)$ is the generator of the kernel of $\phi : \mathbb{C}[x,y]\to\mathbb{C}[t]$ defined by $f(x,y)\mapsto f(t^2-t, t^3-t^2)$. I was able to show that $(x^3-y^2+xy)\subseteq \ker(\phi)$, but I am getting stuck with proving the other inclusion. I tried using division algorithm but I got stuck and didn't know how to proceed.
 A: Any polynomial can be written as 
$$f(y)x^2+g(y)x+h(y)$$ by using the given polynomial to eliminate anything bigger than $x^2$. Let the lowest powers of $y$ in $f,g,h$ be $a,b,c$ then the lowers power of $t$ in the three term, upon substitution are, $$2a+2, 2b+1, 2c.$$ Since the middle term is odd we must have that it is zero and the polynomial reduces to $$f(y)x^2+h(y)$$
And we have $2a+2=2c$, dividing by a power of $y$ we can assume that $a=0$, $c=1$
Now look at the lowest terms are 
$$(t^2-t)^2=t^4-2t^3+t^2$$ and 
$$t^3-t^2$$ we see that if the two terms are to cancel, we must have the same constant coefficient for these terms, and thus it is impossible to cancel the $t^3$ terms. As the higher order terms do not give another $t^3$ term this polynomial must be zero.
A: Here is another approach, avoiding computations:
You have shown $(x^3-y^2+xy) \subset \ker \phi$ and both are prime ideals.
Since $(x^3-y^2+xy)$ is of height one and $\mathbb C[x,y]$ is two-dimensional, the only chance for the inclusion to be proper, is that $\ker \phi$ is maximal. Then the image would be a field, but the only field between $\mathbb C$ and $\mathbb C[t]$ is $\mathbb C$ itself.
