Kernel of ring homomorphism $k[X,Y] \rightarrow k[t^2, t^3]$ Let $k$ be a field, $f: k[X,Y] \rightarrow k[t^2, t^3]$, 
$X \mapsto t^3, Y\mapsto t^2$. 
I would like to verify that 
$\ker f = (X^2 - Y^3)$. It is easy to see that $X^2 - Y^3 \in \ker f$ and therefore $(X^2 - Y^3) \subseteq \ker f$, but I struggle with proving $\ker f \subseteq (X^2- Y^3)$. 
context: I want to show that $(X^2-Y^3)$ is prime in $k[X,Y]$. 

Suppose that $p$ is in the kernel. Then by polynomial division you can write $p=q(X^2-Y^3)+r$ where no term of $r$ has degree $\ge 2$ in $X$, and clearly $r$ is then also in the kernel.
Now $r$ consists of terms of the form $aY^n$ and terms of the form $aXY^n$. Under $f$ the former terms become $at^{2n}$ and the latter terms become $at^{2n+3}$. But $t^{2n}$ and $t^{2n+3}$ are all linearly independent over $k$ in $k[t]$, so the only way for $f(r)$ to be zero is if all the coefficients of $r$ vanish.
In other words, $r=0$ and $p=(X^2-Y^3)q$.
