I'm trying to show that the kernel of the homomorphism $$\varphi: \mathbb{K}[X,Y]\rightarrow \mathbb{K}[t^2,t^3],$$ $$X\mapsto t^2,$$ $$Y\mapsto t^3,$$ is the ideal $I=\langle Y^2-X^3\rangle$ in $\mathbb{K}[X,Y]$, where $\mathbb{K}$ is a field.
I can show that $I$ is contained in $\ker \varphi$, but not the other way around. For a given polynomial $p(X,Y)$ such that $\varphi(p(X,Y))=0$, I wrote it as
$$ p(X,Y)= \sum_{i=1}^n a_iX^i + \sum_{j=1}^mb_jY^j + \sum_{r,s>0}c_{rs}X^rY^S +d, $$
but that didn't help me as much as I hoped. At least I can show that $d=0$, but that was it. The change of variables made by $\varphi$ suggested some integer relationships between $i$, $j$ and $r,s$ but I could not conclude anything.
This was a part of an example seen in a Commutative Algebra online course under the section of Integral Extensions with a spoiler from what owuld be seen in an Algebraic Geometry course (I can provide link for the Youtube video, if needed, though I believe it's not necessary)
Assume that $I \neq \operatorname{Ker}(\varphi)$. Then there exists some $p(X,Y) \in \operatorname{Ker}(\varphi) \setminus I$ (as you have already shown one inclusion). As $Y^2-X^3$ is a monic polynomial in $(\mathbb{K}[X])[Y]$ we may assume that $p$ has degree $0$ or $1$ in $Y$ (otherwise add suitable element of $I$). This means we have only to consider $$ p(X,Y) = a(X) Y + b(X), $$ where $a,b\in \mathbb{K}[X]$. Now note that the degree of $\varphi(a(X)Y)$ is odd (if $a$ is not the zero polynomial) and the degree of $\varphi(b(X))$ is even. Hence, this can only vanish if $a$ is the zero polynomial. But then we get that $\varphi(b(X))$ vanishes and hence, $b$ must be the zero polynomial as well. Then we get the contradiction $0 = p(X,Y) \in \operatorname{Ker}(\varphi) \setminus I.$