Show that $\mathbb{F}_p[x,y]/\langle x^py-xy^p\rangle$ is not finitely generated over $\mathbb{F}_p[x-ay]$ [The current bounty is for a specific answer which I can reward in 24 hours from now. Please ignore this bounty. An answer has been already given.]

Let $p$ be a prime. Prove that $\forall a\in\mathbb{F}_p$ the algebra $\mathbb{F}_p[x,y]/\langle x^py-xy^p\rangle=:K$ is not finitely generated as a module over $F:=\mathbb{F}_p[x-ay]$.

Attempt:
Suppose the oposite. Then there's a surjective homomorphism $\phi:K\to F$. Thus $\exists f+I\in K, x-ay=\phi(f+I)$. Hence, (suppose for the moment that $p>2$),
$$
x^p-a^py^p=(x-ay)^p=\phi(f^p+I)
$$
where $$I=\langle x^py+xy^p\rangle$$
We can observe that $f^p\notin I$ because all the monoms in $f$ (as a polynomial in $x$ and as a polynomial in $y$) are from degree that is divided by $p$ which is impossible in the elements of $I$.
I'm not sure how to proceed.

First consider the ring automorphism $\varphi:\mathbb{F}_p[X,Y]\to \mathbb{F}_p[X,Y] $, $(X,Y)\mapsto (X+aY,Y)$. We observe $\varphi(X^pY-XY^p)=(X+aY)^pY-(X+aY)Y^p=X^pY-XY^p$ because of $a^p=a$ and $\varphi(\mathbb{F}_p[X-aY])=\mathbb{F}_p[X]$. Therefore we only need to show that $K:=\mathbb{F}_p[X,Y]/\left<X^pY-XY^p\right>$ isn't a finitely generated $\mathbb{F}_p[X]$-module. Because $K$ is finitely generated as $\mathbb{F}_p[X]$-algebra by $\overline{Y}$ this is equivalent to: $\overline{Y}$ isn't integral over $\mathbb{F}_p[X]$.
Let's assume $\overline{Y}$ is integral. Then there are $f_0,\ldots  f_n\in \mathbb{F}_p[X]$ such that $Y^{n+1} +f_nY^n+\cdots +f_0\in \left<X^pY-XY^p\right>\subset \left<XY\right>$. By canceling all mixed monomials we get $g(Y)+h(X)\in \left<XY\right>$ what is absurd.
