I'm having problems calculating the Krull dimension of $A=K[x,y]/\langle x^3,x^2-y^2,xy\rangle$, where $K$ is a field. I've heard the solution must be 3 or 4, but as I see it we have the extension: $$K \subseteq A,$$ understanding $K$ as the image of the function $f:K \longrightarrow K[x,y]/\langle x^3,x^2-y^2,xy\rangle, k \mapsto k + \langle x^3,x^2-y^2,xy\rangle$, which is injective. Let me use $\overline{x},\overline{y}$ to represent the classes of $x$ and $y$ in $A$, respectively. That extension is integral since $\overline{x}$ is the root of $T^3$ (monic polynomial with coefficients in $K$) and $\overline{y}$ is the root of $T^2 - \overline{x}^2$ (we can use $\overline{x}$ since we already know that is integral over $K$ and being integral is "transitive"). Since the extension is integral, the dimension must match, and in that case the dimension of $A$ is $0$, since $K$ is a field.
As I said, I believe this is wrong, but I don't know why. Can someone shed some light on this? Thanks!