I am trying to compute $\mathrm{gr}_m(P)$ where $m=\langle X,Y\rangle $ and $P=\langle X^2-Y^3\rangle$ in the power series ring $R=\mathbb C[[X,Y]]$ with the $m$-adic filtration and show that it is not a prime ideal.
I know $\mathrm{gr}_m(R)=\mathbb C[X,Y]$ and the filtration induced on $P$ is $$P_d=P\cap m^d=\{\text{multiples of }X^2-Y^3\text{ having order}\geq d \}=\{(X^2-Y^3)f:\text{ord }f\geq d-2 \}=(X^2-Y^3)m^{d-2}$$ for $d\geq 2$.
This shows $\frac{P_d}{P_{d+1}}$ is generated as a $\mathbb C$-module by $(X^2-Y^3)X^kY^l$ with $k+l=d-2$. The image of this in $\frac{m^d}{m^{d+1}}$ is $X^{k+2}Y^l$. So the ideal I end up with is $(X^2)$. Thus $\mathrm{gr}_m(P)$ corresponds to the ideal $(X^2)$ in $\mathbb C[X,Y]$ which is clearly not prime. We also get $$\mathrm{gr}_m(R/P)\cong \frac{\mathrm{gr}_m(R)}{\mathrm{gr}_m(P)}\cong \mathbb C[X,Y]/(X^2).$$
I hope the argument above is correct. Is there any other way I can see this?