Is there any isomorphism between these two local rings?

Question

Are the following two local rings isomorphic for any prime number $p$ and field with p elements, $\mathbb{Z}_p$ : $$\frac{\mathbb{Z}_p[X,Y]}{(X^3,XY,Y^2)}\ \text{ and } \frac{\mathbb{Z}_p[X,Y]}{(X^3,X^2-XY,Y^2)}\ \ ?$$

If there is, why is that?

first I have expected the answer to be no, but I might be wrong, and in any case I don't know how to prove it. Here's what I've tried to prove that these two rings are isomorphic.

I suppose that there is homomorphism like $\varphi$ such that $$\varphi: \mathbb{Z}_p[X,Y]\ \rightarrow \frac{\mathbb{Z}_p[X,Y]}{(X^3,X^2-XY,Y^2)}\ \ $$ $$X\mapsto f(X,Y) + I $$ $$Y\mapsto g(X,Y) + I $$ such that $f$ and $g$ are polynomials of X and Y in $\mathbb{Z}_p[X,Y]$ and $I = (X^3,X^2-XY,Y^2).$

I want to find right $f$ and $g$ so that $\varphi$ be a surjective ring homomorphism and finaly find the kernel of $\varphi$ to be $J=(X^3,XY,Y^2)$

Answer 1

The two rings are not isomorphic. An easy way to see this is to note that both rings have a unique maximal ideal generated by (the residue classes of) $X$ and $Y$. Then consider for both rings the ideal of elements that are annihilated by all elements of the maximal ideal. For the first ring it is a 2-dimensional vector space generated by $X^2$ and $Y$ whereas for the second ring it is a 1-dimensional vector space generated by $X^2$. But this property is clearly preserved under isomorphism so the two rings cannot be isomorphic.

Answer 2

I’ll be using $\mathbb{F}_p$ instead of $\mathbb{Z}_p$ and I will show these two rings are not isomorphic.

A morphism $\psi$ from the second ring to the first one coresponds to the datum of two polynomials $f,g \in \mathbb{F}_p[x,y]$ such that $f^3, g^2, f(f-g)\in (x^3,xy,y^2)$.

This means that $f \in (x,y)$ and $g \in (x^2,y)$. Thus $fg \in (x^3,yx,y^2)$, so $f^2 \in (x^3,xy,y^2)$. As we can write $f(x) \in ax+(x^2,y)$, we get $a^2x^2 \in (x^3,xy,y^2)$, thus $a=0$ and $f \in(x^2,y)$.

It follows that the image of $\psi$ is contained $(\mathbb{F}_p+(x^2,y)\mathbb{F}_p[x,y])/(x^3,xy,y^2)$ and in particular $x \notin \mathrm{im}\,\psi$, so that $\psi$ isn’t onto.