$\mathbb{R}[x,y]/(x^3,y) \cong \mathbb{R}[x,y]/(x^2,y^2)$

Question

I have to prove this ring isomorphism as an exercise. What I can say is that $\mathbb{R}[x,y]/(x^3,y)$ is made of the polynomials of the form $a_0 + a_1 \bar x + a_2 \bar x^2$ (with $\bar x^3 = 0$) and $\mathbb{R}[x,y]/(x^2,y^2)$ is made of the polynomials of the form $a_0 + a_1 \bar x + a_2 \bar y + a_3 \bar x \bar y$ (with $\bar x^2 = \bar y^2 = 0$). Going further, I could say that these two rings are isomorphic respectively to $\mathbb{R}^3$ and $\mathbb{R}^4$ (where the multiplication is the one induced on the coefficients by the polynomial multiplication). But I fail to see how these two rings can be isomorphic. Can anyone point me to the right direction? Thank you.

Answer 1

These rings are in fact not isomorphic!

Note that $\mathbb{R}[X,Y]/(X^3,Y) \cong \mathbb{R}[X]/(X^3) = \mathbb{R}[x]$, with $x^3=0$.
Also, $\mathbb{R}[Z,W]/(Z^2,W^2) = \mathbb{R}[z,w]$ with $z^2 = w^2 = 0$.

Suppose that there exists some isomorphism of rings $\varphi:\mathbb{R}[x] \rightarrow \mathbb{R}[z,w]$. Since $\varphi$ is surjective, there exists some $a,b,c \in \mathbb{R}$ such that $\varphi(a+bx+cx^2) = z$. Since $z^2=0$ and $\varphi$ is injective (i.e. its kernel is trivial) we obtain: $$a^2 + 2abx + (2ac+b^2)x^2 = 0$$ and hence $a=b=0$.

Therefore, $\varphi(cx^2) = z$ for some $c\neq 0$. Similarly, $\varphi(dx^2) = w$ for some $d \neq 0$. Thus, $zw = \varphi(cx^2) \varphi(dx^2) = \varphi(cdx^4) = 0$, which is a contradiction!