TITLE: Proving that $\mathbb{R}[x,y]/(x^2,y^2)$ is not isomorphic as a ring to $\mathbb{R}[x,y]/(xy,x^2-y^2)$.
QUESTION [7 upvotes]: How do I prove that $\mathbb{R}[x,y]/(x^2,y^2)$ and $\mathbb{R}[x,y]/(xy,x^2-y^2)$ are not isomorphic as rings? It is easy to show that they are not isomorphic as $\mathbb{R}$-algebras; but an arbitrary ring homomorphism could send an element of $\mathbb{R}^*$ to some other invertible element. The usual properties do not distinguish the two rings (reducedness, dimension, normality, etc).

REPLY [2 votes]: $R_1=\Bbb R[x,y]/(x^2,y^2)$ is four-dimensional with $\Bbb R$-basis $1$, $x$, $y$, $xy$.
$R_2=\Bbb R[x,y]/(xy,x^2-y^2)$ is four-dimensional with $\Bbb R$-basis $1$, $x$, $y$, $x^2$.
Both are local rings: the maximal ideal of $R_1$ is $M_1=(x,y)$ and the
maximal ideal of $R_2$ is $M_2=(x,y)$. A homomorphism $\phi:R_1\to R_2$
must take $M_1$ to $M_2$. Therefore $\phi(x)=ax+by+cx^2$
where $a$, $b$, $c\in\Bbb R$. As $x^2=0$ in $R_1$ then $(ax+by+cx^2)^2=0$ in $R_2$.
But
$$(ax+by+cx^2)^2=a^2x^2+2abxy+b^2y^2=(a^2+b^2)x^2$$
in $R_2$ and so $a^2+b^2=0$, that is $a=b=0$, and so $\phi(x)=cx^2$.
Likewise $\phi(y)=dy^2$ where $d\in \Bbb R$. Some  non-trivial
$\Bbb R$-linear combination of $x$ and $y$ must be sent to zero by $\phi$.
Therefore $\phi$ cannot be an isomorphism.