Is it true that $$ \mathbb{R}[x,y]/(x^2-y^2) \cong \mathbb{R}[x,y]/(x-y) \times \mathbb{R}[x,y]/(x+y) $$ as rings?
Feels like the answer is No.
Chinese remainder theorem fails since the ideals $(x-y)$ and $(x+y)$ are not comaximal.
If we assume that there is an isomorphism $$\phi: \mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}[x,y]/(x-y) \times \mathbb{R}[x,y]/(x+y)$$ still I can't reach any contradiction.
How should I proceed?
Using the comment of @user26857, RHS has idempotent zero divisors, namely $(1,0)$ and $(0,1)$, but LHS does not.
Assuming LHS and RHS are isomorphic, LHS must have idempotent zero divisors. If $\overline{f(x,y)}$ is an idempotent zero divisor, then there is $g(x,y)\in\mathbb{R}[x,y]$ s.t. $$f(x,y)g(x,y) = (x-y)(x+y)h(x,y)$$ in $\mathbb{R}[x,y]$ for some polynomial $h(x,y)$. WLOG $x-y\mid f$ and $x+y\mid g$. Then $$f(x,y) = (x-y)f_0(x,y)$$ and $$g(x,y) = (x+y)g_0(x,y)$$ for some polynomials $f_0$ and $g_0$ satisfying $x+y\nmid f_0$ and $x-y\nmid g_0$ , otherwise both will be $\overline{0}$.
By assumption $\overline{f^2} = \overline{f}$. This means $$ (x-y)^2f_0^2(x,y) - (x-y)f_0(x,y) = (x^2-y^2)h_0(x,y) $$ for some polynomial $h_0$. This gives $$ x+y \mid (x-y)f_0(x,y) - 1. $$
Which is a contradiction because $x+y$ has no constant term.