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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?

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    $\begingroup$ Would you have an easier time if you were considering instead the question: Is there an isomorphism $$\Bbb R[u,v]/(uv) \cong \Bbb R[u]\times \Bbb R[v]?$$ $\endgroup$ – Ted Shifrin Apr 27 '18 at 21:28
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    $\begingroup$ Set $x=y=0$. ${}$ $\endgroup$ – Pierre-Yves Gaillard Apr 27 '18 at 21:54
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    $\begingroup$ The answer is negative and the key word is idempotents. $\endgroup$ – user26857 Apr 28 '18 at 8:21
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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.

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