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?

  • 8
    $\begingroup$ 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
  • 2
    $\begingroup$ Set $x=y=0$. ${}$ $\endgroup$ – Pierre-Yves Gaillard Apr 27 '18 at 21:54
  • 3
    $\begingroup$ The answer is negative and the key word is idempotents. $\endgroup$ – user26857 Apr 28 '18 at 8:21

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.