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I have proved that $\mathbb{R}[x]/(x^2-1) \simeq \mathbb{R} \oplus \mathbb{R}$. This fact is a corollary of the generalized C.R.T.

I have proved also that $\mathbb{R}[x]/(x^2+1) \simeq \mathbb{C}$. The isomorphism is given by a map $ax+b \mapsto b + ia$.

I can see why $\mathbb{R}[x]/(x^2+1) \not\simeq \mathbb{R}[x]/(x^2-1)$. This is because a homomorphism always maps $0$ to $0$, and $x^2-1=(x+1)(x-1)$ but $x^2+1$ is irreducible over $\mathbb{R}$.

For the same reason $\mathbb{R}[x]/(x^2) \not\simeq \mathbb{R}[x]/(x^2+1)$. But I still need to prove that $\mathbb{R}[x]/(x^2) \not\simeq \mathbb{R}[x]/(x^2-1)$. I would appreciate any help.

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    @DylanMoreland now I see that the map $ax+b \mapsto -a+ib$ is not suitable. The right map is $ax+b \mapsto b+ia$. Thank you for pointing that.2012-02-07

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Hint: $\ $ If $ char(F) \ne 2\ $ then $\:F[x]/(x^2-1)\ \cong\ F[x]/(x-1) \oplus F[x]/(x+1)\ \cong\ F^2\:$ has nontrivial idempotents, e.g. $\rm\:(0,1)\:,\:$ but $\rm\:F[x]/(x^2)\:$ does not (as one easily verifies).

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    thank you. It's my stupid. $(0,1)$ is not trivial.2012-02-07