Prove that $\mathbb{R}[x]/(x^2) \not\simeq \mathbb{R}[x]/(x^2-1) \not\simeq \mathbb{R}[x]/(x^2+1)$

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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There is a nilpotent element in $\mathbb R[x]/(x^2)$, not in $\mathbb R[x]/(x^2-1)$. Do you see it ? – Lierre Feb 7 '12 at 12:19
@Lierre that is $x$. I think I got it. $\mathbb{R}/(x^2)$ has nilpotent, but $\mathbb{R}/(x^2-1)$ has no, thus there is no map with zero kernel between them. Am I right? – Sergey Filkin Feb 7 '12 at 12:25
Yes you are ! More generally, isomorphic rings share all property expressible with the ring operations. And the property $$\exists x : x^2 = 0$$ is such a property. And by the way, you should right $\mathbb R[x]/(x^2)$ instead of $\mathbb R/(x^2)$, which, as is, is meaning less. – Lierre Feb 7 '12 at 12:33
Also: the "maps $0$ to $0$" argument seems like it could use some filling out. Another way to talk about the difference between these two rings is to note that $\mathbb R[x]/(x^2 - 1)$ has non-zero zerodivisors, which is obvious from your use of the Chinese remainder theorem. – Dylan Moreland Feb 7 '12 at 12:34
@DylanMoreland thank you for your comments and for your edits to my post. I had $\mathbb{R}[x]$ in mind of course. – Sergey Filkin Feb 7 '12 at 12:41

1 Answer

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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I don't see why $(0,1)$ is idempotent in $F[x]/(x-1) \oplus F[x]/(x+1)$ – Sergey Filkin Feb 7 '12 at 16:34
@Sergey $(0,1)^2 = (0,1)$ since $(a,b)(a',b') = (aa',bb')$ – Math Gems Feb 7 '12 at 16:45
thank you. It's my stupid. $(0,1)$ is not trivial. – Sergey Filkin Feb 7 '12 at 17:28