Prove that $\mathbb{R}[x]/{(x^2-1)}\cong\mathbb{R}\times \mathbb{R}$
As my attempt: since $x^2-1=(x-1)(x+1)$
this means $\mathbb{R}[x]/(x-1)\cong \mathbb{R},\mathbb{R}[x]/(x+1)\cong \mathbb{R},$
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1Perhaps this is beyond the scope of your knowledge, but the Chinese Remainder Theorem seems relevant. Check out the section titled "Generalization to arbitrary rings". – Ken Duna Feb 07 '17 at 05:37
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1Also, you really ought to be specifying whether this isomorphism is an isomorphism of rings or of some other algebraic object. – Ken Duna Feb 07 '17 at 06:05
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You can define the map \begin{array}{ccc} \alpha:R[x]/(x^2-1)&\longrightarrow&R\times R\\ f(x)&\mapsto&(f(1),f(-1)) \end{array}
You can check $\alpha$ is an isomorphism, i.e. bijection and homomorphism.
Aolong Li
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3This hits the nail on the head. To add context to this answer, this isomorphism is exactly what applying the Chinese Remainder Theorem gives (after the proper identifications of $\frac{R[x]}{(x-1)} \cong R$ and $\frac{R[x]}{(x+1)} \cong R$). – Ken Duna Feb 07 '17 at 05:47
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Note that the ideals $I_1=\langle x-1\rangle$ and $I_2=\langle x+1 \rangle$ are co-prime since $1=\dfrac{x+1}{2}-\dfrac{x-1}{2}\in I_1+I_2$.
Also $I=I_1\cap I_2=\langle x^2-1\rangle$.
Hence by Chinese Remainder Theorem,
$\Bbb R[x]/\langle x^2-1\rangle=\Bbb R[x]/I\cong \Bbb R[x]/I_1\times \Bbb R[x]/I_2=\Bbb R[x]/\langle x-1\rangle \times \Bbb R[x]/\langle x+1\rangle \cong \Bbb R\times \Bbb R$
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I didn't downvote, because it was an honest mistake. But, the polynomial in question was $x^2-1$. – Ken Duna Feb 07 '17 at 05:53
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$\mathbb R[x]/(x^2-1)=\mathbb R[x]/(x-1)(x+1)=\mathbb R[x]/(x-1) \times \mathbb R[x]/(x+1) \cong \mathbb R \times \mathbb R $ where
$(x-1)$ is maximal ideal in $\mathbb R[x]$, then $R[x]/(x-1)\cong \mathbb R[1]=\mathbb R$, and similar way we have $R[x]/(x+1)\cong \mathbb R[-1]=\mathbb R$
we have $x^2+1=(x-1)(x+1)$, then $\mathbb R[x]/(x-1)(x+1)=\mathbb R[x]/(x-1) \times \mathbb R[x]/(x+1)$
Mustafa
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2Are you doing anything but copying the asker's own attempt? Without explanations? – Jyrki Lahtonen Feb 07 '17 at 06:40
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