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I'm asked to show $$ \frac{\mathbb{R}[x]}{(x - 1)(x^2 + 1)} \cong \mathbb{R} \times \mathbb{C}. $$

My first idea was to use the Chinese Remainder theorem to get $$ \frac{\mathbb{R}[x]}{(x^2 + 1)(x - 1)} \cong \frac{\mathbb{R}[x]}{(x - 1)} \times \frac{\mathbb{R}[x]}{(x^2 + 1)}. $$ But, I don't think these ideals are comaximal, unless I'm mistaken. Am I just not seeing it, or do I need to take another route?

ZombieNerd123
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  • That should work. You can also use the mapping $e:\Bbb{R}[x]\to \Bbb{R}\times \Bbb{C}$, $f(x)\mapsto (f(1), f(i))$. Easy enough to show that $e$ is a homomorphism of rings. And also easy enough to show that it is surjective, and has the desired kernel. The letter 'e' stands for evaluation. – Jyrki Lahtonen Nov 23 '22 at 05:01
  • They are comaximal, having sum $(x^2+1,x-\color{#c00}1) = (\overbrace{\color{#c00}1^2!+1}^{\rm \color{#0a0}{unit}\in\Bbb R},x-1) = \color{#0a0}{(1)}$ by the Euclidean algorithm and $f(x)\bmod x-a = f(a)$ (Polynomial Remainder Theorem). Generally in any PID the ideal sum is generated by the gcd of the generators - see the linked dupe. – Bill Dubuque Nov 23 '22 at 08:28

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The requirement for the Chinese Remainder Theorem is that $1 \in (x^2 + 1) + (x - 1)$. This is the case, since we have $1 = \frac{1}{2}((x^2 + 1) - (x + 1)(x - 1))$. So we can indeed apply the Chinese Remainder Theirem.

Mark Saving
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