-1

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
  • 13
  • 3
  • 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](https://math.stackexchange.com/q/94728/242)). 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

1 Answers1

2

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
  • 29,779
  • 1
  • 13
  • 48
  • Ah, I wasn't looking hard enough. I think my brain is so used to working on integer rings, I wasn't thinking about 1/2 :-) – ZombieNerd123 Nov 23 '22 at 05:17
  • @ZombieNerd123 The technique here is to use the Euclidean algorithm to find the gcd of the two polynomials. No guesswork is needed. – Mark Saving Nov 23 '22 at 05:19
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement [here](https://math.meta.stackexchange.com/q/33508/242). – Bill Dubuque Nov 23 '22 at 08:29