TITLE: Why $\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$?
QUESTION [15 upvotes]: There is this isomorphism in my notes but there is no explanation. So I tried to reason myself but still not convincing enough, or my reasoning may even be wrong. I will appreciate if anyone is willing to lend some helps.
$\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$.
I have heard that we can achieve that by putting $i$ into $X$ in the denominator, but if we do that than the denominator will be $0$, so $\mathbb{R}[X]/(0)=\mathbb{R}[X]$? Then how can it be isomorphic to $\mathbb{C}$?
Thanks so much!

REPLY [5 votes]: I was surprised that nobody added an argument using the first isomorphism theorem and elementary means.
Define $$f: \mathbb{R}[X] \to \mathbb{C}: P \mapsto P(i)$$
It is easy to see that this is a ring epimorphism.
It is also easy to see that $(X^2 +1)\subseteq ker f$. For the other inclusion, assume $P\in ker f$ and consider $P$ as a complex polynomial. Then $P(i) =0$ means that $(X-i)|P$. Note that $P(-i) =0$, so $(X+i)|P$ as well.
It follows that $(X^2+1)|P$ in $\mathbb{C}[X]$. A moment of consideration tells us that then $(X^2+1)|P$ in $\mathbb{R}[X]$ as well which means $P \in (X^2 +1)$.