## May 4, 2017

### RR[x]/(x²+1) isomorphic to CC

It is a very basic problem in ring theory, but I keep forgetting the key ideas involved. The trick is to identify the following evaluation homomorphism: $$\varphi_i: \mathbb{R}[x] \to \mathbb{C},\quad \varphi_i(p(x)) = p(i) .$$ To me, this tells a lot. First, it maps all the real coefficients of $\mathbb{R}[x]$ into $\mathbb{C}$, reducing $\mathbb{R}^n$ to $\mathbb{R}^2$. Second, it maps $x$ into $i$ by essentially evaluating each polynomial $p(x)\in\mathbb{R}[x]$ at $i\in\mathbb{C}$. Although $i \notin \mathbb{R}$, it seems okay because the codomain $\mathbb{C}$ is an extension field of $\mathbb{R}$. Finally, it is surjective because for any $a+bi \in \mathbb{C}$ there exists $a+bx \in \mathbb{R}[x]$ such that $\varphi(a+bx) = a+bi$. The kernel captures the very motive for going to this trouble, i.e. no roots of $x^2+1$ in $\mathbb{R}$. $$\ker(\varphi) = \{p(x)\in\mathbb{R}[x] \mid p(i)=0\} = \{(x^2+1)q(x) \mid q(x)\in\mathbb{R}[x]\} = \langle x^2+1 \rangle .$$ Thus, by the first isomorphism theorem, $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.