# A exercise on ismorphisms of rings of polynomials and complex numbers

How to prove that $\,\,\mathbb{C}\cong \mathbb{R}[x]/(x^2+1)$ with $(x^2+1)=\{(x^2+1)f : f\in\mathbb{R}[x]\}$

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Can you find a surjective homomorphism $\mathbb{R}[x] \to \mathbb{C}$ whose kernel is $(x^2 + 1)$ ? – Joel Cohen Oct 30 '12 at 23:27
Joel I have not found the function :( – Roiner Segura Cubero Oct 30 '12 at 23:29
Ok Jujian did not know about the points, my official language is Spanish. Thanks. – Roiner Segura Cubero Oct 30 '12 at 23:30
@Andres : You know such a function must map the polynomial $x^2+1$ to $0$ (because $x^2 + 1$ should be in the kernel), so what could the image of $x$ be ? – Joel Cohen Oct 30 '12 at 23:38
@Andres, What is your definition of $\mathbf{C}$? – Keenan Kidwell Oct 30 '12 at 23:40

Let $\psi\colon \mathbb{R}[x] \rightarrow \mathbb{C}$ be the map defined by $\psi(f(x)) = f(i)$. Clearly $\psi$ is a $\mathbb{R}$-homomorphism of $\mathbb{R}$-algebras. Suppose $\psi(f(x)) = 0$. By the division formula, there exists $g(x), h(x) \in \mathbb{R}[x]$ such that
$f(x) = (x^2 + 1)g(x) + h(x)$, where deg $h(x) < 2$.
Since $f(i) = 0$ and $i^2 + 1 = 0$, $h(i) = 0$. Suppose $h(x) = ax + b$. Then $ai + b = 0$. Hence $a = b = 0$. Hence $f(x) = (x^2 + 1)g(x)$. Hence Ker $\psi = (x^2 + 1)$. Therefore $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$.