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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 $(x^2 + 1)$ ?2012-10-30
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    Joel I have not found the function :(2012-10-30
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    Ok Jujian did not know about the points, my official language is Spanish. Thanks.2012-10-30
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    @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 ?2012-10-30
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    @Andres, What is your definition of $\mathbf{C}$?2012-10-30
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    Keenan $\mathbb{C}$ is the set of complex numbers2012-10-30
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    @Andres, I mean, what is your definition of the set of complex numbers? Is it the set of expressions $a+bi$ with $a,b\in\mathbf{R}$, a set of ordered pairs, or something else?2012-10-30
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    Keenan any of the two definitions2012-10-30
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    Joel I do not know the function to send $x^2+1$ to O :(2012-10-30
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    @Andres What about $x\mapsto i$?2012-10-30
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    The function is $\varphi:\mathbb{R}[x]\rightarrow \mathbb{C},\,\,\, \varphi(ax+b)=a+bi$??2012-10-30
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    @Andres: Not quite. You need the evaluation map $p(x) \mapsto p(i)$.2012-10-31
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    The function is $\varphi:\mathbb{R}[x]\rightarrow \mathbb{C},\,\,\, \varphi(a_0 +a_1 x+a_2 x^2+a_3 x^3+...)=a_0+a_1 i$??2012-10-31

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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 $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$.