Surjective in the proof of $\Bbb R[x] / \langle x^2 + 1 \rangle \simeq \Bbb C$

Question

In want to show that $\Bbb R[x] / \langle x^2 + 1 \rangle \simeq \Bbb C$, so note that $\langle x^2 + 1 \rangle$ is irreducible in $\Bbb R$, and maximal in $\Bbb R[x]$, so $\Bbb R[x] / \langle x^2 + 1 \rangle$ It is isomorphic to a field.

To prove that it is isomorphic to $\Bbb C$ consider the aplication $\varphi(a+bi)=\overline{a+bx}$. We have that

$\varphi((a+bi)+(c+di))=\overline{(a+c)+(b+d)x}=\overline{(a+bx)}+\overline{(c+dx)}=\varphi(a+bi)+\varphi(c+di)$
$\varphi((a+bi)(c+di))=\overline{(ac-bd)+(ad+bc)x}=\overline{(x^2+1)bd}+\overline{(ac-bd)+(ad+bd)x}=\overline{(a+bx)(c+dx)}=\varphi(a+bi) \varphi(c+di)$

Now my doubt is how to conclude, I think there are two ways but I am not sure.

Either I show that it is injective and by first isomorphism theorem I can state the isomorphism,

Or I show that it is bijective, which is the ideal one?

For injectivity I can find the kernel of the application $\ker(\varphi)=\{z\in \Bbb C:\varphi(z)=\overline{0} \}=\{z\in \Bbb C:\overline{a+bx}=\overline{0} \}=0$

But I don't see clear over connectivity, any suggestions?

I would feel that it was more natural to take the map $\Bbb R[x]\to\Bbb C$ that sends $x$ to $i$ and show that the kernel is $\langle x^2 + 1\rangle$. — Arthur, Mar 03 '24 at 20:52
I think it's slightly more usual to consider a homomorphism $\Bbb R[x] \to \Bbb C$ if you want to make use of the isomorphism theorem. In your case you'll have use polynomial division to prove surjectivity. — Izaak van Dongen, Mar 03 '24 at 20:52

score 3 · Accepted Answer · answered Mar 03 '24 at 20:54

You've proved that $\varphi$ is a field homomorphism, so it remains to show that it is bijective. You've already proved that $\varphi$ is injective, for surjectivity : if $f\in\mathbb{R}[X]$, the euclidean division of $f$ by $x^2+1$ gives $a,b\in\mathbb{R}$ such that $f(x)\equiv ax+b\pmod{x^2+1}$ so $\overline{f(x)}=\varphi(a+ib)$ and it's done.

You can alse use the first isomorphism theorem as follow : let $\psi:\mathbb{R}[x]\longrightarrow\mathbb{C}$ defined by $\psi(f)=f(i)$, it is clearly a surjective ring homomorphism and its kernel is $(x^2+1)\mathbb{R}[x]$. The first isomorphism theorem implies that the induced morphism $\mathbb{R}[x]/(x^2+1)\longrightarrow\mathbb{C}$ is bijective.

To clarify the second approach, you can apply the euclidean division to write every polynomial as $p(x) = (a+bx) + (x^2+1)q(x)$. $p(i) = 0 \iff a+bi = 0 \iff a = b = 0$, so $p(i) = 0 \iff p \in (x^2+1)$. It is clearly surjective because $a+bx \mapsto a+bi$. — Lucas Henrique, Mar 03 '24 at 22:12

Surjective in the proof of $\Bbb R[x] / \langle x^2 + 1 \rangle \simeq \Bbb C$

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