I am reading "Introduction to Algebra" (in Japanese) by Makoto Ishida.
The following example is in this book:
Example 2:
Let $g$ be a mapping from $\mathbb{R}[x]$ to $\mathbb{C}$ such that $\mathbb{R}[x]\ni f(x)\mapsto f(i)\in\mathbb{C}$.
Then, $g$ is a surjective ring homomorphism.
Let $I:=\ker g$.
$f(x)\in I\Leftrightarrow f(i)=0\Leftrightarrow f(x)=q(x)(x^2+1) (q(x)\in\mathbb{R}[x])$.
So, $I=(x^2+1)$.
By the First Ring Isomorphism Theorem, $$\mathbb{R}[x]/(x^2+1)\cong\mathbb{C}.$$
I understand the following fact holds:
$f(x)\in I\Leftrightarrow f(i)=0\Leftrightarrow f(x)=q(x)(x^2+1) (q(x)\in\mathbb{C}[x])$.
Then, I proved that $q(x)\in\mathbb{R}[x]$ as follows:
Assume that $q(x)=a_nx^n+\dots +a_1x+a_0\in\mathbb{C}[x]-\mathbb{R}[x]$.
Then, $a_j\in\mathbb{C}-\mathbb{R}$ for some $j\in\{0,\dots,n\}$.
$\mathbb{R}[x]\ni f(x)=q(x)(x^2+1)=a_nx^{n+2}+a_{n-1}x^{n+1}+(a_{n-2}+a_n)x^n+(a_{n-3}+a_{n-1})x^{n-1}+\dots+(a_0+a_2)x^2+a_1x+a_0.$
Let $k:=\min\{j\mid a_j\in\mathbb{C}-\mathbb{R}\}$.
Since $a_k\in\mathbb{C}-\mathbb{R}$ and $a_k+a_{k+2}\in\mathbb{R}$, $a_{k+2}\in\mathbb{C}-\mathbb{R}$ must hold.
Since $a_{k+2}\in\mathbb{C}-\mathbb{R}$ and $a_{k+2}+a_{k+4}\in\mathbb{R}$, $a_{k+4}\in\mathbb{C}-\mathbb{R}$ must hold.
$\cdots$
So, $a_{n-1}\in\mathbb{C}-\mathbb{R}$ or $a_n\in\mathbb{C}-\mathbb{R}$ must hold.
But $a_n\in\mathbb{R}$ and $a_{n-1}\in\mathbb{R}$.
This is a contradiction.
I am not sure my proof is ok or not.
If my proof is not ok, then please tell me a proof.
If my proof is ok, then please tell me a standard proof of the above fact.