Kernel of the map $\phi :\mathbb{R}[x]\to\mathbb{C}$ defined by $f(x)\mapsto f(2+i)$

Question

The problem is to find the generator of the kernel of the homomorphism map $\phi :\mathbb{R}[x]\to\mathbb{C}$ defined by $f(x)\mapsto f(2+i)$

I think the ideal generated by $(x-2-i)$ is the kernel of $\phi$.
Suppose $g\in((x-2-i))$, then $g=q(x)(x-2-i)$ for some $q(x)\in\mathbb{R}[x]$. Thus, $\phi(g(x))=g(2+i)=q(2+i)(2+i-2-i)=0$. Therefore, $((x-2-i))\subseteq ker(\phi)$.
Now suppose $g\in ker(\phi)$. By division algorithm, we can write: $$g(x)=q(x)(x-2-i)+r(x)$$ where $deg(r(x))<deg(x-2-i)=1$. Thus, $deg(r(x))=0$. Hence, $r(x)=c$ for some constant $c\in\mathbb{R}$. Now, $$\phi(g(x))=\phi(q(x))\phi(x-2-i)+\phi(c)$$ $$\Rightarrow 0=q(2+i)(2+i-2-i)+c$$ Therefore, $c=0$. Hence, $ker(\phi)\subseteq ((x-2-i))$ and thus $ker(\phi)=((x-2-i))$.

But in the solutions it is given that $ker(\phi)=((x^2−4x+5))$. Where did I go wrong?

Note: $(x^2−4x+5)=(x-(2+i))(x-(2-i))$

Edit: The proof I gave is very silly as $(x-2-i)\notin\mathbb{R}[x]$.

Answer 1

$(x-2-i)\notin\mathbb{R}[x]$ because $i$ is a complex number. So, the answer is wrong.

Since $(x^2-4x+5)=(x-(2+i))(x-(2-i))$, hence $((x^2-4x+5))\subseteq ker(\phi)$. Conversely, let $f\in ker(\phi))$. By division algorithm we can write $f(x)=g(x)+r(x)$ where $g(x)\in((x^2-4x+5))$ and $r(x)=r_0+r_1 x$ and $r_0,r_1\in\mathbb{R}$. Now, $\phi(f(x))=\phi(g(x))+\phi(r(x))=r_0+r_1(2+i)$ which is zero only if $r_0=r_1=0$. So, $f=g\in((x^2-4x+5))$. Thus $ker(\phi)\subseteq((x^2-4x+5))$.

So, $ker(\phi)=((x^2-4x+5))$.