Kernel of the homomorphism $F\colon\mathbb{R}[x]\rightarrow \mathbb{C}$

Question

Let $F:\mathbb{R}[x]\rightarrow \mathbb{C}$ be the evaluation function at $x= i$, the imaginary unit. I have shown that $F$ is a homomorphism, but to show that $\ker F= (x^2+1)\mathbb{R}[x]$ I am using some logical argument. It goes like this:

$\text{ker }F$ will contain only those $f\in\mathbb{R}[x]$ that contain $x^2+1$ as factor as $x=i\rightarrow x^2+1 = 0$. Again the functions in $\mathbb{R}[x]$ those had a factor of $x^2+1$ in them have already merged with the zero function of $\mathbb{R}[x]$. Therefore, to regenerate those equivalent forms of zero functions we have to take the product $(x^2+1)\mathbb{R}[x]$. Therefore, $\text{ker }F= (x^2+1)\mathbb{R}[x]$.

Could you please enlighten me with a better way of proving it?

Answer 1

Hint $\ $ The kernel is a principal ideal, by $R[x]$ is Euclidean $\Rightarrow$ PID. $ $ But if a nontrivial principal ideal contains an irreducible element $\,p\,$ then it is generated by $\,p\,$ (else the generator would be a proper divisor of $\,p).\,$ Thus the irreducible $\ x^2+1$ generates the kernel.

Remark $\ $ If you don't know about Euclidean domains or PIDs then you can see it more elementarily as follows. Note that the kernel is closed under gcd, since if $\,x=i\,$ is a root of $\,f,g\,$ then it is a root of $\,\gcd(f,g) = rf+sg\,$ (by Bezout). So if the kernel contains $g$ not divisible by $\,f = x^2+1,\,$ then it contains $\, \gcd(f,g),\,$ which is a proper factor of $\,f = x^2+1\,$ (since $\,f\nmid g).\,$ Thus we infer that $\,i\,$ is a root of a polynomial $\in \Bbb R[x]\,$ of degree $< 2 = \deg(f)$, contradiction.

More generally, a nonzero ideal in a PID is generated by any nonzero element having the least number of prime factors.

Answer 2

I am not sure your first step is correct. Here is how this works :

1. You know that $(x^2+1) \in \ker(F)$.

2. If $f \in \ker(F)$, then by Euclidean division, you can write $$ f(x) = q(x)(x^2+1) + r(x) $$ where $\deg(r(x)) \leq 1$. Then $$ 0 = f(i) = 0 + r(i). $$ Hence, $r(i) = 0$, but $r(x) = ax+b$ for some $a,b\in \mathbb{R}$, which is only possible (why?) if $r(x) \equiv 0$. Hence, $$ f \in (x^2+1). $$ Hence, $\ker(F) = (x^2+1)$.