Given a polynomial ring $\mathbb{Z}$[x], suppose we have the substitution homomorphism $\phi: \mathbb{Z}[x]$ -> $\mathbb{R}$ where x -> $\sqrt{2}$ and is the identity on constant polynomials. Then we want to find the kernal of $\phi$.
Then any function $f$ with integer coefficients that is divisible by $(x - \sqrt{2})$ is clearly in the kernal when we consider functions over $\mathbb{R}[x]$. However, I believe that the principle ideal generated by $f$ = x$^2 - 2$ in $\mathbb{Z}$[x] is the kernel of $\phi$.
However, I'm trying to get the following type of argument to work in order to justify this claim:
Suppose there is a $g$ $\in$ $\mathbb{Z}[x]$ such that $f$ = $x^2 -2$ does not divide $g$. Then since $f$ is monic, we have that we can use the division algorithm to get the following:
$$fq + r = g$$
Where the degree of $r$ is less than 2, and hence must be $1$ or $0$. If it is degree $1$ there is a linear factor of the form $(x-a)$ for some integer $a$. Is there a way to wrap this up to conclude $f$ must generate the kernel?
Yes. You know that $f(\sqrt2)=0$, assume that $g(\sqrt2)=0$ and you have the relation $$g(x)=q(x)f(x)+r(x),$$ where $r(x)$ is either linear or a constant.