Find the kernel of a ring homomorphism $f(x)\to f(\sqrt 2)$ Let $$\phi:\mathbb Z[x]\longrightarrow \mathbb R$$, where $$\phi(f(x))=f(\sqrt 2)$$
find $$\operatorname{Ker}(\phi)=\{f(x)\in \mathbb Z[x]\mid f(\sqrt2)=0\}$$
1st) I wanted to use isomorphism theorem since we have $\mathbb R$-field so $Z[x]/\ker\phi\simeq \mathbb R$ so we find that $\ker\phi$ maximal ideal so it must be irreducible minimal polynomial which has root $\sqrt2$ that is $x^2-2$ but $\phi$ is not surjective.
2nd) Let $f=f_0+f_1x+f_2x^2+f_3x^3+f_4x^4+..$
since $\\f(\sqrt2)=0$ we have $$\\f_0+2f_2+4f_4+8f_6+...=0\\ f_1+2f_3+4f_5+...=0$$but how to show $x^2-2\mid f(x)$ 

$\newcommand\Ker{\operatorname{Ker}}$First note that $x^2-2\in\Ker\varphi$.
Conversely, if $f\in\Ker\varphi$, write $f(x)=(x^2-2)q(x)+r(x)$ with $r(x)=0$ or $\deg r(x)<2$.
Since $f(\sqrt 2)=0$ you get $r(\sqrt 2)=0$, but since $\sqrt 2$ is irrational and $\deg r<2$ we get $r=0$.
This proves $(x^2-2)|f(x)$.
Consequently, $\Ker\varphi=(x^2-2)\Bbb Z[x]$.
