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 $x^2-2\mid f(x)$ 

Let $p(x) \in \ker(\varphi)$. Since $x^2-2$ is a monic polynomial, we use the division in $\Bbb{Z}[x]$ to write
$$p(x)=(x^2-2)q(x)+(ax+b), \qquad \text{ where } a,b \in \Bbb{Z}.$$
Now $\varphi(p(x))=0$ implies that $\varphi(ax+b)=a\sqrt{2}+b=0$. If $a,b \neq 0$, then this is a contradiction because this would imply that $\sqrt{2} \in \Bbb{Q}$. Thus $p(x)=(x^2-2)q(x)$. 
