Find the kernel of a ring homomorphism $f(x)\to f(\sqrt 2)$

Question

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

Answer 1

$\newcommand\Ker{\operatorname{Ker}}$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, $\Ker\varphi=(x^2-2)\Bbb Z[x]$.

Answer 2

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 $p(x)=(x^2-2)q(x)$.

Answer 3

The first isomorphism theorem tells you that $\Bbb{Z}[x]/\ker\phi\cong\operatorname{im}\phi$. This is not isomorphic to $\Bbb{R}$, because as you say $\phi$ is not surjective. The rest of your argument is on the right track, though.

What is true is that $\operatorname{im}\phi$ is isomorphic to a subring of $\Bbb{R}$, so in particular it is an integral domain. This means $\ker\phi\subset\Bbb{Z}[x]$ is a prime ideal (note that it is not maximal), and so $\ker\phi$ contains an irreducible polynomial.

There are still a lot of details missing from your solution; why does $\ker\phi$ contain an irreducible polynomial? Why is $\ker\phi$ generated by this irreducible polynomial? And why must it be $x^2-2$? Most of these are simple details to mention, but depending on the context, should be mentioned.