Show that contraction to $\mathbb Z$ of the prime ideal $\langle \sqrt 2 \rangle \subset \mathbb{Z}[\sqrt 2]$ is the prime ideal of even integers.

Question

Consider the mapping $f: \mathbb{Z} \to \mathbb{Z}[\sqrt 2]$. Show that contraction of the prime ideal $\langle \sqrt 2 \rangle \subset \mathbb{Z}[\sqrt 2]$ is the prime ideal of even integers of $\mathbb{Z}$.

Any element of $\mathbb{Z}[\sqrt 2]$ is of the form $a+b \sqrt 2,\ a,b \in \mathbb{Z}$. Also $ \langle \sqrt 2\rangle =\{\sqrt 2,2, 2\sqrt 2,2^2,2^2\sqrt 2, 2^3, \cdots \}$.

How to define $f:\mathbb{Z} \to \mathbb{Z}[\sqrt 2]$ so that $f^{-1}(\langle \sqrt 2 \rangle)=\langle 2 \rangle=\{ 2^r:\ r \in \mathbb{Z}^{+}\}$ ???

If I define \begin{align} f(2^r) &=0+(\sqrt 2)^r, \ r \geq 2 \\ f(2) &=2 \end{align}

would it suffice ??

Answer 1

Define $f:\Bbb Z\to\Bbb Z [\sqrt2] $ by $a\mapsto a $ for all $a\in \Bbb Z$. Now every even integer is in $\langle \sqrt 2\rangle$. So $f^{-1}(\langle \sqrt 2\rangle)$ is a proper ideal containing $\langle 2\rangle$. Since $\langle 2\rangle$ is a maximal ideal, $f^{-1}(\langle \sqrt 2\rangle)=\langle 2\rangle$.