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Prove $\sqrt{2}$ is an irreducible element of $R= \mathbb{Z} [\sqrt{2} ] $ $$


\begin{aligned} \sqrt{2}&= (a_1+b_1\sqrt{2})(a_2 +b_2 \sqrt{2}) \\&= a_1(a_2 +b_2 \sqrt{2}) + b_1\sqrt{2}(a_2 +b_2 \sqrt{2}) \\&=a_1 a_2 +a_1 b_2 \sqrt{2} + b_1 \sqrt{2} a_2 + b_1 \sqrt{2} b_2 \sqrt{2} \\&=a_1 a_2 +(a_1 b_2 +b_1 a_2 ) \sqrt{2}+b_1 b_2 2 \\&= a_1 a_2 +(a_1 b_2 +a_2 b_1) \sqrt{2} +2 b_1 b_2 \\& a_1 a_2 +b_1 b_2 *2 +\sqrt{2}(a_1 b_2 +a_2 b_1) \end{aligned} $$

that is $$\begin{aligned} a_1 a_2 +b_1 b_2 *2 &=0 \\a_1 b_2 +a_2 b_1 &=1 \end{aligned} $$

Let us look at the norm now

$$ \begin{aligned} 2= N(z_1)N(z_2) &= (a_1^2 -2 b_1^2 )(a_2^2 -2b_2 ^2 ) \\&= a_1^2 (a_2 ^2 -2b_2 ^2 )- 2b_2^2 (a_2^2 -2b_2^2 ) \\&=a_1^2 a_2^2 -2 a_1^2 b_2^2 -2 b_1^2 a_2^2 +4b_1^2 b_2^2 \\&=a_1^2 a_2 ^2 -2 (a_1^2 b_2 ^2 +b_1^2 a_2 ^2 )+4b_1^2 b_2^2 \\2&= a_1^2 a_2^2 +4b_1^2 b_2^2 -2(a_1^2 b_2^2 +b_1^2a_2^2) \end{aligned} $$

using $a_1 a_2 =-2 b_1 b_2 $

$$\begin{aligned} 2 &=a_1^2 a_2^2 +4b_1^2 b_2 ^2 -2 (a_1^2 b_2^2 +b_1^2 a_2 ^2) \\&=4b_1^2 b_2^2 +4b_1^2 b_2 ^2 -2 (a_1^2 b_2^2 +b_1^2 a_2 ^2) \\&=8b_1^2 b_2^2 -(a_1^2 b_2^2 +b_1^2 a_2 ^2) \end{aligned} $$

there should be a contradiction or $z_1$ ,$z_2$ is a unit that is its norm is $\pm 1 $ but cannot find it appreciate a hint

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    $\begingroup$ Hint: Since prime elements are irreducible, it suffices to show that the ideal $(\sqrt{2})$ is prime. But what is $\mathbb{Z}[\sqrt{2}]/(\sqrt{2})$ isomorphic to? $\endgroup$ – Richard D. James Nov 27 '17 at 16:57
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This is not a good approach to take. Instead, use the fact that $2=N(z_1)N(z_2)$. This means that $N(z_1)|2$ and $N(z_1)$ is an integer, so it takes on one of four values. Figure out what they are and you've pretty much solved the problem.

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These chains of equations are not going to help you much.

Hint: You have that $2= N(z_1)N(z_2)$. $N(z_1)$ and $N(z_2)$ are integers and $2$ is special kind of integer.

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