Exercise 2.1.57 (Multiplicativity of the norm in $\mathbb{Q}(\sqrt{-2})$)

Question

Show that $(a+b \sqrt{-2})(c+d\sqrt{-2}) = (ac -2bd) +(bc+ad)\sqrt{-2}$. Thus or otherwise show that 
$(a^2 + 2b^2)(c^2 + 2d^2) = (ac - 2bd)^2 + 2(bc+ad)^2$.

Richard Ganaye · Accepted Answer

\begin{proof} Let $a,b,c,d$ be integers (or rational numbers). We choose $\sqrt{-2} = i \sqrt{2}$ (same result if we choose $\sqrt{-2} = -i \sqrt{2}$). Then
\begin{align*}
(a+b \sqrt{-2})(c+d\sqrt{-2}) &= (a + i \sqrt{2} b)(c +  i \sqrt{2}d)\
&= (ac -2bd) + i \sqrt{2} (bc + ad)\
&= (ac -2bd) +(bc+ad)\sqrt{-2}.
\end{align*}
Then
\begin{align*}
(a^2 + 2b^2)(c^2 + 2d^2) &= [(a + i \sqrt{2} b)(a - i \sqrt{2} b)][ (c +  i \sqrt{2}d) (c -  i \sqrt{2}d)]\
&=[(a + i \sqrt{2} b)(c +  i \sqrt{2}d)][  (a - i \sqrt{2} b)(c -  i \sqrt{2}d)]\
&= [(ac -2bd) + i \sqrt{2} (bc + ad)][ (ac -2bd) - i \sqrt{2} (bc + ad)]\
&= (ac - 2bd)^2 + 2(bc+ad)^2.
\end{align*}
This shows
$$(a^2 + 2b^2)(c^2 + 2d^2) = (ac - 2bd)^2 + 2(bc+ad)^2.$$
\end{proof}

Exercise 2.1.57 (Multiplicativity of the norm in $\mathbb{Q}(\sqrt{-2})$)

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Exercise 2.1.57 (Multiplicativity of the norm in $\mathbb{Q}(\sqrt{-2})$)

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