Show that $b \in N(\mathbb{Q}(\sqrt{a}))$ if and only if $b' \in N(\mathbb{Q}(\sqrt{a}))$

Question

Consider a quadratic form $f= z^2 - ax^2 - by^2$ ; $a,b \in \mathbb{Z}$.

Let $bb' = t^2 - a$ for some $b'$. Then I need to show that $b \in N(\mathbb{Q}(\sqrt{a}))$ if and only if $b' \in N(\mathbb{Q}(\sqrt{a}))$.

Here, $N(\mathbb{Q}(\sqrt{a}))$ denoted the group of norms of the elements of $\mathbb{Q}(\sqrt{a})$. This norm is defined as $N(u + v\sqrt{a}) = (u + v\sqrt{a})(u - v\sqrt{a}) = u^2 - av^2$

Therefore, (for $\Rightarrow$), I can express $b$ as equal to $u^2 - av^2$ ($u,v \in \mathbb{Q}$). So, $$(u^2 - av^2)b' = t^2 - a$$ But from this I am not able to express $b'$ as equal to some $x^2 - ay^2$ ($x,y \in \mathbb{Q}$). Any help here!

I suppose the other direction $\Leftarrow$ is also somewhat similar. So can anyone help me in proving $\Rightarrow$ ?

Answer 1

Here $t^2-a=N(t+\sqrt a)$ where $t+\sqrt a\in\Bbb{Q}(\sqrt a)$. If $b=N(z)$ for some $z\in\Bbb{Q}(\sqrt a)$ then $$ b'=\frac{t^2-a}{b}=\frac{N(t+\sqrt a)}{N(z)}=N(\frac{t+\sqrt a}{z}). $$ Because $\Bbb{Q}(\sqrt a)$ is a field we have $(t+\sqrt a)/z\in\Bbb{Q}(\sqrt a)$.