<s>[INST] Norms in quadratic fields This should be well-known, but I can't find a reference (or a proof, or a counter-example...). Let $d$ be a positive square-free integer. Suppose that there is no element in the ring of integers of $\mathbb{Q}(\sqrt{d})$ with norm $-1$. Then I believe that no element of $\mathbb{Q}(\sqrt{d})$ has norm $-1\ $
(in fancy terms, the homomorphism $H^2(G,\mathscr{O}^*)\rightarrow H^2(G,\mathbb{Q}(\sqrt{d})^*)$, with $G:=\operatorname{Gal}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})=$ $\mathbb{Z}/2 $, is injective). Is that correct? If yes, I'd appreciate a proof or a reference.
 [/INST] Dirichlet's version of Gauss composition is in the book by Cox, (page 49 in first)  with a small typo corrected in the second edition.
For our purpose, duplication, it has a better look to equate $a=a'$ from the start, with $\gcd(a,b) = 1$ sufficing,
$$ \left( ax^2 +bxy+ acy^2 \right)  \left( aw^2 +bwz+ acz^2 \right) = c X^2 + b XY + a^2 Y^2 $$
where
$$  X = axz + ayw+byz \; \; , \; \; \;  Y = xw - c yz  $$
so that the square of $\langle a,b,ac \rangle$   is $\langle c,b,a^2 \rangle.$
Today's question concerns $c=-1$
$$ \left( ax^2 +bxy -ay^2 \right)  \left( aw^2 +bwz -az^2 \right) = - X^2 + b XY + a^2 Y^2 $$
where
$$  X = axz + ayw+byz \; \; , \; \; \;  Y = xw + yz  $$
so that  $$\langle a,b,-a \rangle^2 =   \langle -1,b,a^2 \rangle.$$
We also see Stanley's fact that the discriminant is the sum of two squares,  $b^2 + 4 a^2$ the way I wrote things.
By the Gauss theorem on duplication, $  \langle -1,b,a^2 \rangle$ is in the principal genus
Furthermore, we now know that  the principal form is $SL_z \mathbb Z$ equivalent to
$$  \langle 1,b,-a^2 \rangle  $$
The principal form may not integrally represent $-1$ but does so rationally.
As to being in the same genus, we can use Siegel's definition of rational equivalence without essential denominator.
$$
\left(
\begin{array}{rr}
 0  & 1 \\
 -a^2 & -b \\
\end{array}
\right)
\left(
\begin{array}{rr}
 1 & \frac{b}{2} \\
 \frac{b}{2} & -a^2  \\
\end{array}
\right)
\left(
\begin{array}{rr}
 0 & -a^2 \\
 1 & -b \\
\end{array}
\right) = \; a^2 \; 
\left(
\begin{array}{rr}
 -1 & \frac{b}{2} \\
 \frac{b}{2} & a^2  \\
\end{array}
\right)
$$
$$
\left(
\begin{array}{rr}
 b  & 1 \\
 -a^2 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rr}
 -1 & \frac{b}{2} \\
 \frac{b}{2} & a^2  \\
\end{array}
\right)
\left(
\begin{array}{rr}
 b & -a^2 \\
 1 & 0 \\
\end{array}
\right) = \; a^2 \; 
\left(
\begin{array}{rr}
 1 & \frac{b}{2} \\
 \frac{b}{2} & -a^2  \\
\end{array}
\right)
$$
</s>