Any binary quadratic $\mathbb{Z}$-form $q$ induces a symmetric bilinear form
$$ B_q(u,v) = q(u+v) - q(u) -q(v) \ \ \forall u,v \ \in\mathbb{Z}^2 $$
and it is considered *non-degenerate* (over $\mathbb{Z}$) if its discriminant

$$ \text{disc}(q) := \det(B_q(e_i,e_j)_{1 \leq i,j \leq 2}) $$
where $e_1 = (1,0)$ and $e_2 = (0,1)$ is invertible in $\mathbb{Z}$, i.e., equals $\pm 1$: see (2.1) in: http://math.stanford.edu/~conrad/papers/redgpZsmf.pdf .

Suppose $q$ **is** degenerate, but still $\text{disc}(q) \neq 0$ (so it is non-degenerate over $\mathbb{Q}$). So its *special orthogonal group scheme* $SO_q$ defined over $\text{Spec} \ \mathbb{Z}$, does not have to be smooth, but it is flat as $\mathbb{Z}$ is Dedekind (loc. sit. Definition 2.8 and right after), and it is closed in the full *orthogonal group* $O_q$, whence the quotient $Q:=O_q/SO_q$ is representable.

My question is: Is $Q$ a finite group of order $2$ over $\text{Spec} \ \mathbb{Z}$ ?

Apparently, when applied to any integral domain $R$ which is an extension of $\mathbb{Z}$, the elements of $O_q(R)$, in some matrix realization, must be of $\det \pm 1$, so we could think of $Q$ as a $\mathbb{Z}$-group of order $2$, but as a functor of points, $O_q$ can be applied to any $\mathbb{Z}$-algebra $R$, for which we may find elements of $O_q(R)$ which are not of $\det = \pm 1$.

For example, let $q(x,y)=x^2+y^2$. One can verify it is degenerate.
We get $SO_q = \text{Spec} \ \mathbb{Z}[x,y]/(x^2+y^2-1)$. Consider its matrix realization $$ \left \{ A=\left( \begin{array}{cc}
x & -y \\
y & x \\
\end{array}\right): \det(A)=1 \right \}. $$
Then the component of $\det = -1$ elements in $O_q$ is obtained by $\text{diag}(1,-1)SO_q$.
So apparently, $Q = \mu_2$ (which unlike the other order $2$ group $(\mathbb{Z}/2)_\mathbb{Z}$, it has a double point at the reduction at $(2)$, not two distinct ones).

So far everything is good. But $A=\text{diag}(3,1)$ belongs to $O_q(R)$ where $R=\mathbb{Z}/8$ (as $A^T \cdot A = I_2$ in $R$, where $I_2$ represents $q$), but $\det(A)\neq \pm 1$ in $R$ !
Does it mean that $O_q$ is more than these two connected components ?

I thought to avoid this problem by considering $O_q$ and $SO_q$ as flat sheaves (in the small site of flat extensions of $\mathbb{Z}$, since what I really care is of $H^1_\text{fppf}(\mathbb{Z},O_q)$), but we may still find extensions such as $\mathbb{Z} \times \mathbb{Z}$ containing a square root of unity other than $-1$ ?!

Could you please help ?

Thank you !

Rony

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