What is $\mathrm{O}_q/\mathrm{SO}_q$ if $q$ is a quadratic $\mathbb{Z}$-form which is degenerate?

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 $$-1$$ ?!

Thank you !

Rony

• I am aware of the fact that $SO_q$ should not necessarily be the kernel of $\det$, it can be also the kernel of the Dickson morphism which leads to $Q = (\mathbb{Z}/2)_{\mathbb{Z}$, this is why I asked generally is it a $2$-order group (the $\det =\pm 1$ elements were mentioned in the context of integral domains that are extensions of $\mathbb{Z}$). – Rony Bitan Nov 8 '18 at 7:08
• This is the continue of the above comment:..leads to $Q = (\mathbb{Z}/2)_{\mathbb{Z}}$,this is why I asked generally is it a $2$−order group (the $\det =\pm 1$ elements were mentioned in the context of integral domains that are extensions of $\mathbb{Z}$). – Rony Bitan Nov 8 '18 at 7:14
• Hi. I don't understand your "counter example". The matrix corresponding to $x^2+y^2$ is the identity matrix, which has determinant $1$, hence your quadratic form IS regular ! Anyway, of course your $A$ has not determinant $\pm 1$, but this is not surprising. Any element $A$ in $O_q(R)$ has a determinant which lies in $\mu_2(R)$. But $\mu_2(R)$ are elements $r$ of $R$ such that $r^2=1$. It is $\pm 1$ when $R$ is a domain, but can get bigger if not (eg $R=\mathbb{Z}/8$). – GreginGre Nov 8 '18 at 9:31
• Dear Gregin. This depends on how you define regularity. According to Conrad's paper above one gets disc(q)=4. The reason why I mention it is because if it were regular, the quotient is described there. My question is: if $O_q$, as a scheme, is a disjoint union of two components: $SO_q$ and $O_q^-$, then A should belong to $SO_q(R)$ or to $O_q^-(R)$, am I write ? So to which one it does ? If: $A \in O_q^-(R)$ how this goes with: $O_q^- : \text{Spec} \mathbb{Z}[x,y] / (x^2+y^2=-1)$ ? – Rony Bitan Nov 8 '18 at 11:08
• I mean even if you tensor $O_q^-$ with $R=\mathbb{Z}/8$, still $(x,y)=(3,1)$ does not solve $x^2+y^2=-1$ in $R$. – Rony Bitan Nov 8 '18 at 11:24

If indeed $$\textbf{O}_q = \textbf{O}_q^+ \cup \textbf{O}_q^-$$ defined over $$\text{Spec} \, \mathbb{Z}$$, then applied as a functor of points to some ring $$R$$ (being a $$\mathbb{Z}$$-algebra), how could exist a point in $$\textbf{O}_q(R)$$ which is neither in $$\textbf{O}_q^+(R)$$ nor in $$\textbf{O}_q^-(R)$$ ?
The answer, more generally, is that given a union of schemes $$X = X_1 \cup X_2$$ which is not disjoint, and a ring $$R$$ which is not a domain (having some zero-divisors), then $$X(R)$$ should not be equal to $$X_1(R) \cup X_2(R)$$, because a union of schemes corresponds to the multiplication of their underlined varieties. This equality holds for domains, however.
The simplest example is the group $$\mu_2$$ defined over $$\text{Spec}\,\mathbb{Z}$$ being the union of the two components whose varieties are $$x=1$$ and $$x=-1$$. These coincide at the prime $$(2)$$. The point $$3$$ belongs to $$\mu_2(R=\mathbb{Z}/8)$$ though not to any of the components $$R$$-points, since $$(3-1)(3+1) \equiv 0 (\text{mod}8)$$. This means that not any point of $$\mu_2(R)$$, considered as a morphism $$\text{Spec}\,R \to \mu_2$$, can be lifted to a morphism from $$\text{Spec} \, R$$ to the disjoint union of $$\text{Spec} \,\mathbb{Z}[t]/(t-1)$$ and $$\text{Spec} \,\mathbb{Z}[t]/(t+1)$$.
Since $$q$$ is assumed to be non-degenerate over $$\mathbb{Q}$$ and invertible matrices over $$\mathbb{Z}$$ can only have determinant $$\pm 1$$, the above union is the all orthogonal group. As $$\textbf{O}_q^+$$ is a closed flat subgroup, $$\textbf{O}_q/\textbf{O}_q^+$$ is a finite $$\mathbb{Z}$$-group of order $$2$$ (can be either $$\mu_2$$ or $$(\mathbb{Z}/2)_\mathbb{Z}$$).