# conics isomorphic to $\mathbb{P}^1$ over $\mathrm{Spec}(\mathbb{Z})$

Let $C \subset \mathbb{P}^2$ be a smooth non-sigular conic over $\mathbb{Q}$, say given by an equation $f(x, y, z)=0$, where $f$ is a homogeneous polynomial of degree two with rational coefficients.

I know that $C$ is isomorphic to $\mathbb{P}^1$ if and only if $C$ has a rational point.

What happens over $\mathrm{Spec}(\mathbb{Z})$? That is, assume that $f$ has integer coefficients, so I can see $C$ as a subvariety of $\mathbb{P}^2_{\mathbb{Z}}$. Is it true that $C$ is isomorphic to $\mathbb{P}^1_{\mathbb{Z}}$ if and only if $C(\mathbb{Z})$ is not empty?

Does it mean that there are conics isomorphic to $\mathbb{P}^1$ over $\mathbb{Q}$ but not over $\mathbb{Z}$ or everything is the same because I can clear denominators?

Let $C$ be a smooth projective scheme over $\mathbb Z$ such that $C_\eta$ is a conic over $\mathbb Q$.
Then $C(\mathbb Z) = C_\eta(\mathbb Q)$ by the valuative criterion for properness.
Therefore, the existence of a section implies that the generic fibre is a split conic. In particular, $C_\eta$ is isomorphic to $\mathbb P^1_{\mathbb Q}$.
Also, every special fibre is split: for all primes $p$ of $\mathbb Z$, the smooth conic $X_p = X\otimes \mathbb F_p$ has a rational point. Therefore all the fibres of $X$ over $\mathbb Z$ are projective lines.
This implies that $X$ itself is isomorphic to $\mathbb P^1_{\mathbb Z}$.