Let $\mathfrak{M}(2)$ be the algebraic stack over $\mathbb{Z}[1/2]$ which classifies the elliptic curves with the $\Gamma(2)$ level structure and let $M(2)$ be its coarse moduli space. Is there an isomorphism between $M(2)$ and $\mathbb{P}^1_{\mathbb{Z}[1/2]}$?
So firstly, you need to compactify $M(2)$ to get $\mathbb{P}^1_{\mathbb{Z}[1/2]}$. Once you do, the answers to your question here:
imply that $\overline{M(2)}_{\mathbb{Q}}\cong\mathbb{P}^1_\mathbb{Q}$. On the other hand by general nonsense, $\overline{M(2)}$ is a smooth scheme over $\mathbb{Z}[1/2]$, and hence all fibers are geometrically of genus 0.
Now, to prove that it's actually $\mathbb{P}^1_{\mathbb{Z}[1/2]}$, I believe it should suffice to produce enough distinct sections on $\overline{M(2)}$ (I think 3 should be enough, though maybe in this case 1 is enough?)
To do this, it suffices to look at the 3 cusps, which correspond via the DeligneRapoport compactification to the three nonisomorphic $\Gamma(2)$structures on a Neron 2gon. These three $\Gamma(2)$structures are defined over $\mathbb{Z}[1/2]$, and are not geometrically isomorphic over any prime $p\ne 2$, and hence the three cuspidal sections do not intersect over any prime $p\ne 2$. This shows that all fibers of $\overline{M(2)}$ are isomorphic to $\mathbb{P}^1$ over the residue field, and my intuition is that the existence of these three sections serve to "rigidify the deformation problem" (or something) to guarantee that the surface is indeed just $\mathbb{P}^1_{\mathbb{Z}[1/2]}$ (I would be grateful if someone could elaborate on this)
Remark: Note that we can always produce a noncuspidal section, namely, the elliptic curve $$E : y^2 = x^3x$$ which has discriminant $\Delta = 64$, and hence defines an elliptic curve over $\mathbb{Z}[1/2]$, together with the points $(0,0),(1,0)$ which define a $\Gamma(2)$structure on $E$ over $\mathbb{Z}[1/2]$. This section together with the 3 cuspidal sections never intersect over any prime $p\ne 2$, and indeed forms a maximal set of nonintersecting sections, since at $p = 3$, $\mathbb{P}^1_{\mathbb{F}_3}$ only has four $\mathbb{F}_3$rational points!
Appendix: A Neron $n$gon is a chain of $n$ copies of $\mathbb{P}^1$'s, where $0$ in one copy is glued to $\infty$ in the next copy. A Neron 1gon is just a nodal cubic. The smooth locus of an $n$gon is isomorphic to $\mathbb{G}_m\times\mathbb{Z}/n\mathbb{Z}$, and is equipped with the obvious group structure, where $\mathbb{Z}/n\mathbb{Z}$ is viewed as a constant group scheme. It's $n$torsion is thus just $\mu_n\times\mathbb{Z}/n\mathbb{Z}$. For $n = 2$, we find that over $\mathbb{Z}[1/2]$, its 2torsion is $\mu_2\times\mathbb{Z}/2\mathbb{Z}$, which is completely decomposed as a finite etale group scheme over $\mathbb{Z}[1/2]$, and thus admits three nonisomorphic bases (ie, three nonisomorphic $\Gamma(2)$structures), each of which corresponds to a cuspidal section of $\overline{M(2)}$.

$\begingroup$ I forget to add in my question '' 2nero polygone ''. In your answer you said that smooth implies irreducible, I think that is not true because we need to know that our space is connected and which is true by a result of Zariski since $(X(2)(\mathbb{C})$ is connected. Exemple $X(3)(\mathbb{C})$ is not connected but $X(3)$ is smooth. $\endgroup$ – Adel BETINA Oct 19 '16 at 21:26

$\begingroup$ @AdelBETINA Where did I say "smooth implies irreducible"? Anyway, $X(2)$ is certainly connected. $\endgroup$ – Will Chen Oct 19 '16 at 21:28

$\begingroup$ I want to say how to prove that the fibre is reduced at $2$? $\endgroup$ – Adel BETINA Oct 19 '16 at 21:35

$\begingroup$ Since in your answer you think that the normalisation of $X(2)$ over the jline is $\mathbb{P}^1_{\mathbb{Z}}$ $\endgroup$ – Adel BETINA Oct 19 '16 at 21:36

$\begingroup$ @AdelBETINA I don't think that's even true, at least if you take the KatzMazur regular model of $M(2)$. $\endgroup$ – Will Chen Oct 19 '16 at 21:36