 Solution

On $D_{+}(x_3)$, write $r=x_0/x_3$, $q=x_1/x_3$, and $s=x_2/x_3$. The equations include $q=\pi s^2$ and $r=s^3$; after making these substitutions, the remaining equations are identities. ThereforeThe other chart isThere is a homomorphism $B\to R[t]$ defined by $a\mapsto \pi t$, $b\mapsto t^2$, $c\mapsto t^3$. To prove injectivity rather than merely exhibit a parametrization, reduce monomials using the four relations. Powers of $a$ reduce to at most one $a$; any remaining product of $a$ with $b$ or $c$ eliminates $a$; powers of $c$ reduce to at most one $c$. Thus every element is an $R$-linear combination ofTheir images are $1,\pi t,t^{2n},t^{2n+3}$, with pairwise distinct degrees. Since $R$ is a domain and $\pi \ne 0$, these images are $R$-linearly independent. Hence they are a basis, the map is injective, andIn particular $B$ is torsion-free, indeed free, as an $R$-module.

The two charts cover $X$ by the prime-ideal argument in part i. Both coordinate rings are torsion-free over the discrete valuation ring $R$, hence flat: over a principal ideal domain a torsion-free module is a filtered union of finitely generated free submodules, and filtered colimits preserve flatness. Thus the morphism $X\to \mathrm{Spec}R$ is flat on an open cover and therefore flat everywhere. It is proper because $X$ is a closed subscheme of the proper projective space ${\mathbb{P}}_R^3$. Consequently $X$ is both proper and flat over $R$.