Let $R$ be the commutative unital ring $\mathbb{Z}[x, y, z]/(x^3-y^2-1728z)$. Let $p$ be a prime number, assume $p>3$. Is it true that $R\otimes_{\mathbb{Z}} \mathbb{F}_p\approx \mathbb{F}_p[u, v]$?
My argument is that when we reduce the relation mod $p$, we get that $z$ is a unit times $x^3-y^2$, so it can be expressed in terms of two generators which satisfy no non-trivial relations among themselves.
Yes, your idea is correct. To make it into a rigorous proof, note that there is a homomorphism $f:R\otimes\mathbb{F}_p\to\mathbb{F}_p[u,v]$ which maps $x$ to $u$, $y$ to $v$, and $z$ to $1728^{-1}(u^3-v^2)$, and there is also a homomorphism $g:\mathbb{F}_p[u,v]\to R\otimes\mathbb{F}_p$ which maps $u$ to $x$ and $v$ to $y$. Clearly $fg=1$ and $gf=1$ by looking at what they do on the generators (for $gf=1$ you use the fact that $z=1728^{-1}(x^3-y^2)$ in $R\otimes\mathbb{F}_p$), and so they are inverse isomorphisms.
