Is this module flat?

Question

Let $k$ be a field. How do we prove that the $k[x]$-module $k[x,y,z]/(yz-x)$ is flat? Where the action is given by sending $p(x)$ to its class.

I've tried to prove that it sends monomorphisms to monomorphisms, but haven't been successful.

Answer 1

This holds for every commutative ring $k$. It suffices to prove that $k[y,z]$ becomes a flat $k[x]$-module when $x$ acts via multipliciation with $yz$. In fact, it is a free module.

We have $x^i \cdot y^j = y^{i+j} z^i$ and $x^i \cdot z^j = y^i z^{i+j}$. This shows that the module is generated by $\{y^i : i \geq 0\} \cup \{z^i : i > 0\}$. The set $\{y^i : i \geq 0\}$ generates the submodule of polynomials where in each monomial the $y$-degree is $\geq $ the $z$-degree, and $\{z^i : i > 0\}$ generates the submodule of polynomials where in each monomial the $y$-degree is $<$ the $z$-degree. Hence, the generated submodules intersect trivially. It remains to show that $\{y^i : i \geq 0\}$ is independent (then likewise for $z$ by symmetry). If $p_i=\sum_{j} a_{ij} x^j \in k[x]$ satisfy $\sum_i p_i(x) \cdot y^i = 0$, then $\sum_{i,j} a_{ij} y^{i+j} z^j = 0$ in $k[y,z] = k[y][z]$, hence $\sum_i a_{ij} y^{i+j}=0$ for each $j$, hence $a_{ij}=0$ for each $i$.

Answer 2

$k[x]$ is a PID and in particular a Dedekind domain, thus flat coincides with torsion-free. On your module, $x$ acts on $k[y,z]$ by multiplication with $yz$, this is clearly torsion-free.