Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 126 / 1 / b / Solution

A morphism of schemes $f:X\to Y$ is a flat morphism when every local-ring map ${\mathcal{O}}_{Y,f(x)}\to {\mathcal{O}}_{X,x}$ makes ${\mathcal{O}}_{X,x}$ a flat module.

In (i), the coordinate map is $k[t]\to k[y]$, $t\mapsto y^2$, andIt is therefore a free module of rank two and the morphism is flat, including in characteristic two.

In (ii), $k[x]$ is finite over the cusp ring $R=k[x^2,x^3]$ and has generic rank one. Were it flat, finite flatness over the local ring at the cusp would make it free of rank one. Its fiber there is insteadwhich has dimension two, so this morphism is not flat.

In (iii), the base coordinate $t$ acts as $x$, and the nonzero element $y$ satisfies $ty=xy=0$. Thus the coordinate ring has torsion as a $k[t]$-module. Since $k[t]$ is a principal ideal domain and a module over it is flat exactly when it is torsion-free, this morphism is not flat. Consequently