(The affine line over the integers)
\(\AA^1_\ZZ = \Spec \ZZ[x] \to \Spec \ZZ\) is flat, because \(\ZZ[x]\) is a free \(\ZZ\)-module with basis \(1, x, x^2, \ldots\), and free modules are flat. More generally \(\AA^n_R \to \Spec R\) is flat for every ring \(R\).
An \(R\)-module \(M\) is faithfully flat if and only if \(M\) is flat and the functor \(- \otimes_R M\) on \(R\)-modules is faithful.
Flatness is needed on both sides: faithfulness of \(- \otimes M\) alone does not give flatness. For \(R = \ZZ\) and \(M = \ZZ \oplus \ZZ/2\), \(- \otimes M\) is faithful because \(M\) has \(\ZZ\) as a direct summand, but \(M\) is not flat, since \(\ZZ \xrightarrow{2} \ZZ\) tensored with \(\ZZ/2\) is the zero map on \(\ZZ/2 \neq 0\).
Show that $\AA^1_\ZZ$ is flat over $\Spec \ZZ$.
Show that $M$ is faithfully flat if and only if $M$ is flat and $- \otimes_R M$ is a faithful functor.
Dependencies
D-DEFFFLATFaithfully flat modules, and descent of exactnessuses