Flatness descends along faithfully flat base change

If $N$ is flat over $R$, then $N{\otimes }_{R}S$ is flat over $S$ by [L1].

Conversely, assume $N{\otimes }_{R}S$ is flat over $S$. Let $A\to B\to C$ be an exact sequence of $R$-modules. Tensoring with $N$ and then with $S$ gives $((A{\otimes }_{R}N){\otimes }_{R}S)\to ((B{\otimes }_{R}N){\otimes }_{R}S)\to ((C{\otimes }_{R}N){\otimes }_{R}S)\mathrm{.}$ Associativity of tensor product identifies this with $A{\otimes }_R(N{\otimes }_{R}S)\to B{\otimes }_R(N{\otimes }_{R}S)\to C{\otimes }_R(N{\otimes }_{R}S),$ which is exact because $N{\otimes }_{R}S$ is flat over $S$ and [L2] transports that flatness back along $R\to S$.

Since $R\to S$ is faithfully flat, [L3] reflects exactness. Therefore the sequence $A{\otimes }_{R}N\to B{\otimes }_{R}N\to C{\otimes }_{R}N$ was already exact, so $N$ is flat over $R$.

Hence flatness descends and ascends along faithfully flat base change.