Syzygies of a base-flat module over a flat algebra stay base-flat and fibrewise exact

Put $K_0=M$ and, for $j\ge 1$, write the resolution as short exact sequences $0\to K_j\to F_{j-1}\to K_{j-1}\to 0.$ The base $K_0=M$ is $R$-flat by hypothesis, and each $F_{j-1}$ is $R$-flat by [F2].

Suppose $K_{j-1}$ is $R$-flat. By [F1], the sequence in step 1.1 stays short exact after tensoring with any $R$-module $N$. To show $K_j$ is flat, take any injection $N^{\mathrm{\prime }}\hookrightarrow N$ and compare the two tensored short exact rows. The vertical map $F_{j-1}{\otimes }_R{N}^{\mathrm{\prime }}\to F_{j-1}{\otimes }_{R}N$ is injective by [F2]. Since $K_j{\otimes }_R{N}^{\mathrm{\prime }}$ injects into the first free-module tensor, an element killed by $K_j{\otimes }_R{N}^{\mathrm{\prime }}\to K_j{\otimes }_{R}N$ must already be zero. Thus this latter map is injective, and [F2] makes $K_j$ $R$-flat. Repeat for $j=1,\dots ,n$.

Each short exact sequence of step 1.1 remains exact after tensoring with $R^{\mathrm{\prime }}$ by [F1], because its quotient $K_{j-1}$ is now known to be $R$-flat. Splicing these tensored sequences yields exactness of the full base-changed resolution. Localization preserves exactness by [F3], so the same is true at every prime of its base-changed algebra. No choice principle is used: the free resolution is part of the given data.