Direct sums and direct summands of flat modules are flat

Let $\{M_i{\}}_{i\in I}$ be flat and put $M={⨁}_{i\in I}{M}_i$. For any exact sequence $A\to B\to C$, [L2] gives $(A{\otimes }_{R}M)\to (B{\otimes }_{R}M)\to (C{\otimes }_{R}M)$ as the direct sum over $i$ of the exact sequences obtained by tensoring with $M_i$. Therefore the displayed sequence is exact, so $M$ is flat by [L1].

Suppose $F\cong N\oplus N^{\mathrm{\prime }}$ is flat. For any exact sequence $A\to B\to C$, tensoring with $F$ gives $(A{\otimes }_{R}N)\oplus (A{\otimes }_R{N}^{\mathrm{\prime }})\to (B{\otimes }_{R}N)\oplus (B{\otimes }_R{N}^{\mathrm{\prime }})\to (C{\otimes }_{R}N)\oplus (C{\otimes }_R{N}^{\mathrm{\prime }})\mathrm{.}$ If an element of $B{\otimes }_{R}N$ maps to zero in $C{\otimes }_{R}N$, then the same element viewed in the direct sum lies in the image of $A{\otimes }_{R}F$ because $F$ is flat. Projecting back to the first summand shows exactness for tensoring with $N$. Thus $N$ is flat.

Steps 1.1 and 1.2 prove the two claims.