Let $X \rightarrow S$ a morphism of schemes with $S$ reduced and Noetherian over a field. Let $\mathcal{F}$ be a coherent sheaf on $X$. To show $\mathcal{F}$ is flat over $S$ does it suffice to show that whenever $R$ is a discrete valuation ring equipped with a morphism $$\mathrm{Spec}(R) \rightarrow S$$ we have $\pi^*\mathcal{F}$ is flat over $\mathrm{Spec}(R)$ where $\pi: \mathrm{Spec}(R) \times_S X \rightarrow X$.
I am aware of a related question: in the case $\mathcal{F}$ is the structure sheaf of $X$:
Flat if restriction to all curves in the base is flat
I would first like to confirm this is true, and second find a reference or proof of this fact. Perhaps we could deduce it from the link above? I am not familiar enough with yoga of sheaves/ spaces to do so, however.
Best Answer
Let $M$ be an $A$ module. We define the $A$ algebra $M_A$ to be the tensor algebra of $A$.
Now observe $M_A$ is flat over $A$ if and only if $M$ is a flat $A$ module. Indeed $M$ is a direct summand of $M_A$ considered an $A$ module. Conversely the tensor product of flat modules is flat.
We now apply the result in the linked question to the tensor algebra $\mathcal{F}_S$.