If $f:X\to S$ is a morphism of schemes, and $S$ is locally Noetherian, then $f$ being finite and flat is equivalent to the sheaf $f_*\mathcal{O}_X$ being a finite locally free $\mathcal{O}_S$-module. (See https://stacks.math.columbia.edu/tag/02K9)
If $S=\text{Spec} k$, then this essentially asks that $\Gamma(X,\mathcal{O}_X)$ is a finite free module over $k$. All modules over a field are free, so as long as $f$ is a finite map this should hold.
But then this means that every finite scheme over a field is flat. This gives me pause because I have often seen people talking about finite flat group schemes. I understand that group schemes over a field are not the only group schemes of interest, but I haven't seen anyone claim or imply somehow that finite group schemes are automatically flat over a field, so I just want to be sure I haven't made a mistake or overlooked anything.
Best Answer
Indeed, all schemes over a field are flat. This is immediate from the definition: to check whether $f:X\to\operatorname{Spec} k$ is flat, you have to check whether the restriction to affine open subschemes $\operatorname{Spec}A\to\operatorname{Spec} k$ is flat, which just means that $A$ is flat as a $k$-module. But since $k$ is a field, every $k$-module is free, and in particular flat.