Let $G$ be a smooth scheme over $S$ of characteristic $p$, do we have that the reltaive Frobenius morphism $F\_{G/S}$ is faithfully flat?

There is an excersice in Liu's book saying that this is true when $S$ is the spectrum of a field, it's [Exercise 3.13](https://www.math.u-bordeaux.fr/~qliu/Book/page-145.pdf). I think it should be possible to generilize it, feel free to add nice properties to $S$ or $G$, e.g. Noetherian, finite type $k$-schemes.

Because in my case, $S=\DeclareMathOperator{\Spec}{Spec} \Spec A/\mathfrak{m}^n$, where $(A,\mathfrak{m},k)$ is a discrete valuation ring, and $G=\mathbb{G}\_{m,S}=\Spec A/\mathfrak{m}^n[T,T^{-1}]$ the multiplicative group scheme over $S$, and it's smooth over $S$ since $\Spec \mathbb{Z}[T,T^{-1}]$ is smooth over $\Spec \mathbb{Z}$, and I want to show that $F\_{G/S}:G\rightarrow G^{(p)}$ is faithfully flat.