Let $X$ be a smooth morphism over a scheme $S$ of positive characteristic $p > 0$. The morphism $F_S: S \to S$ induced by the ring homomorphism $O_S \to O_S : a \mapsto a^p$ the absolute Frobenius of $S$. We let $X^{(p)} $ denote the fibered product $X \times_S S$, where the second factor $S$ is endowed with the structure of an S-scheme via the absolute Frobenius $F_S : S \to S$. The relative Frobenius $F_{X/k} : X \to X^{(p)}$ fits in following diagram in Definition 33.36.4 from Stacks Project.
Exercise 4.3.13 in Qing Liu's Algebraic Geometry and Arithmetic Curves asks of us to prove
(a) Show that $X^{(p)} \to S $ is smooth (use Proposition 3.38). (b) Let us suppose that $S$ is the spectrum of a field $k$. Let $F_{X/k} : X \to X^{(p)}$ be the relative Frobenius. Show that $F_{X/k}$ is flat at the rational points of $X$ (use Proposition 2.27). (c) Show that $F_{X/k}$ is faithfully flat, that is flat and surjective
and I'm having trouble in solving some parts there.
(a) follows simply from smoothness of $X \to S$ by base change.
(b) could work like this: $X$ is smooth and therefore at every point rational point $x$ the $\mathfrak{m}$-adic completion of the stalk $O_{X,x}$ at it's maximal ideal $\mathfrak{m}$ is
$$\hat{O}_{X,x} \cong k[[T_1,..., T_d]] $$
with $d = \dim O_{X,x}.$
Therefore at $x$ the map on stalks $F_{X/k}: O_{X,x} \to O_{X,x}$ induces a map on the completions $k[[T_1,..., T_d]]$ which should be determined by $T_j \mapsto T_j^p$.
Why is the last map flat and how can I deduce the flatness of $F_{X/k}$ at $x$? If $d=1$ then $k[[T_1]]$ is principal ideal domain and since the map $k[[T_1]] \to k[[T_1]], T_1 \mapsto T_1^p$ is torsion free therefore flat and we win. What about case $d > 1$ und how to deduce flatness of $F_{X/k}$ in $O_{X,x}$.
(c) No idea how to prove that $F_{X/k}$ flat. Flatness commutes base change so that it's sufficient to show it at closed points. But not all closed points are rational, so I not see how (b) helps.