Let $A$ be an abelian variety over a field $k$ with $\operatorname{char}k = p$. I have a couple of questions about the proof that the $p$-torsion of $A$, i.e. $A[p^\infty]$, is Zariski dense in $A$. My reference is Edixhoven, Geer, and Moonen's book on abelian varieties (Theorem (5.30) of their working draft). Let me first sketch their argument:
They start by noting that $[p^m] = V^m \circ F^m$ so that $A[F^m]\subset A[p^m]$, so it's enough to consider a closed subscheme $Y$ of $A$ such that all inclusions $A[F^m]\to A$ factor though $Y$. Of course, we need to show that $Y = A$. Choose an affine open neighborhood $U = \operatorname{Spec}R$ of the neutral element $e\in A$, say $R = k[T_1,\dots,T_n]/(f_1,\dots,f_r)$ with $e$ corresponding to $\mathfrak m:=(T_1,\dots,T_n)$, and let $J$ be the ideal of $Y\cap U$. Basic theory of the Frobenius morphism implies that $A[F^m]$ is defined by the ideal $\mathfrak m^m = (T_1^{p^m},\dots,T_n^{p^m})$ in $R$. At this point, they consider the completion $\widehat{R}$ of $R$ at $\mathfrak m$; since $R$ is regular of dimension $g$, we can identify $\widehat{R}$ with $k[[T_1,\dots,T_g]]$ with $T_i$ mapping to generators $x_i$ of the maximal ideal of $\widehat{R}$. That $A[F^m]\to A$ factors though $Y$ implies $\operatorname{Spec}\widehat{R}/\mathfrak m^m\widehat{R}\to \operatorname{Spec}\widehat{R}$ factors though $\operatorname{Spec}\widehat{R}/J\widehat{R}$, so $J\widehat{R}\subset\mathfrak m^m\widehat{R}$ for all $m\geq 0$, and the intersection of the latter is zero (Krull's intersection theorem), so $J\widehat{R}= 0$. But then the complete local ring $\widehat{\mathcal O}_{Y,e} = \widehat{R}/J\widehat{R}$ of $Y$ has dimension $g$, hence $Y = A$.
Here are my questions.
(1) Is the purpose of considering the completion simply because $U$ might not contain all the $A[F^m]$, so we can't just compare the ideals of $A[F^m]$, $Y\cap U$, and $U$ directly (in other words, we're zooming in around $e$)?
(2) From the conclusion that $J\widehat{R} = 0$, can't we immediately conclude $J = 0$ because $R\to\widehat{R}$ is faithfully flat?
(3) I would like to propose the following alternate proof that avoids completions. To get $U$ to contain all the $A[F^m]$, one could take $U_m$ containing each $A[F^m]$ (since the $A[F^m]$ are finite, hence affine), then take a finite subcover of the $U_m$ (since $A$ is quasi-compact), and then the union $U$ of the $U_m$ would still be affine, say $U = \operatorname{Spec}R$. The coordinate ring of $A[F^m]$ is still given by $R/\mathfrak m^m$ which is the surjective image of the natural map $R/J\to R/\mathfrak m^m$ (since $A[F^m]\to U$ factors though $Y\cap U\to U$), so $J\subset\mathfrak m^m$ for all $m\geq 1$ and hence $J = 0$. Is this a correct alternate proof?