Zeroth degree part of localization of a flat graded $A$-module is flat. This question arises from the proof of Thm 3.9.9 in Hartshorne. Let $T$ be an integral noetherian scheme and $X \subseteq \mathbb{P}^n_T$ be a closed subscheme. The proof shows that the following are equivalent

(i) $\mathscr{F}$ is flat over $T$;
(ii)$H^0(X,\mathscr{F}(m))$ is a free $A$-module of finite rank, for all $m \gg 0$.

In showing that (ii) implies (i), he let $M = \oplus_{m\ge m_0}H^0(X,\mathscr{F}(m))$ where $m_0$ is a large integer s.t.  $H^0(X,\mathscr{F}(m))$ are all free. Then $\mathscr{F} = \tilde{M}$. But then he says that since $M$ is a free hence flat $A$-module, $\mathscr{F}$ is also flat over $A$. However, the stalk of $\mathscr{F}$ is not obtained simply by localization. We must localize $M$ and take the $0$th degree part. So how can we show that zeroth degree part of localization of a flat graded $A$-module is flat.

Lemma (Stacks 00JR): Let $S$ be a graded ring. Let $M$ be a graded $S$-module. Let $\mathfrak{p}$ be an element of $\operatorname{Proj} S$. Let $f\in S$ a homogeneous element of positive degree so that $f\notin\mathfrak{p}$, i.e., $\mathfrak{p}\in D_+(f)$. Let $\mathfrak{p}'\in S_{(f)}$ be the element of $\operatorname{Spec}(S_{(f)})$ corresponding to $\mathfrak{p}$. Then $S_{(\mathfrak{p})} = (S_{(f)})_{\mathfrak{p}'}$ and compatibly $M_{(\mathfrak{p})} = (M_{(f)})_{\mathfrak{p}'}$.

Hartshorne's reduced to the case $T=\operatorname{Spec} A$ for $A$ a local noetherian domain. Let $S=A[x_0,\cdots,x_n]$, $\mathfrak{p}\subset S$ a graded prime, and $M$ a graded flat $S$-module. Since $D_+(x_i)$ cover $\Bbb P^n_A$, we may choose $x_i$ so that $\mathfrak{p}\in D_+(x_i)$. Applying the lemma, we see that $M_{(\mathfrak{p})} = (M_{(x_i)})_{\mathfrak{p}'}$.
Next, we note that $S_{x_i}=S_{(x_i)}[x_i,x_i^{-1}]$, so $S_{(x_i)}\to S_{x_i}$ is a flat extension of rings. Therefore as $M_{x_i}$ is a flat $S_{x_i}$-module, it is also a flat $S_{(x_i)}$-module. Since a direct sum of modules is flat iff each summand is and$M_{x_i}$ is the direct sum of its graded pieces, we see that $M_{(x_i)}$ is a flat $S_{(x_i)}$-module. Localizing at $\mathfrak{p'}$, we see that $(M_{(x_i)})_{\mathfrak{p}'}$ is a flat $(S_{(x_i)})_{\mathfrak{p}'}$-module, and applying the lemma again we see that this means that $M_{(\mathfrak{p})}$ is a flat $S_{(\mathfrak{p})}$-module.
