Yoga on coherent flat sheaves $\mathcal{F}$ over projective space $\mathbb{P}^n$

Question

I'm reading Mumfords's Lectures on Curves on an Algebraic Surface (jstor-link: https://www.jstor.org/stable/j.ctt1b9x2g3) and I found in Lecture 7 (RESUME OF THE COHOMOLOGY OF COHERENT SHEAVES ON $\mathbb{P}^n$; p 47) dealing with yoga on coherent sheaves $F$ over pojective space $\mathbb{P}^n$ I found on page 52 a proof I not understand:

Corollary 3: Given a coherent sheaf $\mathcal{F}$ on $\mathbb{P}^n \times S$, $\mathcal{F}$ is flat over $S$ if and only if there exists an $m_0$ such that if $m \ge m_0$, $p_* \mathcal{F}(m)$ is locally free. Hence, in this case, the Hilbert polynomial of on $\mathbb{P}^n _s$ is locally constant.

Proof: If $F$ is flat over $S$, then let $m_0$ be large enough so that derived image $R^i p_*(\mathcal{F}(m))=(0)$, if $i>0, m \ge m_0$. Using Corollary $1$ and $1 \frac{1}{2}$ one deduces that $p_*(\mathcal{F}(m)) \otimes k(s)$ maps onto $H^0(\mathbb{P}^n _s, \mathcal{F}_s(m))$ for all $s\in S, m \ge m_0$. Then by (iii), $p_* \mathcal{F}(m)$ is locally free. As for converse [...]

In original:

Problem: The "...Then by (iii), $p_* \mathcal{F}(m)$ is locally free..." part I not understand. (iii) (on page 51) states:

By above we know $p_*(\mathcal{F}(m)) \otimes k(s) \to H^0(\mathbb{P}^n _s, \mathcal{F}_s(m))$ is surjective, that is we can apply (iii) to $i=1$ and deduce $R^1p_*(\mathcal{F}(m))$ is locally free sheaf. But Mumford claims this for $p_* \mathcal{F}(m)= R^0 p_* \mathcal{F}(m)$.

Is this an error in the proof or do I miss something?

Answer 1

I think you can apply $(iii)$ with $i = 0$ to obtain that $p_*\mathcal{F}(m)$ is locally free. Since the surjectivity of the base change map in degree $i-1 = -1$ is trivial, you only need the surjectivity in degree $i=0$ required by the condition in $(ii)$.

In summary, you use the base change theorem several times. From Serre's vanishing theorem, you have $H^1(\mathbb{P}_s^n,\mathcal{F}(m)) = 0$ for $m$ large enough and all $s\in S$. From $(ii)$ with $i=1$, you get that $R^1p_*\mathcal{F}(m) = 0$. From $(iii)$ with $i=1$, you get that $R^0p_*\mathcal{F}(m) \otimes k(s) \rightarrow H^0(\mathbb{P}_s^n,\mathcal{F}(m))$ is surjective for every $s$ in $S$ ($R^1p_*\mathcal{F}(m)$ is zero so locally free). Then, you use $(iii)$ with $i=0$ as I explained above to deduce that $p_*\mathcal{F}(m)$ is locally free.

I hope everything is correct.