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

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?

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.