# A question on Castelnuovo-Mumford regularity

Consider a short exact sequence $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ of coherent sheaves on $\mathbb{P}^n$. Assume that $\mathcal{F}''$ (resp. $\mathcal{F}$) is $m-1$ (resp. $m$)-regular. Why does this imply that $\mathcal{F}'$ is $m$-regular?

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As Chen's comment makes clear, this seems to be false. For instance, on $\mathbb{P}^1$, consider the short exact sequence where $\mathcal{F}$ is $\mathcal{O}\oplus \mathcal{O}$ and $\mathcal{F}''$ is $\mathcal{O}(d)$ for $d\gg 0$. Then $\mathcal{F}'$ is $\mathcal{O}(-d)$. – Jason Starr Nov 23 '13 at 18:20
@Starr: I think the statement is true as it is an exercise in FGA explained page 116 after Remark 5.2. – Chen Nov 23 '13 at 18:57
@Chen: No, as abx points out, there is a difference between modules and sheaves when it comes to Castelnuovo-Mumford regularity. Incidentally, we seem to have formed a "triangle": your comment on abx's answer motivated my answer. My answer based on your comment motivated abx's comment. That comment motivated your comment, and now I have responded again to your comment. If we continue a few more times, then according to the rules of chess, one of us can declare a draw! – Jason Starr Nov 24 '13 at 1:31
@Starr: Sorry for the mess. I am a bit confused. The exercise in FGA that I cite deals with coherent sheaves, not modules. It will also be very helpful if you could elaborate a bit on your morphism from $\mathcal{F}$ to $\mathcal{F}''$. – Chen Nov 24 '13 at 3:09
@Chen: I looked, and it is true that Nitsure lists this as an exercise. However, the exercise is wrong. The exercise would be correct if the induced map $\Gamma_*(\mathbb{P}^n,\mathcal{F})\to \Gamma_*(\mathbb{P}^n,\mathcal{F}'')$ were surjective, i.e., if we still had a short exact sequence after we pass from coherent sheaves on $\mathbb{P}^n$ to graded modules over the polynomial ring. However, as written, the exercise is wrong. Also my first sheaf map sends the first generator to $x^d$ and the second generator to $y^d$. The second map comes by Koszul (and twisting). – Jason Starr Nov 24 '13 at 12:53