Let $f:X \to Y$ be a projective morphism of complex Noetherian schemes. Assume $Y$ is smooth and for all $y \in Y$, $f^{-1}(y)$ is of pure dimension $1$. Let $\mathcal{F}_1, \mathcal{F}_2$ and $\mathcal{F}_3$ be coherent sheaves on $X$ flat over $Y$ satisfying the following short exact sequence:

$$0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0.$$

Assume $R^0f_*(\mathcal{F}_1)$ and $R^1f_*(\mathcal{F}_3)$ are zero. We then have the associated long exact sequence of right derived functor:

$$0 \to R^0f_*(\mathcal{F}_2) \to R^0f_*(\mathcal{F}_3) \to R^1f_*(\mathcal{F}_1) \to R^1f_*(\mathcal{F}_2) \to 0.$$

Assume $R^0f_*(\mathcal{F}_3)$ and $R^1f_*(\mathcal{F}_1)$ are locally free $\mathcal{O}_Y$-modules. Then,

$1)$ Is it possible that one of $R^0f_*(\mathcal{F}_2)$ or $R^1f_*(\mathcal{F}_2)$ is locally free $\mathcal{O}_Y$-module?

$2)$ For any $y \in Y$ closed, denote by $k(y)$ the constant sheaf $k(y)$ on the point $y$ (as mentioned in Hartshorne's Algebraic geometry Corollary, III.$9.4$). Is it then true that $- \otimes k(y)$ (as mentioned in loc. cit. Theorem III.$12.11$) is exact when applied to the above long exact sequence. In particular, is the following sequence exact?

$$0 \to R^0f_*(\mathcal{F}_2)\otimes k(y) \to R^0f_*(\mathcal{F}_3)\otimes k(y) \to R^1f_*(\mathcal{F}_1)\otimes k(y) \to R^1f_*(\mathcal{F}_2)\otimes k(y)$$

  • $\begingroup$ 1): Yes, for instance when $Y$ is a point... 2): No. Why should that be true? $\endgroup$
    – abx
    Feb 13, 2014 at 17:22
  • $\begingroup$ @abx: Could $(1)$ be true if Y is a smooth projecctive curve? I would think this is true since a submodule of a free module over a PID is free (the local ring of $Y$ is going to be a PID) $\endgroup$
    – user46578
    Feb 13, 2014 at 19:22

1 Answer 1


Take the universal extension of $O$ by $O(-2)$ on $P^1$. By definition it is a vector bundle $E$ on $X = P^1\times A^1$ which is an extension $$ 0 \to p_1^*O(-2) \to E \to O \to 0. $$ Let $Y = A^1$ and $f = p_2$. Then after taking the pushforward you will get $$ 0 \to R^0f_*(E) \to O \stackrel{t}\to O \to R^1f_*(E) \to 0, $$ where $t$ is the coordinate on $A^1$. This shows that $R^0f_*(E) = 0$ while $R^1f_*(E)$ is the structure sheaf of the point $0$. Of course, after tensoring with the structure sheaf of $0$ the sequence is not exact anymore.


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