Let $f\colon X\to Y$ be a flat morphism between two smooth projective varieties. Let $L$ be a locally free sheaf on $X$ and $\mathcal{F}$ a coherent sheaf on $Y$. How to prove $f_*(L\otimes f^*\mathcal{F})\cong f_*L\otimes\mathcal{F}$? I think it is a wellknown result but I couldn't find a reference. You can assume $f$ is smooth and surjective if you want. Thank you.

2$\begingroup$ Tag 0B54 gets pretty close, but not quite all the way there. Because $f$ is flat, you may write $f^*$ for $Lf^*$; because $Y$ is smooth quasiprojective, $\mathscr F[0]$ is perfect (you can find a finite locally free resolution). Because $L$ is locally free, you can write $\otimes$ on the RHS instead of $\otimes^{\mathbf L}$. This only leaves the derived tensor product on the LHS, but I don't see a reason why that should be usual tensor product in this situation. (This could be a way to look for a potential counterexample.) $\endgroup$ – R. van Dobben de Bruyn May 9 '20 at 1:19

1$\begingroup$ By the way, this is true with no assumption on $f$ and $L$ (in fact, for any morphism of ringed spaces) if $\mathcal{F}$ is locally free. $\endgroup$ – abx May 9 '20 at 4:46
Based on my comment, I constructed the following counterexample (which I believe is standard):
Example. Let $(E,O)$ be an elliptic curve, let $Y = E$ and $X = E \times E$, with $f \colon X \to Y$ the first coordinate projection. Let $\mathscr L = \mathcal O_{E \times E}(\Delta  E \times O)$, and let $\mathscr F = \mathcal O_O$.
Then $f_* \mathscr L = 0$, since $H^0(U \times E, \mathcal O_{U \times E}(\Delta_U  U \times O)) = 0$ for every open $U \subseteq E$ as $\mathscr L_U$ is a (fibrewise) degree $0$ line bundle that is not trivial.
On the other hand, $\mathscr L \otimes f^* \mathscr F = \mathcal O_{O \times E}$ since $\Delta_{O \times E} = (O,O) = (E \times O)_{O \times E}$. The short exact sequence $$0 \to \mathcal O_X(O \times E) \to \mathcal O_X \to \mathcal O_{O \times E} \to 0$$ gives a long exact sequence $$0 \to \mathcal O_E(O) \to \mathcal O_E \to f_*\mathcal O_{O \times E} \to \mathcal O_E(O) \to \mathcal O_E \to R^1f_* \mathcal O_{O \times E} \to 0$$ since $R^if_* \mathcal O_X(O \times E) = \mathcal O_E(O)$ for $i \in \{0,1\}$ by the usual (derived) projection formula. Thus, \begin{align*} & & & & f_*\big(\mathscr L \otimes f^*\mathscr F\big) = \mathcal O_O \neq 0 = f_*\mathscr L \otimes \mathscr F. & & & & \square \end{align*} Remark. What's going on is that $$Rf_* \mathscr L = \mathcal O_O[1]$$ is not flat over $Y$ even though $\mathscr L$ is. There is a $\mathscr Tor_1$ term interfering in the derived projection formula of [Tag 0B54]. To see the above formula for $Rf_* \mathscr L$, use the short exact sequence $$0 \to \mathcal O_X(E \times O) \to \mathscr L \to \mathscr L_{\Delta} \to 0.\tag{1}\label{1}$$ Since $\mathcal O_X(\Delta)_\Delta = T_E = \mathcal O_E$, we get $\mathscr L_\Delta = \mathcal O_E(O)$. Note that $f$ induces an isomorphism $\Delta \to E$, so the long exact sequence of \eqref{1} reads $$0 \to f_*\mathscr L \to \mathcal O_E(O) \to \mathcal O_E \to R^1f_* \mathscr L \to 0.$$ Above we computed $f_* \mathscr L = 0$, so the map $\mathcal O_E(O) \to \mathcal O_E$ is the natural inclusion, so $R^1f_* \mathscr L = \mathcal O_O$. $\square$
On the other hand, $Rf_* \mathcal O_X = \mathcal O_E \oplus \mathcal O_E[1]$ is a complex of free modules, so the derived tensor product of the LHS of [Tag 0B54] is just a usual tensor product, as we saw implicitly in the computation of $Rf_* \mathcal O_{O \times E} = Rf_* f^* \mathcal O_O$ above.

$\begingroup$ Dear Remy, I think your counterexample is right and thank you. However, I am actually in a special situation that I demand the $L$ to be a torsion line bundle on an abelian variety. The $L$ in your example is not a torsion line bundle. In this case if $f_*L\neq0$, then the projection formula is true by base change theorem. If $f_*L=0$, then i don't know how to do it. I wish to prove that $h^0$ of $L$ restricted to each fibre would be 0 but I couldn't. $\endgroup$ – Rick Sanchez May 9 '20 at 3:04

$\begingroup$ Even if $\mathscr L = \mathcal O_X$ I am not sure what to expect. For a smooth morphism in characteristic $0$ it is a highly nontrivial theorem that $R^if_* \mathcal O_X$ is locally free for all $i$ (uses Hodge theory), but in positive characteristic this is unknown as far as I'm aware. (Warning: in the case above the derived pushforward $Rf_* \mathcal O_X$ splits because this is true for $R\Gamma(E,\mathcal O_E)$, but in general it will be a little more complicated and you need to think about a spectral sequence for Tor as you filter $Rf_* \mathscr L$ by cohomological degree.) $\endgroup$ – R. van Dobben de Bruyn May 9 '20 at 3:13

$\begingroup$ That said, if your interest is specifically in characteristic $0$, then maybe the Hodge theory literature might be of help ― sometimes results for $\Omega^i_{X/Y}$ can be extended to torsion line bundles. I'm not super familiar with this, so I don't know any good references off hand. $\endgroup$ – R. van Dobben de Bruyn May 9 '20 at 3:19

$\begingroup$ Let's work with a fibration betweem abelian varieties over $\C$. It needs not to be complicated. When $L=\mathcal{O}_X$, then $h^0$ is always 1. And you will have that the projection formula is true for any quasicoherent $\mathcal{F}$ by base change. See Qing Liu's book algebraic geometry and arithmetic curves Remark 5.3.21(c) for example. When $L$ is only torsion and $f_*L\neq0$, it's similar since in this case $L$ restricted onto fibre is trivial. $\endgroup$ – Rick Sanchez May 9 '20 at 3:23

$\begingroup$ I think a more precise version well deserves a new question, so that other people can share their thoughts! $\endgroup$ – R. van Dobben de Bruyn May 9 '20 at 3:29