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 well-known result but I couldn't find a reference. You can assume $f$ is smooth and surjective if you want. Thank you.



Projection formula for flat morphisms
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](https://stacks.math.columbia.edu/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](https://stacks.math.columbia.edu/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.


