Projection formula for flat morphisms 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.
 A: 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.
