In proposition 6.12 Hartshorne, it's said that if $\mathcal L,\mathcal M$ are invertible sheaves on a ringed space $X$, then $\mathcal L\otimes_{\mathcal O_X}\mathcal M$ is an invertible sheaf. Here invertible means locally free of rank 1, which means there is an open cover of $X$ such that $\mathcal L_{|U}\cong \mathcal O_X$, same for $\mathcal M$.
In the proof Hartshorne says that it's clear, probably, but not for me. What open cover of $X$ are we taking exactly ? There are open covers $\{U_i\}_{I\in J},\{V_j\}_{j\in J}$ such that $\mathcal L_{|U_i}\cong \mathcal O_X$ and $\mathcal M_{|V_j}\cong \mathcal O_X$.
Does taking $\{U_i\cap V_j\}_{i,j}$ work ? For me there is no reason for $\mathcal L_{|U_i\cap V_j}\cong \mathcal O_X $ or $\mathcal M_{|U_i\cap V_j}\cong \mathcal O_X$ and so no reason for $(\mathcal L\otimes_{\mathcal O_X}\mathcal M)_{|U_i\cap V_j}\cong \mathcal O_X$.
Best Answer
This is fairly straightforward if we formulate things with the internal logic of sheaves. Note that a sheaf $F$ is invertible if and only if “there exists an isomorphism $F \cong \mathcal{O}_X$“ holds in the internal logic.
We now begin our proof in the internal logic. We know that $\mathcal{L} \cong \mathcal{O}_X$, so take some isomorphism $f : \mathcal{L} \to R$. This induces an isomorphism $f \otimes 1_M : \mathcal{L} \otimes \mathcal{M} \to \mathcal{O}_X \otimes \mathcal{M}$. And there is a canonical isomorphism $\mathcal{O}_X \otimes \mathcal{M} \to \mathcal{M}$ - the one sending $r \otimes x \mapsto rx$. Finally, we can also take some isomorphism $g : \mathcal{M} \to \mathcal{O}$. We compose these isomorphisms to get an isomorphism $\mathcal{L} \otimes \mathcal{M} \cong \mathcal{O}$, so the former is invertible.
We can translate this proof to one that doesn’t use the internal logic. Given some $x \in X$, take some $U, V$ and some isomorphisms $f : \mathcal{L}|_U \to \mathcal{O}_X|_U$, $g : \mathcal{O}_X|_V \to \mathcal{M}_X|_V$. Let $W = V \cap U$. Then we can build up a sequence of isomorphisms $\mathcal{L}|_W \otimes \mathcal{M}|_W \to \mathcal{O}|_W \otimes \mathcal{M}|_W \to \mathcal{M}|_W \to \mathcal{O}|_W$, so the tensor product is of rank 1 in a neighbourhood $W$ of $x$. Note how this closely mirrors the internal logic proof.