This is a question about the proof of Proposition 6.13 (b) in Chapter 2 of "Algebraic Geometry" by Hartshorne.
Let $ \mathcal {L} (D) $ be the invertible sheaf associated with the Cartier divisor $ D $.
Proposition 6.13 (b) says that for Cartier divisors $ D_1, D_2 $ on scheme $ (X, \mathcal {O}) $,
$ \mathcal {L} (D_1-D_2) = \mathcal {L} (D_1) \otimes \mathcal {L} (D_2) ^ {-1} $.
When $ D_1 = {(f_x, U_x) }$ and $D_2 = {(g_x, U_x) } $,
$ D_1 -D_2 ={(f_x g_x ^ {-1}, U_x) } $ holds.
$ \mathcal {L} (D_1-D_2) $ is generated by $ f_x ^ {-1} g_x $ over $ U_x $.
$ \mathcal {L} (D_1) $ is generated by $ f_x ^ {-1} $ over $ U_x $.
$ \mathcal {L} (-D_2) $ is generated by $ g_x $ over $ U_x $.
In the text of the proof,
$ \mathcal {L} (D_1-D_2) ＝ \mathcal {L} (D_1) \cdot \mathcal {L} (D_2) ^ {-1} $, so $ \mathcal {L} (D_1) \cdot \mathcal{L} (D_2) ^ {-1} $ is clearly isomorphic to $ \mathcal {L} (D_1) \otimes \mathcal {L} (D_2) ^ {-1 } $.
I think $ \mathcal {L} (D_2) ^ {-1} $ is different from the inverse element $ \mathcal{H}om (\mathcal {L} (D_2), \mathcal {O}) $ of the Picard group.
What does $ \mathcal {L} (D_2) ^ {-1} $ mean?
And what is "$ \cdot $" of $ \mathcal {L} (D_1) \cdot \mathcal{L} (D_2) ^ {-1} $?
Best Answer
$\cdot$ means product inside $\mathcal{K}$: the sheaves $\mathcal{L}(D)$ are defined to be invertible subsheaves of the sheaf of rings $\mathcal{K}$, so it makes sense to take their product.
Secondly, no, the $^{-1}$ in $\mathcal{L}(D_2)^{-1}$ means exactly the same thing as the inverse in the Picard group. Sheaves of the form $\mathcal{L}(D)$ for $D$ a Cartier divisor are locally isomorphic to $f\cdot\mathcal{O}_X\subset \mathcal{K}_X$, and it's not too hard to see that on an open set where $\mathcal{L}(D)$ is generated by $f$, the dual $\mathcal{L}(D)^{-1}= \mathcal{H}om(\mathcal{L}(D),\mathcal{O}_X)$ restricts to $\mathcal{H}om(f\cdot\mathcal{O}_X,\mathcal{O}_X)\cong \mathcal{H}om(\mathcal{O}_X,f^{-1}\cdot\mathcal{O}_X)$, so $\mathcal{L}(D)^{-1}$ is locally generated by $f^{-1}$ on the same open set where $\mathcal{L}(D)$ is generated by $f$.