Line bundle of point is twisting sheaf of Serre on projective line

Question

I think I misunderstood something fundamental, but I cannot figure out where. Let $X = \mathbb{P}^1$ and let $D$ be the principal divisor $[0:1]$. I believe that $\mathcal{L}(D) = \mathcal{O}(1)$, the twisting sheaf of Serre? But when computing this, I get the following: let $U_x$ and $U_y$ be the open sets where $x$ (resp. $y$) do not vanish. Then on $U_x$, the restriction of $D$ is empty and on $U_y$ it is given by the the section $x/y$ via the canonical identification $U_y \cong \operatorname{Spec} k[x/y]$, so in particular, we have $\Gamma(U_x,\mathcal{L}(D)) = k[y/x]\cong k[y]$ and $\Gamma(U_y, \mathcal{L}(D)) = k[x/y](y/x)\cong k[x]x^{-1}$. In particular, $x^{-1}$ is a section of $\mathcal{L}(D)$ over $D(y)$; which doesn't make sense to me, since it blows up on $[0:1]\in D(y)$. Also, how does one define the isomorphism $\mathcal{L}(D) \cong \mathcal{O}(1)$?

Answer 1

Ok I think I answered my own question. So as modules, obviously we have an isomorphism $\Gamma(D(y),\mathcal{O}_X) \cong k[ x/y](y/x) \cong k[x]x^{-1}\cong \Gamma(D(y),\mathcal{L}(D))$ and my interpretation of $x^{-1}$ as being problematic on $[0:1]$ is inappropriate, because $x$ is nothing more than a symbol.

To define the isomorphisms, we have the map $\Gamma(D(y), \mathcal{L}(D)) = k[x/y] (y/x)\rightarrow k[x, y, y^{-1}]_1 = \Gamma(D(y), \mathcal{O}(1))$ given by $(y/x)f \mapsto yf$ which is an isomorphism of modules. Similarly, we have an isomorphism of modules $\Gamma(D(x), \mathcal{L}(D)) = k[x/y] \rightarrow k[x, y, x^{-1}]_1=\Gamma(D(x), \mathcal{O}(1))$ given by $f\mapsto xf$ and to check that this is an isomorphism of sheaves simply observe that these two morphisms agree on the overlap, and hence is well-defined.