Question about twisting sheaves: Twisting by two points on $\mathbb{P}^1$ and basepoints of $\mathscr{O}(-1)$

Question

I was trying to do some computations to get familiar with twisting sheaves and I ran into some questions. I would greatly appreciate any help! Thanks in advance!

1. For line bundles, being globally generated is the same as being basepoint free. The bundle $\mathscr{O}(-1)$ on $\mathbb{P}^n$ has no global sections which means that it must have a basepoint. But I am not sure how to go about finding this basepoint. Aren't the sections of $\mathscr{O}(-1)$ just degree -1 homogeneous elements of $\mathscr{O}_{\mathbb{P}^n}$? If so, on each open $U_{x_i}$, the sections are elements of the form $\frac{f}{x_i^n}$ where $f$ is a degree $n-1$ homogeneous element of $k[x_0,\dots,x_n]$. In this case, I don't see why all sections of $\mathscr{O}(-1)$ must vanish at the same point. (And what does it mean to be a basepoint of a line bundle that has no global sections? In Vakil's notes, a base point is defined to be where all sections vanish. To find a basepoint, does it suffice to check where all the global sections vanish?)

2. I would like to compute the global sections of $\mathscr{O}(p+q)$ where $p=[0,1],q=[1,0]\in \mathbb{P}^1$ where $\mathbb{P}^1=$Proj$k[x,y]$. We can describe $p+q$ as the Cartier divisor $(U_x,y),(U_y,x)$ (I'm using Hartshorne's definition of a Cartier divisor). Then, the sections of the line bundle $\mathscr{O}(p+q)$ are: on $U_x$, elements of the form $\frac{1}{y}\frac{f}{x^n}$ where $f\in k[x,y]$ is homogeneous of degree $n$; on $U_y$, elements of the form $\frac{1}{x}\frac{f}{y^n}$ where $f\in k[x,y]$ is homogeneous of degree $n$. Then, global sections are $g\in k[x,y]$ such that $g=\frac{1}{x}\frac{f_1}{y^n}=\frac{1}{y}\frac{f_2}{x^n}$ for some $f_1,f_2\in k[x,y]$ degree $n$. This implies that global sections must be of the form $\frac{ax+by}{xy}$ which tells us that $h^0(\mathbb{P}^1,\mathscr{O}(p+q))=2$. I am a bit confused because I thought that by degree count, $\mathscr{O}(p+q)\cong \mathscr{O}(2)$ and so must have 3-dimensional global sections.

3. Similarly I tried to compute $\mathscr{O}(p)$ where $p=[1,0]$ by describing $p$ as the Cartier divisor $(U_x,y),(U_y,y)$. This just gave me the trivial line bundle. I think the way I am describing/corresponding divisors to Cartier divisors might be incorrect but I haven't been able to figure out how.

Answer 1

1. "All sections" means all global sections here. The only global section of $\mathscr O(-1)$ is the trivial one $0$, and it vanishes on all of $\mathbb P^n$.
2. The Cartier divisor should be $\{(U_x,y/x), (U_y,x/y)\}$. In this case, a global section would look locally like $\dfrac{x}{y}\dfrac{f_1}{x^m}$ in $U_x$ or $\dfrac{y}{x}\dfrac{f_2}{y^n}$ in $U_y$. Notice that on the intersection $U_x\cap U_y$ we get $\dfrac{f_1}{x^m} = \left(\dfrac{y}{x}\right)^2\dfrac{f_2}{y^n}$, which is exactly how we would compute a global section via the transition function $(\dfrac{y}{x})^2$ of $\mathscr O(2)\cong\mathscr O(p+q)$.
3. Again, the Cartier divisor must be modified. You want $\{(U_x,y/x),(U_y,1)\}$. The coordinate on $U_x\cong\mathbb A^1$ is $y/x$, not $y$.