Show that $\mathcal{O}(p)$ on $\mathbb{P}^1$ is $\mathcal{O}(1)$ Show that $\mathcal{O}(p)$ on $\mathbb{P}^1$ is $\mathcal{O}(1)$.
I think I would like to show that there exist a rational section of this sheaf with order equal to $1$. 
I am splitting $\mathbb{P}^1$ into its affine covering $U_0(z_0); \ x_0 \neq 0$ and $U_1(y_1); \ x_1 \neq 0$. 
Suppose $p=[0:1] \in U_1(y_1)$. 
Let us look at the rational section $y_1$ which has a zero at $[0:1]$, thus I imagine it is a perfectly valid section for $\mathcal{O}(p)(U_1)$. Let us look at what happens on $U_1 \cap U_0$, we then have that $y_1=1/z_0$. 
We would now say that the section $y_1$ has order zero. However, this can't be correct. Is $1/z_0$, with having a pole at $[1:0]$ not a valid rational section? 
If not, how do I find a rational section of this sheaf?
 A: Here is the description of a rational section  of $\mathcal O_{\mathbb P^1}(p)$ which is even regular.
 The sheaf $\mathcal O_{\mathbb P^1}(p)$ is the subsheaf $\mathcal O_{\mathbb P^1}(p)\subset \mathcal K$ of the constant sheaf $\mathcal K$ of rational functions defined by the specification: $$\Gamma(U,\mathcal O_{\mathbb P^1}(p))=\{g\in \Gamma(U,\mathcal K)\vert \operatorname {div}(g)+1\cdot p\geq 0\}$$ where $U\subset \mathbb P^1$ is open and $\operatorname {div}(g)$ is the  divisor associated to $g$.
The constant function $1$ is clearly a global regular section $f=1 \in  \Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(p))$ since $$ \operatorname {div}(f)+1\cdot p=0+1\cdot p\geq 0$$  Notice that the divisor of $f$ seen as a section of  $\mathcal O_{\mathbb P^1}(p)$ is $$\operatorname {div}^{\mathcal O_{\mathbb P^1}(p)}(f)=\operatorname {div}(f)+1.p=1.p$$ The subtle fact that for $n\gt  0$ the constant function $1$  is a regular section of $\mathcal O_{\mathbb P^1}(n\cdot p)$, and that this section has  a zero of order $n$ at $p$ (and no other zero) is unfortunately not sufficiently emphasized in textbooks.                    
