# Regarding the Cartier divisor of zero and Hartshorne II 7.7

I cannot quite understand Hatshorne's wording in the Cartier divisor of zero, which is given as follows.

I would like to have some further explanation over the definition and ananalogy in terms of Weil divisor. Base on this definition, I cannot quite follow the proof of one of the statements in Proposition 7.7. Which is:

Explicitly, I cannot understand why the given $$f$$ will satisfy the desired property. Namely, why the divisor of zero of $$f$$ is $$D$$.

------Edited 17/10/2019------------

I now have a more explicit idea on what I would like to ask. My question's still surrounding 7.7 (b). I perfectly understand by linear equivalence, we can write $$D=D_0+(f)$$ and $$f$$ will define a global section $$s\in\Gamma(X,\mathscr{L})$$. By result of (a), the divisor of zero of $$s$$, $$(s)_0=D_1$$, where $$D_1$$ is certain effective Cartier divisor.

Now, my question is: How can I show $$D=D_1$$, the exact equivalence?

• You might find 'my' answer to math.stackexchange.com/questions/1994463/… helpful: – peter a g Oct 13 at 3:54
• Sorry that my earlier question statement didn't address my concern clearly. Now I hope this will give a better picture. – IvanSo Oct 17 at 5:45

Now Apparently I have an idea. $$D=D_0+(f)$$ gives the linear equivalence of divisors and $$f$$ define a global section $$s\in\Gamma(X,\mathscr{L})$$ as mentioned in the first statement of the proof of (b).
What's missing is just the isomorphism to $$\mathscr{O}_{U_i}$$, which actually had been defined (a) to be a morphism $$\varphi$$ gluing the set $$\{\varphi_i:f\in\Gamma(X,\mathscr{L})\mapsto ff_i\in\Gamma(X,\mathscr{O}_{U_i})\}$$.