In this short note, we present a constructive proof for the existence of the integral Zariski decomposition as defined in Enokizono (2024).
Definition 1 Let be a pseudo-effective divisor on a normal complete surface . A decomposition is said to be an integral Zariski decomposition if the following conditions hold:
- is a -positive divisor on .
- or is a negative definite divisor on .
- for any irreducible component of .
Definition 2 Let be a –divisor on a normal complete surface . We say that is –positive if is not nef over for any negative definite integral divisor , i.e. there exists an irreducible component of such that .
The following is a useful characterization of –positive divisors.
Proposition 1 (Enokizono (2024), Proposition 3.16) Let be a pseudo-effective –divisor on a normal complete surface with the Zariski decomposition . Then is –positive if there is a chain such that is an irreducible reduced curve and . In particular, is –positive divisor.
The reason that is –positive is that the nef part of the Zariski decomposition must be at least and the negative definite part is then fractional, i.e. less than $.
Applying the method in the proof of Ye, Zhang, and Zhu (2018), Theorem 1.2, which should be attributed to Sakai (1990), we can construct an integral Zariski decomposition for any pseudo-effective -divisor.
Theorem 1 Let be a pseudo-effective –divisor on a normal complete surface and the Zariski decomposition of . Then it has an integral Zariski decomposition with , where is –positive.
Proof. Write and .
If , then is an integral Zariski decomposition.
Suppose that .
If for any irreducible component of , then for any irreducible curve with . Therefore, is an integral Zaiski decomposition.
We now construct a chain of divisors inductively as follows.
If be an irreducible component such that , then we let and . If or has no irreducible component such that , then we let and . Otherwise, repeating this procedure until we get a chain of –divisors such that is an irreducible curve, , and or for any irreducible component of . Let and .
We claim that is –positive.
By the construction of , we know that . Since , is a Zariski decomposition of , where . Therefore, by Proposition 1, is –positive.
We can then conclude that is an integral Zariski decomposition.