§ Proof that is a sheaf [TODO ]
- Give topology for by defining the base as --- sets where does not vanish.
- Note that the base is closed under intersection: .
- To check sheaf conditions, suffices to check on the base.
- To the set , we associate the locally ringed space . That is, we localize at the multiplicative monoid .
- We need to show that if , and given solutions within each , we need to create a unique solution in .
§ Reduction 1: Replace by
- We localize at . This allows us to assume that [ideal blows up as it contains unit ], and that [localization makes into a unit, can rescale? ]
- So we now have that cover the spectrum . This means that for each point , there is some such that . This means that .
- Look at ideal . For every prime (maximal) ideal , there is some such that . This means that the ideal is not contained in any maximal ideal, or that .
- This immediately means that for arbitrary .
- Recall that in a ring, the sums are all finite, so we can write as a sum of FINITE number of , since only a finite number of terms in the above expression will be nonzero. [ is quasi-compact! ]
- This is a partition of unity of .
- Given , if is zero in all , then in .
- being zero in each means that in . This means that , because something is zero on localization iff it is killed by the multiplicative set that we are localizing at.
- On the other hand, we also know that since cover .
- We can replace by , since . So if the cover , then so too do .
§ Check sheaf conditions
- Suppose is equal to