A Universe of Sorts

§ Proof that $Spec(R)$ is a sheaf [TODO ]

Give topology for $Spec(R)$ by defining the base as $D(f)$ --- sets where $f \in R$ does not vanish.
Note that the base is closed under intersection: $D(f) \cap D(g) = D(fg)$ .
To check sheaf conditions, suffices to check on the base.
To the set $D(f)$ , we associate the locally ringed space $f^{-1}(R)$ . That is, we localize $R$ at the multiplicative monoid $S \equiv \{ f^k \}$ .
We need to show that if $D(f) = \cup D(f_i)$ , and given solutions within each $D(f_i)$ , we need to create a unique solution in $D(f)$ .

We localize at $f$ . This allows us to assume that $D(f) = Spec(R)$ [ideal blows up as it contains unit ], and that $f = 1$ [localization makes $f$ into a unit, can rescale? ]
So we now have that $\{ D(f_i) \}$ cover the spectrum $Spec(R)$ . This means that for each point $\mathfrak p$ , there is some $f_i$ such that $f_i \not \equiv_\mathfrak p 0$ . This means that $f_i \not \in \mathfrak p$ .
Look at ideal $I \equiv (f_1, f_2, \dots, f_n)$ . For every prime (maximal) ideal $mathfrak p$ , there is some $f_i$ such that $f_i \not in \mathfrak p$ . This means that the ideal $I$ is not contained in any maximal ideal, or that $I = R$ .
This immediately means that $1 \in R: = \sum_i f_i a_i \in I$ for arbitrary $a_i \in R$ .
Recall that in a ring, the sums are all finite, so we can write $1$ as a sum of FINITE number of $f_i$ , since only a finite number of terms in the above expression will be nonzero. [ $Spec(R)$ is quasi-compact! ]
This is a partition of unity of $Spec(R)$ .

Given $r \in R = O(Spec(R))$ , if $r$ is zero in all $D(f_i)$ , then $r = 0$ in $R$ .
$R$ being zero in each $D(f_i)$ means that $r = 0$ in $R[f_i^{-1}]$ . This means that $f_i^{n_i} r = 0$ , 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 $a_1 f_1 + \dots + a_n f_n = 1$ since $D(f_i)$ cover $R$ .
We can replace $f_i$ by $f_i^{n_i}$ , since $D(f_i) = D(f_i^{n_i})$ . So if the $D(f_i)$ cover $R$ , then so too do $D(f_i^{n_i})$ .