## § 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 $D(f)$.

#### § Reduction 1: Replace $R$ by $R[f^{-1}]$

• 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 $Spec(R)$.

#### § Separability

• 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 $D(f_i^{n_i})$.

#### § Check sheaf conditions

• Suppose $r_i/f_i^{n_i} \in R[f_i^{-1}]$ is equal to $r_j/f_j^{n_j}$