Let $A$ be a ring, and take $f_1, ..., f_n$ in $A$.
Geometrically, it seems $A_{f_i}$ cover $A$ when $\langle f_1, ..., f_n \rangle = 1$ (in that case, for each point $\mathfrak{p}$, there must be at least one $f_i$ not contained in $\mathfrak{p}$, so it seems $\cup D(f_i) = \text{Spec}(A)$).
Now algebraically, I want to show that the equilizer of the two maps from $\prod_i A_{f_i}$ to $\prod_{(i, j)} A_{f_i f_j}$ is $A$. So that would mean that, for each $(a_i/f^i)_{i}$ such that $a_i/f^i = a_j / f^j$ in $A_{f_i f_j}$, there is a unique $a \in A$ such that $a = a_i / f^i$ in $A_{f_i}$. However I can't seem to show this. Can anyone give me a source for this?
Categorical arguments are particularly appreciated.
Interestingly, if $\langle f, g \rangle = 1$ then $\langle f^i, g^j \rangle = 1$ (take a relation $af + gb = 1$ and take it to a high enough power).
I guess we can treat the case for two elements first. I hope we don't end up using induction though:
We seek the pullback of the diagram with morphisms $A_{f} \rightarrow A_{fg}$ and $A_g \rightarrow A_{fg}$. Take $a/f^i$ in $A_f$ and $b/g^j$ in $A_g$ such that $a/f^i = b/g^j$ in $A_{fg}$.Then $a g^j = b f^i$ in $A_{fg}$. After multiplying with $c$, we can replace $c g^j$ in $ac g^j = c b f^i$ with $d f^i$ so that $a d f^i = cb f^i$. Then $ad - cb = 0$ is annihilated by a power of $f$. Similarly it is annihilated by a power of $g$, so that it must be $0$ (it is annihilated by any linear sum of them, in particular $1$). Of course, this means $ad = cb$...
An alternative method would be to define $\text{Spec}(A)$ as a sheaf of certain sections of the set map $\pi : \amalg_{\mathfrak{p} \in \text{Spec}(A)} A_{\mathfrak{p}} \rightarrow \text{Spec}(A)$. Sections are the ones which are locally of the form $s_{a/b}: D(b) \rightarrow \amalg_{\mathfrak{p} \in \text{Spec}(A)} A_{\mathfrak{p}}$ sending $\mathfrak{p}$ to $a/b$ in $A_{\mathfrak{p}}$. That is, sections $s : U \rightarrow \amalg_{\mathfrak{p} \in \text{Spec}(A)} A_{\mathfrak{p}}$ such that $U$ has an open cover by $D(b_i)$ and $s = s_{a_i / b_i}$ on $D(b_i)$.
Now this satisfies the sheaf condition, so the goal is to show that $\text{Spec}(D(f)) \cong A_{f}$. Actually this is pretty standard I think.
For completeness, here is a proof:
Claim: Let $\mathcal{O} = \text{Spec}(A)$ be the sheaf defined above. Then $\mathcal{O}(D(f)) \cong A_f$.
There is a canonical map $A \rightarrow \mathcal{O}(D(f))$ sending $a \in A$ to the function $D(f) \rightarrow \amalg_{\mathfrak{p} \in \text{Spec}(A)} A_{\mathfrak{p}}$ that it defines (it sends $\mathfrak{p}$ to $a/1$ in $A_{\mathfrak{p}}$). The universal property of localization shows that this map factors through a map $A_{f} \rightarrow \mathcal{O}(D(f))$, as $f$ is sent to a unit under this map. We show that this map is an isomorphism.
Surjectivity: Take $a \in \mathcal{O}(D(f))$. We aim to show that $a$ is contained in the image of $A$. Take $(A : a) = \{ b \in A : ba \in A \}$. Suppose for a contradiction that $1 \notin (A : a)$...
Injectivity:...