Let $A$ be a ring, then structure sheaf on $\text{Spec }A$ is given by a sheaf on the base of distinguished opens, $\mathcal{O}(U_f)=A_f$, where $A_f=f^{-1}A$, the localization of $A$ by the multiplicative closure of $\{f\}$ for some $f\in A$.
To have a sheaf on a base, we need to verify the various sheaf on a base axioms. First and foremost we need to have restriction maps $\theta^g_f:A_g\rightarrow A_f$ whenever $U_f\subset U_g$, and these restriction maps must satisfy $\theta^f_f=\text{Id}$ and $\theta^f_h\circ\theta^g_f=\theta^g_h$. I believe we should define these restriction maps by the universal property of localization, that is $F:A\rightarrow R$ is a ring homomorphism, and $U$ is a set, then if $F(u)$ is a unit for all $u\in U$ we have a unique morphism $S^{-1}A\rightarrow R$, where $S$ is the multiplicative closure of $U$.
Ok, so lets set it up. We have that $U_f\subset U_g$, so we want a map $A_g\rightarrow A_f$. Let $\pi_f$ and $\pi_g$ denote the projections $A\rightarrow A_f$ and $A\rightarrow A_g$ respectively. To get the map, we set $U=\{g\}$, then we must have that $\pi_f(g)$ is a unit in $A_f$. We have that $U_f\subset U_g$, so that implies that $f^n=a\cdot g$ for some, so $g$ has inverse $a/f^n$ in $A_f$. The conditions of the universal property are satisfied, and it follows that the restriction map is given by:
$$
\begin{align}
\theta^g_f:A_g&\longrightarrow A_f\\
\frac{b}{g^k}&\longmapsto \frac{b\cdot a^k}{f^{n\cdot k}}
\end{align}
$$
The properties of the restriction maps are then easily verified by the uniqueness of the map.
I want to prove sheaf axiom $1$, so suppose that $U_g$ is covered by finitely many $\{U_{f_i}\}$, which we can do because every distinguished open is quasi compact. Further note that $g$ is a unit in each $A_f$ with inverse given by $a_i/f_i^{n_i}$. Now suppose that $s\in A_g$ such that $\theta^g_{f_i}(s)=0$. Write $s$ as $\frac{b}{g^k}$ for some $k$, then we have that:
$$\theta^g_{f_i}(s)=\frac{b\cdot a_i^k}{f^{n_i\cdot k}_i}=0$$
It follows that there exists an $m_i$ such that:
$$f_i^{m_i}(a_i^k\cdot b)=0$$
Now we have that:
$$\sqrt{\langle g\rangle}=\sqrt{\sum_i\langle f_i\rangle}=\sqrt{\sum_i\langle f_i^{m_i}\rangle}$$
so we have that:
$$g^l=\sum_if_i^{m_i}\cdot c_i$$
for some $c_i\in A$. I want to be able to write that $c_i=a_i^k$ so that I can multiply by $1$ and get that $s$ is zero but I am not sure how…
Is this the right approach? I am honestly not sure what to do here to show that $s$ is zero…
Edit: I know there is a way to do prove this by noting that $U_f=\text{Spec }A_f$, but I would prefer to also be able to do this without making that identification, as it doesn't feel very illuminating.
Best Answer
Your approach is mostly ok, but you're making it harder than you have to. I'll address how to make things simpler at the end - for now, let me explain how we can get your approach to work without changes.
First, let's recall what you're trying to show, which is that there's some positive integer (say) $l$ such that $g^{l} \cdot b = 0$, which will imply that $s = 0$ in $A_{g}$. For each $i$, we have an equation that looks like $$f_{i}^{m_{i}}a_{i}^{k}b = 0.$$ But $a_{i}g = f_{i}^{n_{i}}$ by definition, so we can multiply this equation by $g^{k}$ for each $i$ to get $$f_{i}^{m_{i}+n_{i}k}b =0.$$ Hence, you can run the same argument you did with $m_{i}+n_{i}k$ in place of $m_{i}$ to furnish $c_{i} \in A$ such that $$g^{l} = \sum_{i} f_{i}^{m_{i}+n_{i}k}c_{i}$$ for some positive integer $l$. But now, we get $$g^{l}b = \sum_{i} c_{i}f_{i}^{m_{i}+n_{i}k}b = 0,$$ since each term is $0$ by the second equation above.
What you should notice in the argument above is that $k$ and the $a_{i}$s didn't really play an important role. And indeed, you can save yourself a lot of grief by noting that $g$ is a unit in $A_{g}$, so your section $s$ is $0$ if and only if $g^{k}s = b/1 = 0$. But your localization maps send $b/1$ to... $b/1$, so you don't get all the symbolic grief that you do with the approach above.