I just finished a section of a course that touched on the spectra $\mathrm{Spec}(R)$ of commutative rings $R$, first as sets, then as topological spaces, and finally as locally ringed spaces. (I understand that elsewhere this is called the structure sheaf on $\mathrm{Spec}(R)$).
While trying to do an exercise about the bijection between nontrivial nilpotent elements of $R$ and clopen sets of $\mathrm{Spec}(R)$ (really about showing the equivalency of $\mathrm{Spec}(R)$ being disconnected, $R$ having orthogonal idempotents, and $R$ being isomorphic to a nontrivial product ring) I found a lot of answers on here and elsewhere that essentially proceeded by taking $\mathrm{Spec}(R) = V(\mathfrak{a}) \sqcup V(\mathfrak{b})$, defining an element $f \in \mathrm{Spec}(R)$ through gluing by $f \restriction_{V(\mathfrak{a})} \equiv 1$, $f \restriction_{V(\mathfrak{b})} \equiv 0$ and then using the isomorphism $\mathrm{Spec}(R) \cong R$ to say that $f$ was our nontrivial idempotent.
Intuitively, it makes sense to me that, given that you can use a nontrivial idempotent to decompose $R$ as a product ring, $f$ would "act like 1" on $V(\mathfrak{a})$ and "act like 0" on $V(\mathfrak{b})$, but I do not actually understand how we can tell that $f$ is idempotent.
From what I understand about the definition of the structure ring, what we know about $f$ as an element of $R$ is this:
For any prime ideal $\mathfrak{p} \supseteq \mathfrak{a}$, $\frac{f}{1} = \frac{1}{s}$ for $s \notin \mathfrak{p}$, which is equivalent to the existence of $r \in R$ such that $r ( f s - 1 ) = 0$. Also, for any prime ideal $\mathfrak{q} \supseteq \mathfrak{b}$, $\frac{f}{1} = \frac{0}{s}$ for $s \notin \mathfrak{q}$, which is equivalent to the existence of $r \in R$ such that $r f s = 0$.
Is there any way to get from these facts or similar to $f$ being idempotent? Was my intuition valid? I didn't quite understand the solutions given that used this strategy, and a lot of them seemed to assume the thing I was trying to prove.
More broadly, this strategy of "gluing together" a desired element of a ring by describing the behavior of the corresponding element of $\mathcal{O}(\mathrm{Spec}(R))$ restricted to different subsets seems really useful, but I don't see how you can actually use it if all you know about the element you have constructed are the facts above.
(To be clear I eventually found a more elementary strategy for that exercise. I'm not just asking for a proof that $f$ is idempotent, I want to know if my intuition here was valid.)
I am assuming you mean idempotent when you wrote nilpotent.
If $\mathfrak{p} \in \mathrm{V}(\mathfrak{a})$, then $\frac{f^2 - f}1 = \left(\frac f1\right)^2 - \frac f1 = 1^2 - 1 = 0$ in $R_\mathfrak{p}$. If $\mathfrak{p} \in \mathrm{V}(\mathfrak{b})$, then $\frac{f^2 - f}1 = \left(\frac f1\right)^2 - \frac f1 = 0^2 - 0 = 0$ in $R_\mathfrak{p}$. In both cases we obtain $s \in R \setminus \mathfrak{p}$ such that $s(f^2 - f) = 0$ in $R$.
Let $\mathfrak{n} = \{g \in R : g(f^2 - f) = 0\}$ be the annihilator of $(f^2 - f)$. This is an ideal. The previous paragraph shows that for all prime ideals $\mathfrak{p}$ there is an element $s \in \mathfrak{n} \setminus \mathfrak{p}$. In particular, $\mathfrak{n}$ is not contained in any prime ideal. Since any ideal $\neq (1)$ is contained in a maximal ideal, this is only possible for $\mathfrak{n} = (1)$.
Now $1$ is contained in the annihilator, so $f^2 - f = 0$.