Intuition of Algebraic Partition of Unity In Vakil's Rising Sea (Nov. 18, 2017 Draft) on p. 130 - 131, he gives a proof that his definition of the structure sheaf on affine schemes indeed defines a sheaf. He coins this proof an argument by "partition of unity". In fact, he writes that the name is due to Serre. (In Vakil's book it appears again, e.g. on p. 161 and p. 203.) My questions are:

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*Why is this called partition of unity? Does it - or in which way does it - resemble the topological/analytical partition of unity?

*Does anyone have a reference where Serre said this?

For those who do not have Vakil's argument in front of their eyes, let me sketch it.

Definition. Let $A$ be a ring and we define $\mathcal{O}_{\operatorname{Spec}{A}}(D(f)) = A_f$ with the obvious restriction map.
Claim. Suppose $\operatorname{Spec}{A} = \bigcup_{i=1}^n D(f_i)$ for this argument. We want to show that this satisfies the base gluability property.
Proof. Let $\frac{a_i}{f_i^{\ell_i}} \in A_{f_i}$ be elements which agree on overlaps $A_{f_i f_j}$. After some algebraic manipulations (please trust me on this or you may read it in Vakil's book), we can rewrite $\frac{a_i}{f_i^{\ell_i}} = \frac{b_i}{h_i}$ with overlap condition being $h_j b_i = h_i b_j$ and $\operatorname{Spec}{A} = \bigcup_{i=1}^n D(h_i)$.
The latter implies $1 = \sum_{i=1}^n r_i h_i$ for some $r_i \in A$. The key step is to define $$r = \sum_{i=1}^n r_i b_i.$$ This is an element in $A$ and restricts to $\frac{b_i}{h_i}$ since the overlap condition gives $$ rh_j = \sum_{i=1}^n r_i b_i h_j = \sum_{i=1}^n r_i h_i b_j = b_j. $$

This leaves me with one last question: What motivates the definition of $r = \sum_{i=1}^n r_i b_i$?
 A: The short answer is that, yes, this is basically the same kind of partition of unity argument as one sees in analysis, with some adaptations to handle the coarseness of the Zariski topology and the fact that one is not working with actual functions taking values in some field.
Suppose, however, that $A$ is a finitely generated integral domain over an algebraically closed field $k$.
Then, for every maximal ideal $\mathfrak{m} \subseteq A$, we have an evaluation homomorphism $\operatorname{ev}_\mathfrak{m} : A / \mathfrak{m} \to k$, and the Jacobson radical of $A$ is trivial, so we can think of $A$ as a ring of $k$-valued functions: simply replace each $f \in A$ with the $k$-valued function $\mathfrak{m} \mapsto \operatorname{ev}_\mathfrak{m} (f)$ defined on the set of maximal ideals of $A$.
In this scenario the analogy between the analytic and algebraic situations is clear to see.
Recall that the problem is to find $r \in A$ such that $r = b_i / h_i$ on each $U_i = \mathrm{D} (h_i)$.
In the analytic situation, we would seek $c_i \in A$ such that the closure of $\mathrm{D} (c_i)$ is contained in $U_i$ and $\sum_i c_i = 1$, so that extending $c_i b_i / h_i$ by $0$ outside $U_i$ defines a continuous (or smooth or analytic etc.) function on the whole space.
This is not reasonable in the algebraic situation because open sets in the Zariski topology are "often" dense!
If we could find such a partition of unity then the formula $\sum_i c_i b_i / h_i$ would be a sum of functions defined on the whole space and it is easy to see that it satisfies the equations we want.
Since extension by zero is not feasible in the algebraic situation, we must find a workaround.
We can clear denominators in the problem statement: it is sufficient to find $r \in A$ such that $h_i r = b_i$ (on the whole space).
If we could find $r_i \in A$ such that $\sum_i r_i h_i = 1$, then
$$r = \sum_i r_i h_i r = \sum_i r_i b_i$$
but we do indeed have such $r_i \in A$ because the $U_i$ constitute a cover.
One then notes that the RHS of the equation is already a well-defined function on the whole space and checks that it satisfies the equations we require.
If we define $c_i = r_i h_i$ then we have $\sum_i c_i = 1$ as in the analytic situation, and we have $r_i b_i = c_i b_i / h_i$ on $U_i$, so in some sense we do get $r = \sum_i c_i b_i / h_i$ – but this should not be taken too literally, and one should note that no extension by zero has happened here.
