The Stacks project

Remark 79.12.5. Let $f : X \to Y$ be a separated morphism of algebraic spaces. The sheaf $(X/Y)_{fin}$ comes with a natural map $(X/Y)_{fin} \to Y$ by mapping the pair $(a, Z) \in (X/Y)_{fin}(T)$ to the element $a \in Y(T)$. We can use Lemma 79.12.4 to define operations

by the rules

The reason this works is that $Z_1 \cap Z_2$ is both open and closed inside $Z_1$ and $Z_2$ (which also implies that $Z_1 \cup Z_2$ is the disjoint union of the other three pieces). Thus we can think of $(X/Y)_{fin}$ as an $\mathbf{F}_2$-algebra (without unit) over $Y$ with multiplication given by $ss' = \star _2(s, s')$, and addition given by

which boils down to taking the symmetric difference. Note that in this sheaf of algebras $0 = (1_ Y, \emptyset )$ and that indeed $s + s = 0$ for any local section $s$. If $f : X \to Y$ is finite, then this algebra has a unit namely $1 = (1_ Y, X)$ and $\star _3(s, s') = s(1 + s')$, and $\star _4(s, s') = (1 + s)s'$.