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Let $A$ be a (commutative) ring. A family $(B_i)_{i\in I}$ of $A$-algebras is said to be an fppf cover if it satisfies three properties: (1) each $B_i$ is flat as an $A$-module, (2) each $B_i$ is finitely presented as an $A$-algebra, and (3) the family is faithful, meaning that any $A$-module map $M\to N$ is an isomorphism if (and only if) each of the induced maps $B_i\otimes_A M \to B_i\otimes_A N$ is.

It is a theorem that for every fppf cover $(B_i)_{i\in I}$, there is a finite subfamily $J\subseteq I$ such that $(B_j)_{j\in J}$ is an fppf cover. In other words, affine schemes are quasi-compact in the fppf topology. The standard proof uses algebraic geometry: combine the fact that any flat finitely presented map of affine (say) schemes is open with the fact that affine schemes are quasi-compact in the Zariski topology.

My question is whether anyone knows a "direct" proof of this fact. Ideally it would not use the theorem that flat finitely presented morphisms are open or maybe not even any algebraic geometry at all. You might also hope that it works in some more general context, for instance for maps of (commutative?) monoids in any symmetric monoidal category.

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    $\begingroup$ It can't hold in the general setting mentioned in the last paragraph, take $\mathsf{Qcoh}(X)$ for some large (in particular not quasi-compact) scheme $X$. $\endgroup$
    – Martin Brandenburg
    Commented Mar 19, 2013 at 14:58

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Since the ring map $A \to B_i$ is of finite presentation, the image of $\text{Spec}(B_i)$ in $\text{Spec}(A)$ is constructible, see Theorem Tag 00FE, usually known as Chevalley's theorem. Then $\text{Spec}(A)$ is the union of these constructible sets. A constructible set is open in the constructible topology on $\text{Spec}(A)$, see Section Tag 08YF. Moreover, the constructible topology is compact (ibid). Thus we need only finitely many indices $i$.

Probably not the answer you were looking for. Bork, bork, bork! (I was going to run this answer through the Swedish chef translator, but that would probably annoy the powers that be. Chicken in the basket, two points!)

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    $\begingroup$ +1, but the Muppet Show reference didn't really do it for me. $\endgroup$
    – Todd Trimble
    Commented Dec 22, 2013 at 14:10
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    $\begingroup$ OK, this is a slightly different proof using algebraic geometry. The kind of direct proof I was looking for was something like the proof that affine schemes are quasi-compact in the Zariski topology. This boils down to the fact that every element of an ideal generated by a set $S$ is in the ideal generated by a finite subset of $S$. I was hoping (and still do) that a similar argument could exist in the fppf topology. $\endgroup$
    – JBorger
    Commented Dec 22, 2013 at 16:01

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