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Let $f : A \to B$ be a finite, finitely presented, flat map of (commutative) rings. It is a known consequence of Chevalley's theorem (on constructible sets) that the induced map $Spec B \to Spec A$ is open.

The Stacks Project (Tag 00FE) proves Chevalley's theorem in various steps, including a reduction to the case mentioned above.

My question: is there a quick, independent argument showing that a finite, finitely presented, flat map is open?

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Here is a proof which, I think, qualifies a ``independent''. As an $A$-module, $B$ is flat and finitely presented and therefore locally free (this is essentially Nakayama's lemma). Thus we may assume $B$ free over $A$, with basis $(e_1,\dots,e_d)$. It suffices to show that for any fixed $g\in B$, the image of $\mathrm{Spec}(B_g)$ in $\mathrm{Spec}(A)$ is open. Let $Z\subset\mathrm{Spec}(A)$ be the complement of this image. The result follows from the following recipe for $Z$:

Claim. Write $g^d=\sum_{i=1}^d a_ie_i$ ($a_i\in A$). Then $Z=\mathrm{Spec}(A/(a_1,\dots,a_d))$.

Indeed, let $\mathfrak{p}\in\mathrm{Spec}(A)$ be a prime, with residue field $k$. Then we have the equivalences: $$\begin{array}{rcl} \mathfrak{p}\in Z & \Leftrightarrow & B_g\otimes_A k=0\\ & \Leftrightarrow & g\text{ is nilpotent in }B\otimes_A k\\ & \Leftrightarrow & g^d =0 \text{ in }B\otimes_A k\\ & \Leftrightarrow & a_1=\dots=a_d=0 \text{ in } k\\ & \Leftrightarrow & a_1,\dots,a_d\in\mathfrak{p}. \text{ QED} \end{array} $$ (For the third equivalence, consider the multiplication by $g$ as an endomorphism of the $d$-dimensional $k$-vector space $B\otimes_A k$).

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  • $\begingroup$ How does local freeness of finitely presented flat modules follow from Nakayama's lemma? I remember an exercise in Atiyah–Macdonald but it assumes Noetherianness. On the other hand, it does follow from Grayson–Raynaud criterion of projectivity (a Mittag–Leffler countably generated flat module is projective), but the proof of this criterion is complicated. $\endgroup$
    – Z. M
    Commented Oct 29 at 18:42
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    $\begingroup$ @Z.M See for example stacks.math.columbia.edu/tag/00NX (proof that (1) implies (6)). $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 30 at 7:32
  • $\begingroup$ Thanks a lot. I guess that I confuse myself with the fact that finitely generated modules over a local ring are necessarily free, which does not seem to be a consequence of Nakayama's lemma (but the Grayson–Raynaud criterion is not necessary for this either). $\endgroup$
    – Z. M
    Commented Oct 30 at 12:05

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