This post and the next is my work on Exercises 7.4 A-7.4.O of section 7.4 in Vakil’s note. We discuss Chevalley’s Theorem and prove it using Grothendieck’s Generic Freeness Lemma.

We first discuss some properties of constructible sets and then we prove Grothendieck’s generic freeness lemma following a sequence of exercises in Vakil’s notes. Then we use Generic Freeness to prove Chevalley’s Theorem. Though there are more direct ways to prove it, such as the proof we did in Thursday’s lecture (06/02/2020) in Applied Scheme Theory (proof of Theorem 2.2.9 in Algebraic Geometry II by Mumford and Oda). We only use Generic Freeness here as we will use it again in the future for generic flatness. Note that except proposition 1.2 the rests of the first section on the properties of constructible sets are not needed later.

We will discuss some applications of Chevalley’s Theorem including its implication of Hilbert’s Nullstellensatz in the next post: Generic Freeness and Chevalley’s Theorem II (Applications). I divided the post into two parts since the next part about applications is to be continued. The pdf file below contains the full article so far. Later more will be added into the next post.

Note that except proposition 1.2 the rests of the first section on the properties of constructible sets are not needed later.

As we remarked that the proof of Lemma 1.1 applied to the case that projective is the induction step of the proof of a generalised result. We include this result here. It’s not related to another part of the post so I add it as an appendix.

The Boolean algebra is generated by all the open sets and closed sets under the operations union, intersection and taking complements(negation). Every element in this Boolean algebra can be written as a finite sequence of these symbols. Given such a sequence, let’s look at where the first union symbol occurs. The subsequence before this union symbol consists of only finitely many intersection and negation symbols thus can be rewritten as an intersection of open and closed sets. The same carries on to the following union symbols.

In proposition 1, how is it “by definition” that a constructible set is a finite disjoint union of locally closed sets?

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The Boolean algebra is generated by all the open sets and closed sets under the operations union, intersection and taking complements(negation). Every element in this Boolean algebra can be written as a finite sequence of these symbols. Given such a sequence, let’s look at where the first union symbol occurs. The subsequence before this union symbol consists of only finitely many intersection and negation symbols thus can be rewritten as an intersection of open and closed sets. The same carries on to the following union symbols.

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