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Can anyone prove, or guide me towards a proof, of the following result?

Let $Y$ be an irreducible variety, $C \subset Y$ an irreducible closed subset of codimension $d$ and $U \subset Y$ a non-empty open set.

Then there is a chain $$C = C_d \subset C_{d-1} \subset \cdots \subset C_0 = Y$$

of closed irreducible subsets such that:

(i) $\ \ \ \mathrm{codim}_Y(C_j) = j \ $ for $ \ j = 0, \ldots, d$;

and

(ii) $\ \ \ \ C_j \cap U \neq \emptyset \ $ for $ \ j < d.$

This is Lemma 4.12 in [E. Dufresne and H. Kraft, Invariants and separating morphisms for algebraic group actions, Math. Z 280 p.231-255 (2015)]. The "proof is easy and left to the reader".

Brendan Darrer
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Jon Elmer
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1 Answers1

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You can without loss of generality assume that $Y$ is affine, say $Y = \operatorname{Spec} A$, and $U = \operatorname{Spec} A_f$ for some $f \in A \setminus \{0\}$. Then $C$ corresponds to a prime ideal $\mathfrak p \subset A$ of height $d$, and you are looking for a chain of prime ideals $$0 = \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \dots\subsetneq \mathfrak p_d = \mathfrak p,$$ such that $$f \notin \mathfrak p_i \qquad (i < d).$$

Claim. If $d>1$, then there exists a prime ideal $0 \subsetneq \mathfrak q \subsetneq \mathfrak p$ with $f \notin \mathfrak q$.

Proof. You have $d = \dim A_{\mathfrak p}$. By this answer, there are infinitely many prime ideals of height $1$ in $A_{\mathfrak p}$, but $f$ can only be contained in finitely many of them. $\square$

Given a $\mathfrak q$ from the claim, set $d' = \operatorname{height} \mathfrak q$. Then you have can easily find a chain of prime ideals $$ 0 = \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \dotsc \subsetneq \mathfrak p_{d'} = \mathfrak q \subsetneq \mathfrak p$$ If $d' < d-1$, there is still a gap to fill between $\mathfrak q$ and $\mathfrak p$. To do this, consider $B = A / \mathfrak q$. Then $\dim B \geq d - d' > 1$ and $f\neq 0$ in $B$, and so you can do induction.

red_trumpet
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    A couple of questions for my own understanding: in the claim we chose a prime ideal of height one in $A_{\mathfrak{p}}$, not containing $f/1$. This would be of the form $S^{-1}\mathfrak{q}$ for some $\mathfrak{q} \subset \mathfrak{p}$, where $S = A \setminus \mathfrak{p}$, with $f \not \in \mathfrak{q}$. Do we not in fact have $\mathrm{ht}(\mathfrak{q})=1$? So we can simplify the proof above a bit. – Jon Elmer Dec 20 '24 at 10:46
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    In addition, turning this step around, does this mean that there exist infinitely many closed sets $C_1 \subset Y$ of codimension one with $C \subset C_1$, but only finitely many with $C_1 \cap U = \emptyset$? – Jon Elmer Dec 20 '24 at 10:50
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    @JonElmer Yes, you can basically choose $\mathfrak q$ to have height one. You are also right with your second point: The codimension one irreducible sets $C_1$ with $C_1 \cap U = \emptyset$ are irreducible components of the closed set $Y \setminus U$, of which there is only a finite number. But in general there are infinitely many $C_1$ with $C \subset C_1$ (this is the answer I quoted). – red_trumpet Dec 20 '24 at 12:06