# Inductive order of vanishing along subvarieties.

(This question is related to some of my most recent questions, so maybe the question setup might be familiar to one or the other. ;) )

I'm in need of some help / hints / notes about a specific setup as follows. I know it is a longer question, but maybe someone has the patience and can help me:

Let $$X = \text{Spec}(A)$$ be a smooth variety (integral separated $$K$$-scheme of finite type) with generic point $$y_0$$ over some algebraically closed field $$K$$. We furthermore suppose we have a regular sequence $$f_1, \dots, f_p \in A$$, such that $$Y_p = V(f_1, \dots, f_p)$$ is a nonempty smooth closed subvariety with generic point $$y_p$$, i.e. $$\overline{\{y_p\}} = Y_p$$. In particular we can assume that the ideal $$(f_1, \dots, f_p) \subseteq A$$ is prime. We always consider closure of points with their integral subscheme structure.

Now as $$Y_p$$ is smooth, also $$V(f_1, \dots, f_i)$$ is smooth along $$Y_p$$ for any $$1 \leq i \leq p$$. This yields for any $$i = 1, \dots, p$$, that $$Y_p$$ is contained in exactly one irreducible component $$Y_i = \overline{\{y_i\}}$$ of $$V(f_1, \dots, f_i)$$ (because otherwise $$\mathcal O_{V(f_1, \dots, f_i), y_p}$$ would not be an integral domain, in contradiction to it being regular and local).

We thus get an inclusion of irreducible components $$Y_i$$ of $$V(f_1, \dots, f_i)$$ by $$Y_p := Z \subsetneq Y_{p-1} \subsetneq \cdots \subsetneq Y_1 \subsetneq Y_0 := X = \overline{\{y_0\}},$$ where $$Y_{i+1}$$ is of codimension $$1$$ in $$Y_i$$.

Hence if we consider any $$f_{i+1}$$ as an element in the function field $$K(Y_{i}) = \mathcal O_{Y_i, y_i}$$, then $$f_{i+1} \in K(Y_{i})^\times$$ (otherwise $$Y_{i+1}$$ would not be of codimension $$1$$). Note that $$K(Y_i) = \text{Frac}(\mathcal O_{Y_i, y_{i+1}})$$ and we have the order of vanishing

$$\text{ord}_{Y_{i+1}}(f_{i+1}).$$

In a proof of something else I read it was used that $$\prod_{i = 0}^{p-1} \text{ord}_{Y_{i+1}}(f_{i+1}) = 1,$$ however I do not see the equality. My professor said the equality should not be that surprising, but could not give an argument on the fly.

My thoughts so far: The question kind of boils down to showing that each factor in the product is $$1$$. Let's take for example the first case, i.e. $$\text{ord}_{Y_1}(f_1) = 1$$. Here we are in the convenient situation that $$\mathcal O_{X, y_1}$$ is regular, hence a DVR. But if $$\text{ord}_{Y_1}(f_1) = 1$$, then this would imply that $$f_1$$ is a uniformizer in the maximal ideal of $$\mathcal O_{X,y_1}$$. If $$\mathfrak p \in \text{Spec}(A)$$ corresponds to $$y_1$$, then it is a minimal prime over $$f_1$$ with $$(f_1) \subseteq \mathfrak p \subseteq (f_1, \dots, f_p)$$. But I see no reason why $$f_1$$ should be a generator of the maximal ideal $$\mathfrak p A_{\mathfrak p}$$.

A great thanks to anyone who has read until here!

• What exactly do you mean by $\mathrm{ord}_{Y_{i+1}}(f_{i+1})$? Order makes sense only if ALL the height one prime ideals containing $f_{i+1}$ are dvrs in general, unless you have a different notion. Jun 26, 2019 at 16:33
• Dear Mohan, thanks for the comment. In the general (so also singular) case for let's say a local domain of codimension one $A = \mathcal O_{X,y}$ with $f \in A$, I'm using the definition $\text{ord}_{\overline{\{y\}}}(f) = l_A(A/(f))$, where the right hand side is the length of the $A$-module $A/(f)$. Equivalenty we can go via the normalization of $X$, if $X$ is not normal (tell me if you are interested in this equivalent definition). Is that what you meant? Jun 26, 2019 at 17:04
• The proof is straightforward if you first localize at the prime ideal generated by the $f_i$s and noting that generic points of $Y_{i+1}\subset Y_i$ is essentially a further localization. Notice that once you localize, each $(f_1,\ldots, f_i)$ is a prime ideal defining $Y_i$. Jun 26, 2019 at 23:41
• Ah, okay, so what you mean is: If we have the localization $(A_{\mathfrak p}, \mathfrak m)$ with $\mathfrak p = (f_1, \dots, f_p)$, then of course the regular sequence stays regular. So for $\mathfrak q = (f_1, \dots, f_i) \in \text{Spec}(A_{\mathfrak p})$, we have that $\text{depth}A_{\mathfrak p}/\mathfrak q = \text{dim}A_{\mathfrak p}/\mathfrak q = p-i$, because $f_{i+1} + \mathfrak q, \dots, f_{p} + \mathfrak q$ is a regular sequence. Also the maximal ideal $\mathfrak m/\mathfrak q$ is generated by $(f_{i+1}, \dots, f_p)$, so $A_\mathfrak p/\mathfrak q$ is regular, local and hence a domain Jun 27, 2019 at 11:03
• Dear @Mohan, thank you again for your helpful comment. I have posted an answer below and would be very grateful if you find the time to skim over it. Thanks so much! Jun 29, 2019 at 10:43

We consider the regular local domain $$\mathcal O_{X, \mathfrak p} = A_\mathfrak p$$, where $$\mathfrak p = (f_1, \dots, f_p) \in \text{Spec}(A)$$. Then $$f_1, \dots, f_p$$ form a coordinate system for $$A_\mathfrak p$$, in particular the sequence stays regular. Consider now the ideal $$\mathfrak q := (f_1, \dots, f_i) \subseteq A_\mathfrak p$$. Then, as $$f_1, \dots, f_p$$ are a coordinate system for $$A_\mathfrak p$$, the local ring $$A_\mathfrak p/\mathfrak q$$ is regular (hence a domain) of dimension $$p-i$$. Note that the prime ideal $$\mathfrak q'$$ in $$A$$ corresponding to $$\mathfrak q = \mathfrak q'A_\mathfrak p$$ is clearly the unique minimal prime over $$(f_1, \dots, f_i)$$ satisfying $$(f_1, \dots, f_i) \subseteq \mathfrak q' \subseteq \mathfrak p$$.
So we now consider the irreducible component $$Y_i$$ and use the notation as above, in particular we assume that $$Y_i = \text{Spec}(A/\mathfrak q')$$. Consider minimal prime $$\mathfrak a$$ over $$f_{i+1} + \mathfrak q' \in A/\mathfrak q'$$ determining $$Y_{i+1}$$, with corresponding prime ideal $$\mathfrak a' \in \text{Spec}(A)$$ satisfying $$(f_{i+1}, \mathfrak q') \subseteq \mathfrak a' \subseteq \mathfrak p$$. We then have (with some abuse of notation) $$\mathcal O_{Y_i, y_{i+1}} \cong (A/\mathfrak q')_\mathfrak a \cong (A_\mathfrak p)_{\mathfrak a'} / \mathfrak q' (A_\mathfrak p)_{\mathfrak a'} \cong (A_\mathfrak p/\mathfrak q)_{\mathfrak a'},$$
which is a regular local domain as localization of a regular local domain. By the argument above, the maximal ideal $$\mathfrak a'(A_\mathfrak p/\mathfrak q)_{\mathfrak a'}$$ is generated by $$f_{i+1} + \mathfrak q$$, so we get $$\text{ord}_{Y_{i+1}}(f_{i+1}) = 1.$$