Inductive order of vanishing along subvarieties.

Question

(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!

Answer 1

With the help of Mohan's comment, I would have proved the statement in the following way:

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