(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!

  • $\begingroup$ 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. $\endgroup$
    – Mohan
    Jun 26, 2019 at 16:33
  • $\begingroup$ 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? $\endgroup$
    – johnnycrab
    Jun 26, 2019 at 17:04
  • $\begingroup$ 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 $Y_i$. $\endgroup$
    – Mohan
    Jun 26, 2019 at 23:41
  • $\begingroup$ 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 $\endgroup$
    – johnnycrab
    Jun 27, 2019 at 11:03
  • $\begingroup$ 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! $\endgroup$
    – johnnycrab
    Jun 29, 2019 at 10:43

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

  • 1
    $\begingroup$ Looks fine. Glad to be of help. $\endgroup$
    – Mohan
    Jun 29, 2019 at 13:00
  • $\begingroup$ Your comment let the pieces fall in the right place. Thank you again. $\endgroup$
    – johnnycrab
    Jun 29, 2019 at 13:18

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