Let $K$ be a perfect field, and let $f_1, \ldots, f_m \in K[X_1,\ldots,X_n]$ be polynomials. Consider the affine scheme $$X = \mathrm{Spec}(K[X_1,\ldots, X_n]/(f_1,\ldots,f_m))$$ and let $N = \dim(X)$. Given a closed point $\mathfrak{m} \in X$, we define the multiplicity of $\mathfrak{m}$ to be $N!$ times the leading coefficient of the Hilbert-Samuel polynomial of the local ring $\mathcal{O}_{X,\mathfrak{m}}$. Since $X$ is an excellent scheme, it is known that the set of all multiplicities occuring among the (closed) points of $X$ is finite.

My question: Is it possible to find an upper bound for the maximal multiplicity of a closed point in $X$ in terms of (e.g. the degrees of) the polynomials $f_1,\ldots, f_m$ ?

Indeed, the special case $m=1$ should be easy: Write $f = f_1$ and assume w.l.o.g. that $f(0,\ldots,0) = 0$. Then the multiplicity of $\mathfrak{m} = \langle X_1, \ldots, X_n \rangle$ equals the minimal degree of a monomial $X_1^{i_1} \cdots X_n^{i_n}$ appearing in $f$ with non-zero coefficient (which is certainly bounded from above by the degree of $f$). The case of general $\mathfrak{m}$ should follow by base change.

Any help is appreciated! In particular, I am not sure whether this is a difficult problem or not. (So if you have an opinion on that, then please let me know as well.)

EDIT Sept. 20th, 06:20 pm: Could it be the case that $$\deg(f_1)+\ldots + \deg(f_m)$$ serves as an upper bound? To motivate this idea, suppose again that $f_j(0, \ldots, 0) = 0$ for all $1 \leq j \leq n$. Then I tend to believe that the singularity of $\mathrm{Spec}(K[X_1,\ldots, X_n]/(f_1 \cdots f_m))$ at $\mathfrak{m} = \langle X_1, \ldots, X_n \rangle$ (which should be bounded by $\deg(f_1)+\ldots + \deg(f_m)$ according to the special case $m=1$) is worse than the singularity of $X$ at $\mathfrak{m}$. But maybe my geometric intuition is mistaken here.

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    $\begingroup$ You have two $n$ appearing. Number of variables and the dimension of $X$. $\endgroup$
    – Mohan
    Commented Sep 20, 2021 at 14:50
  • $\begingroup$ @Mohan: Thank you, I fixed it. $\endgroup$
    – Algebrus
    Commented Sep 20, 2021 at 15:04
  • $\begingroup$ When $m = n$, Bezout's theorem implies that the upper bound is the product of the degrees, which is sharp since there are systems of polynomials of arbitrarily high degrees which simultaneously vanish at only one point, say the origin. In general for $m \geq n$ it should be the product of the $n$ largest degrees $\endgroup$
    – pinaki
    Commented Sep 20, 2021 at 18:21
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    $\begingroup$ @Algebrus: one version of Bezout's theorem is that the sum of multiplicities of isolated zeroes of $n$ polynomials in $n$ variables is less than or equal to the product of their degrees. See e.g. Corollary IX.3 of arxiv:1806.05346 (written by me, to be clear). $\endgroup$
    – pinaki
    Commented Sep 21, 2021 at 10:40
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    $\begingroup$ @Algebrus: it is Corollary VIII.3, my apologies. I will try to address your comments in an answer. $\endgroup$
    – pinaki
    Commented Sep 21, 2021 at 14:13

1 Answer 1


Edited: the proof below assumes $k$ is algebraically closed. The proof for the multiplicity inequality has been added.

Given $x \in X := V(f_1, \ldots, f_m)$, let $k$ be the local dimension of $X$ at $x$ (i.e. $k$ is the maximum of the dimension of all irreducible components of $X$ containing $x$).

Claim: $mult_x(X) \leq $ the product of the largest $n-k$ elements of the sequence $\deg(f_1), \dots, \deg(f_n)$.

If $k = 0$, then the claim holds due to Bézout's theorem. This estimate is also sharp: given degrees $d_1, \ldots, d_n$, take generic homogeneous polynomials of the specified degrees in $n$-variables - they intersect only at the origin, and the multiplicity at the origin is precisely $\prod_i d_i$ due to Bézout's theorem.

In the case that $n > k \geq 1$, we will prove the following

Sub-claim 1: For each $j = 1, \ldots, n-k$, there are $g_1, \ldots, g_j$ such that each $g_j$ is a linear combination of the $f_i$, and $V(g_1, \ldots, g_j)$ is a complete intersection on a neighborhood of $x$.

Proof: Proceed by induction on $j$. For $j = 1$, set $g_1 := f_1$. If $k = n - 1$, then stop. Otherwise, for $2 \leq j \leq n - k$, we can assume by induction we found $g_1, \ldots, g_{j-1}$ such that $V(g_1, \ldots, g_{j-1})$ is a complete intersection on a neighborhood of $x$. We claim that there is a linear combination of the $f_1, \ldots, f_n$ which does not identically vanish on any irreducible component of $V(g_1, \ldots, g_{j-1})$. Indeed, otherwise since $k$ is infinite, we can find $m$ linearly independent linear combinations of $f_1, \ldots, f_m$ which vanish identically on one of the irreducible components of $V(g_1, \ldots, g_{j-1})$ containing $x$, which would mean that the local dimension of $X$ at $x$ is $n - j + 1 > k$, a contradiction. Now let $g_j$ be that linear combination of the $f_i$ which does not identically vanish on any irreducible component of $V(g_1, \ldots, g_{j-1})$, and repeat. $\square$

Once you find $g_1, \ldots, g_{n-k}$ as above, let $Y := V(g_1, \ldots, g_{n-k})$. The second observation is that $mult_x(X) \leq mult_x(Y)$, which follows from the following general fact:

Sub-claim 2: Let $X \subseteq Y \subseteq k^n$ be a chain of closed subschemes and $x$ be a closed point of $X$ such that $X$ and $Y$ have the same local dimension at $x$. Then $mult_x(X) \leq mult_x(Y)$.

Proof: For each $q \geq 0$, there is a surjection $$O_{x,Y}/m_x^qO_{x,Y} \to O_{x,X}/m_x^qO_{x,X}$$ where $m_x$ is the ideal of $x$. Treating this surjection as an $O_{x,Y}$-module homomorphism, it follows that $$l(O_{x,Y}/m_x^qO_{x,Y}) \geq l(O_{x,X}/m_x^qO_{x,X})$$ where $l$ is the "length" (see e.g. Atiyah-Macdonald, Proposition 6.9). By the assumption on dimension, for $q \gg 1$, both of these lengths are given by polynomials in $q$ of degree $d$, where $d := \dim_x(X) = \dim_x(Y)$. It follows that the coefficient of $q^d$ in the polynomial providing the values of $l(O_{x,Y}/m_x^qO_{x,Y})$ can not be smaller than the corresponding coefficient of the polynomial for $l(O_{x,X}/m_x^qO_{x,X})$. $\square$

The third observation is that if one of the $f_i$ appears in linear combinations defining more than one $g_j$, say $g_{j_1}, g_{j_2}$, then replacing $g_{j_2}$ by an element of the form $g_{j_2} + ag_{j_1}$ for an appropriate $a \in k$ we may ensure that $f_i$ does not appear in the linear combination defining $g_{j_2}$. In this way we can find appropriate "invertible" linear combinations $g'_1, \ldots, g'_{n-k}$ of $g_1, \ldots, g_{n-k}$ such that

  • $V(g_1, \ldots, g_{n-k}) = V(g'_1, \ldots, g'_{n-k})$, and
  • there is a reordering of $f_1, \ldots, f_n$ such that $\deg(g'_j) \leq \deg(f_j)$, $j = 1, \ldots, n-k$.

The final observation is that $mult_x(Y) = mult_x(Y \cap H)$ for a generic affine subspace of dimension $k$ containing $x$. On the other hand, if $h_1, \ldots, h_k$ are linear polynomials such that $x$ is an isolated point of $V(g'_1, \ldots, g'_{n-k}, h_1, \ldots, h_k)$, then we are done by the $k = 0$ case.

  • $\begingroup$ Thank you very much for your answer. I have not yet checked every detail, but maybe I can ask the following two follow-up questions: (1) What do you mean by a generic linear combination? (2) I am wondering whether your inductive procedure yielding the polynomials $g_1,\ldots.g_{n-k}$ boils down to some standard result from commutative algebra (maybe Noether normalization or something related?). $\endgroup$
    – Algebrus
    Commented Sep 22, 2021 at 8:50
  • $\begingroup$ Thanks for your edit, but I still do not believe that $mult_x(X) \leq mult_x(Y)$ is correct. $\endgroup$
    – Algebrus
    Commented Sep 22, 2021 at 14:37
  • $\begingroup$ I agree that the multiplicity Inequality needs to be verified properly - will think about it as time permits. Btw, the Inequality you claimed from Bennett can not be true in general, e.g. consider the closed subscheme $X := V(x)$ of $Y := V(x^q)$ for $q \geq 2$. $\endgroup$
    – pinaki
    Commented Sep 22, 2021 at 15:16
  • $\begingroup$ Yes, I just figured out that I applied Bennett's result incorrectly. But still: Isn't is counter-intuitive that $mult_x(X) \leq mult_x(Y)$? I mean, we have arrived at $Y$ (roughly speaking) by deleting equations from $X$, so the multiplicity of the point should have decreased (at least this is my geometric intuition). Anyway, I will also continue thinking about it. $\endgroup$
    – Algebrus
    Commented Sep 22, 2021 at 15:37
  • $\begingroup$ Thank you for your patience, now I am convinced! $\endgroup$
    – Algebrus
    Commented Sep 23, 2021 at 13:21

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