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.



Bound for multiplicities of closed points on scheme
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.


