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Let $x$ be a (closed) point on an algebraic variety $X$ (of dimension $n$) defined over an algebraically closed field $k$. What is the multiplicity $mult_x(X)$, and how to compute it?

While having a hard time recently to compute the multiplicity of some surface singularities, I thought it might be useful to have a list of equivalent definitions. These are the ones I know of:

Notations: let $m_x$ be the maximal ideal of $x$ at $X$. For definitions 1 to 3 below assume (a neighborhood of $x$ in) $X$ is embedded in a projective space $\mathbb{P}^N(\mathbb{k})$.

Geometric Definitions:

  1. (Copying from Mumford, Algebraic Geometry I: This is valid only in the case $k = \mathbb{C}$.) For every linear subspace $L$ of dimension $N - n$ such that $x$ is a component of $L \cap X$, define a number $i(x;L \cap X)$ as follows: it is the unique number such that for every sufficiently small neighborhood $U$ of $x$ (in the classical topology), there is a neighborhood $U'$ of $L$ (in the space of $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$) such that if $L' \in U'$ and $L'$ intersects $X$ transversally, then $i(x;L \cap X) = |L' \cap X \cap U|$. Then $mult_x(X)$ is the minimum of $i(x;L \cap X)$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$ for which $x$ is an isolated point of $L \cap X$.

  2. (Corollary of (1), but holds also over positive characteristic - or at least so I think. For my problem this turned out to be the right definition to use) For every linear subspace $L$ of dimension $N - n$ such that $L \cap X$ is discrete, let $s(x;L \cap X)$ be the number of points of intersection (counted with intersection multiplicity) of $L \cap X$ other than $x$. Then $mult_x(X) = \min\lbrace \deg(X) - s(x;L \cap X)\rbrace$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(k)$ for which $L \cap X$ is discrete.

  3. (From Ramanujam's "On a geometric interpretation of multiplicity") Take a proper birational map $\phi: Y \to X$ such that the pull back of the maximal ideal of $P$ defines an effective Cartier divisor $D$ on $Y$. Then $mult_x(X) = (-1)^{n-1}D^n$.

Algebraic Definitions:

  1. $mult_x(X)$ is $(n-1)!$ times the leading coefficient of the Hilbert–Samuel polynomial of $m_x$.

  2. If $X$ is a hypersurface in a neighbourhood of $x$ defined by a single equation $f$, then $mult_x(X)$ is the integer $q$ such that $f \in m_x^q \setminus m_x^{q+1}$.

What other definitions are out there?

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    $\begingroup$ Your condition 2) does work in all characteristics, but requires an algebraically closed field; this is in Samuel ( amazon.com/… ). $\endgroup$ Commented May 4, 2013 at 22:29

2 Answers 2

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I think more fundamental and interesting than the notion of "multiplicity" is the notion of "tangent cone". Given $x \in X = Spec \ R$ defined by an ideal $I$, it's an interesting fact that $I^\infty$ is small, so that $gr_I R := \oplus_n I^n/I^{n+1}$ is interesting to look at.

(You can't do much with this if say, $R$ is the continuous functions on a compact Hausdorff space, as $I=I^2$.)

I think of the degeneration of $R$ to $gr_I R$ (given by the Rees algebra) as follows: we stretch $X$ away from the point $x$, and take the stretch factor to $\infty$. When we're done, there's lots to ask. Is the limit reduced, irreducible, etc.? But in particular, since the ring is now graded, we can compute its growth, and the leading coefficient thereof is the "multiplicity".

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Here is another equivalence: Let $\pi\colon Z\to \mathbb{P}^n$ be the blow-up of the point $x$, let $E\simeq \mathbb{P}^{n-1}$ be the exceptional divisor and let $\tilde{X}\subset X$ be the strict transform of $X$. Then, the intersection of $\tilde{X}$ with $E$ is a variety of degree $\mathrm{mult}_x(X)$ in $E\simeq \mathbb{P}^{n-1}$.

In the case where $X$ is a curve, you obtain finitely many points and the multiplicity is then just the number of points (that you need to count with multiplicity).

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  • $\begingroup$ Nice. I guess it's a geometric take of the answer by Allen Knutson. $\endgroup$
    – pinaki
    Commented Jun 18, 2020 at 18:02
  • $\begingroup$ The answer of @‍AllenKnutson referenced by @pinaki. $\endgroup$
    – LSpice
    Commented Oct 29, 2023 at 23:15

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