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Let $A$ be a commutative ring. Let $f\in A\setminus\{0\}$ and $I\subseteq A$ any ideal. I would like to define the multiplicity of $f$ at $I$ as $$\mu_f(I):= \max\{\, d\ge 0 \mid f\in I^d\,\},$$ where $I^0:= A$. In the case where $A$ is Noetherian and either local or an integral domain, the Krull Intersection Theorem (Eisenbud, Corollary 5.4) implies that $\mu_f(I)$ is well-defined. Main scenario: $A$ is the local ring of a locally Noetherian scheme $X$ at some point $P$, $I$ is the corresponding maximal ideal and $f$ is locally representing a Cartier divisor on $X$.

I have only seen this in Hartshorne, Page 388, for surfaces, but I do not see why the definition should be limited to surfaces. In general, I only know the following definition of geometric multiplicity, for locally Noetherian schemes $X$ and points $P\in X$ of codimension one: $$\bar\mu_f(P):=\mathrm{length}_{\mathcal O_{X,P}}(\mathcal O_{X,P}/(f))$$ Does this coincide with the above definition? If yes, why is $\bar\mu$ so prominent? After all, $\mu$ is more general.

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My guess is that $\mu_f$ misbehaves, but I don't know enough to say how it misbehaves. –  Qiaochu Yuan Apr 22 '13 at 19:54
Yea, that was exactly my suspicion. I would like to know how (or rather when) it misbehaves, though - most of the time I am dealing with very forgiving kinds of schemes anyway. –  Jesko Hüttenhain Apr 22 '13 at 20:45
With the notation in the definition of $\overline{\mu}$, assume $P$ is a singular point. Then in general there is an $f\in\mathcal{O}_{X,P}$ such that $\mu_f(I)=1$, but none such that $\overline{\mu}_f(I)=1$. –  Laurent Moret-Bailly Apr 23 '13 at 15:41
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3 Answers

up vote 3 down vote accepted

In addition to two good answers, maybe one case the question has a positive answer is when $(R,m)$ is a regular local ring and $f$ is a nonzero element. For a Noetherian local ring $A$ and ideal $I$ which is primary to the maximal ideal, let $e(I, A)$ denote the Hilbert-Samuel multiplicity of $A$ with respect $I$. Then $e(m/(f),R/(f)) = \hbox{ord} (f) e(m, R) = \hbox{ord} (f)$ where $\hbox{ord} (f) = \sup \{ i \mid f \in m^i \}$. Here $e(m, R) = 1$ since $R$ is a regular local ring.

The special case is when the dimension of a ring (not necessarily regular) $R$ is $1$ and $f$ is a non zero-divisor (but not a unit) of $R$. Assume that $R$ is Cohen-Macaulay (domain or reduced implies Cohen-Macaulyness in dim $1$). Then we have that length $(R/(f)) = e((f), R) \ge \hbox{ord} (f) e(m,R)$ where the last inequality follows from Theorem 14.10 in Matsumura's commutative ring theory. A positive answer to your question implies an equality in the formula and $e(m,R) = 1$. The latter condition is equivalent to $R$ being regular if $R$ is unmixed. Note that the associated graded ring is a polynomial ring over the residue field if a ring is regular. In particular, it is a domain. I believe that once $R$ is a regular local ring, then the equality follows.

I believe in Hartshorne's algebraic geometry book, a surface is a nonsingular (locally regular) projective surface over an algebraically closed field. This probably is a reason why it works well. I hope someone can add some geometric point of view.

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I suppose $\mathrm{ord}(f)=\sup\{ i\mid f\in m^i \}$, yes? So basically, I can use my definition in the case where $X$ is a regular scheme, which might just be good enough for me. +1 and thanks! –  Jesko Hüttenhain Apr 23 '13 at 7:22
The definition of ord is what I used. It is in the second last sentence of the first paragraph. One good fact that might help you is that for a local ring $(R,m)$, $e(m,R) = e(gr_m(R)_+, gr_m(R))$ where $gr_m(R)_+ = \oplus_{i \ge 1} [gr_m(R)]_i$. Maybe it is worth looking at the definition of the Hilbert-Samuel multiplicity in commutative algebra, the degree in algebraic geometry. Also, the associativity formula in commutative algebra can give a connection between geometric multiplicity in algebraic geometry. –  Youngsu Apr 23 '13 at 18:28
Your definition of $\mathrm{ord}(f)$ confuses me because for one thing, it does not depend on $f$ at all and second, $I\in m^i$ looks a lot like you ment to write $f\in m^i$. –  Jesko Hüttenhain Apr 23 '13 at 18:42
Sorry about that. You are absolutely correct. Let me edit that part. One example is when $R = \mathbb{C}[[x]]$. For $f \in R$ One can not talk about a the highest degree in general, but there is well defined lowest degree. This is the order of $f$. Or where the Taylor series of $f$ at $0$ starts. –  Youngsu Apr 23 '13 at 21:15
Thanks a bunch. I accepted this answer because it helps me most for what I am doing, but thanks to Filippo Edoardo and Will Savin for the very helpful explanations. –  Jesko Hüttenhain Apr 24 '13 at 13:00
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The definitions do not coincide for $P$ a cusp. That is, consider the ring $k[t^2,t^3]$ and $I=(t^2, t^3)$. Then if $f=t^4$ the length is $3$ but the power is true. It agrees when the local ring is a DVR.

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a true two. ${}{}$ –  Mariano Suárez-Alvarez Apr 22 '13 at 22:13
I think the length is $4$ since $t^5 \notin (t^4)$. As mentioned in Filippo Alberto Edoardo's answer the failure might be coming from the point that the associated graded ring (tangent cone) is not a domain. In this example the associated graded ring is isomorphic to $k[X,Y]/(X^2)$ which is not a domain. Theorem 14.8 in Matsumura's book may give you some algebraic hint on the question and example. –  Youngsu Apr 23 '13 at 4:50
I meant to say Theorem 14.10 which is essentially a corollary of Theorem 14.9. –  Youngsu Apr 23 '13 at 5:31
Not that it really matters, but I thought $3$ sounded reasonable: We have the ring $k[t^2,t^3]/(t^4)=k[x,y]/(x^2,x^3-y^2)$ where $x=t^2$, $y=t^3$ and $xy=t^5$. One more question, though: You say that the Definitions agree when the local ring is a DVR: Do you have a reference? –  Jesko Hüttenhain Apr 23 '13 at 7:14
The ring $k[x,y] /(x^2, x^3-y^2)$ is isomorphic to $k[x,y]/(x^2,y^2)$. The monomials of $k[x,y]$ that are not zero in this rings are $1,x,y,xy$. As you mentioned $xy <-> t^5$ is not zero. –  Youngsu Apr 23 '13 at 18:13
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One possible issue is that your multiplicity function is not in general a valuation. Indeed, for a noetherian ring $A$ and an ideal $I$ with respect to which $A$ is separated, $\mu_I$ is a discrete valuation of rank one if and only if the associated graded ring $\mathrm{gr}_I(A)$ is an integral domain. May be, one reason why the function becomes less interesting if this is not the case is that often one would like to understand the localization $A_I$ as the valuation ring with respect to $\mu_I$ (at least if $I$ is prime), but this is not always possible: given a regular ring $A$ and a maximal ideal $\mathfrak{m}$ in it, your function $\mu_\mathfrak{m}$ is a discrete valuation of rank 1 whenever $\mathfrak{m}\neq 0$ but if you drop these hypothesis this fails (as can be seen in Will Sawin's answer): moreover, the valuation ring is described explicitly as $A[\mathfrak{m}/x]_{(x)}$ for any $x\in\mathfrak{m}\setminus\mathfrak{m}^2$.

I suggest giving a look at chapter 6.7 of I. Swanson, C. Huneke Integral Closure of Ideals, Rings and Modules, London Math. Soc. Lecture Notes Series 336, which can be found here . The above results are Theorems 6.7.8 and 6.7.9, respectively.

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