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Let $X$ be a smooth projective complex variety. Assume that $Z$ is a subvariety. Let $T$ be a generic complete intersection of codimension $\dim Z-1$. Assume that $p$ is a point in $Z_T:=T\cap Z$. Is there a formula relating $mult_p(Z)$ and $mult_p(Z_T)$?

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I don't understand the role of $X$. In order for "complete intersection" to make sense, you need to embed everything in $\mathbb{P}^n$. So why not take $X=\mathbb{P}^n$? –  Laurent Moret-Bailly Apr 28 '11 at 7:20
You can take $X=\mathbb{P}^n$. For general $X$, you can think the complete intersection as the intersection of general hyperplane sections. –  Fei YE Apr 28 '11 at 22:07
If $P$ is generic enough, then $mult_P(Z) = mult_P(Z_T) = 1$. –  auniket Jul 18 '11 at 12:10
Dear Auniket, Thank you very much. Can you give me some hints on how to prove it? –  Fei YE Jul 19 '11 at 9:19
I am sorry to reply so late. If by multiplicity you mean what is defined e.g. in here: (or chapter 5 of Mumford's Algebraic Geometry I), then a point on a variety has multiplicity one iff it is non-singular. My claim in the preceding comment follows from this. But then it is true for ALL subvarieties of Z (i.e. not necessarily complete intersections). Therefore I suspect you used some other notions of multiplicity, e.g. the one in Qing Liu's answer. –  auniket Jul 26 '11 at 0:33

1 Answer 1

up vote 2 down vote accepted

The answer is yes (auniket's comment meant probably $T$ is generic and not $p$). This is a classical result : if $A$ is a noetherian local ring of dimension $d>0$, with infinite residue field, then there exists a system of parameters $(f_1,...,f_d)$ such that $\mathrm{mult}(A)=\mathrm{mult}(A/(f_1,...,f_{d-1}))$, see Zariski-Samuel vol. II, Chap. VIII, §10, Theorem 22 and Remark page 296.

Note that the result is false if the residue field is finite. But it is true that $\mathrm{mult}_p(Z)$ is a finite integral combination (with possibly negative coefficients) of multiplicities at $p$ of $Z\cap T_i$'s. See Proposition 5.9 in here.

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