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How to obtain a relation between the Hilbert-Samuel function of the local ring at a point of a reduced, but not necessarily irreducible variety, and the Hilbert-Samuel functions of the corresponding local rings of its irreducible components?

More concrete. R, a regular local Noetherian ring, complete if you wish, I an ideal in R that is the intersection of some prime ideals I_k such that there are no embedded primes.

I am looking for a formula relating the HS function of R/I and those of R/I_k? The filtration is with respect to the maximal ideal of R.

Edit: Info: There is also an analogous formula for Hilbert functions (analogous to the associativity formula for multiplicity). Proposition 3.2 in Equimultiplicity and blowing up, by Herrmann, Ikeda and Orbanz. $H^{(i)}[\underline{x},a,M]=\sum_{p∈Assh(M/aM)}e(\underline{x},R/p)H^{(i)}[aR_p,M_p]$, where $M$ is finitely generated $R$-module, $a$ and ideal in $R$, and $\underline{x}$ a multiplicity system for $M/aM$. If I put $R$ as my ring $R/I$, $a$ as the maximal ideal, and $M:=R$, if I understood their definition of $Assh$ this only gives me information about those components $p$ having $dim(R/p)=dim(M)=dim(R)$.

edited Apr 17 2012 at 21:06

asked Apr 17 2012 at 16:20

Franklin
420●6●14

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You're going to need to think about how those components are glued together. Consider the ring $R \leq k[[x,y]]/(xy)$ generated by $x+y$ and $x^n$, two lines glued together to order $n$. Once you get to multiple components there's going to be lots of schemy inclusion-exclusion to think about. – Allen Knutson Apr 17 2012 at 17:19

@Knutson. I don't understand the example. Is $R:=k[[x+y,x^n]]/(xy)$? How does one sees the two lines? – Franklin Apr 18 2012 at 19:15

A friend told me: To take $k[[s,t]]\mapsto k[[x,y]]/(xy)$ by sending $s\mapsto x+y$ and $t\mapsto x^n$. The image is $R$ and the kernel seems to be $t(s^n-t)$, which gives us that $R=k[[s,t]]/(t(s^n-t))$. But this is a hypersurface singularity. The Hilbert-Samuel function is only going to depend on the order, in this case $2$. Perhaps that is not the ring you meant. – Franklin Apr 18 2012 at 22:06

Youngsu · Answer 1 · 2012-04-17 19:48:23Z UTC

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There is an associativity formula for Hilbert-Samuel multiplicity which says, the multiplicity is the sum of the multiplicities of the irreducible components in your concrete case, i.e. $e_0(m,R/I) = \sum e_0(m,R/I_k)$. You may want to take a look at Theorem 14.7 of Matsumura, commutative ring theory for the general statement. But I don't know the answer to the relationship between Hilbert-Samuel functions.

answered Apr 17 2012 at 19:48

Youngsu
202●7

	There is also an analogous formula for Hilbert functions. Proposition 3.2 in Equimultiplicity and blowing up, by Herrmann, Ikeda and Orbanz. $H^{(i)}[\underline{x},a,M]=\sum_{p\in Assh(M/aM)} e(\underline{x},R/p)H^{(i)}[aR_p,M_p]$, where M is finitely generated R-module, a and ideal in R, and $\underline{x}$ a multiplicity system for M/aM. If I put R as my ring, a as my ideal I, and M:=R. But if I understood their definition of Assh this only gives me information about those components p having dim(R/p)=dim(M)=dim(R). – Franklin Apr 17 2012 at 20:54
	Perhaps I should put the comment above as info in the question. – Franklin Apr 17 2012 at 20:55
	Not $a$ as my ideal, $R$ is already my $R/I$ and a the maximal ideal. – Franklin Apr 17 2012 at 21:06

Hilbert-Samuel function and that of the irreducible components.

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