1
$\begingroup$

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 $dim(R/p)=dim(M)=dim(R)$.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Allen Knutson Apr 17 '12 at 17:19
  • $\begingroup$ @Knutson. I don't understand the example. Is $R:=k[[x+y,x^n]]/(xy)$? How does one sees the two lines? $\endgroup$ – O.R. Apr 18 '12 at 19:15
  • $\begingroup$ 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. $\endgroup$ – O.R. Apr 18 '12 at 22:06
2
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – O.R. Apr 17 '12 at 20:54
  • $\begingroup$ Perhaps I should put the comment above as info in the question. $\endgroup$ – O.R. Apr 17 '12 at 20:55
  • $\begingroup$ Not $a$ as my ideal, $R$ is already my $R/I$ and a the maximal ideal. $\endgroup$ – O.R. Apr 17 '12 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.