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The length of a $R$-module $M$ is the least length of a composition series $$M=M_0\supset M_1\supset ... \supset M_n = 0,$$ where $M_j/M_{j+1}\cong R/P$ for some maximal ideal $P$ for all $j$. If $R$ is a local ring $(R,m)$, then this is equivalent to a power of $m$ annihilates $M$, and the number of power should be the length of $M$ by the structure above.

Now, I am trying to compute the length of a kind of module. Given a ring $R=k[s^4, s^3t, st^3, t^4]_{(s^4, s^3t, st^3, t^4)}$, which is obtained by localizing a subring of $k[s,t]$ at the maximal ideal $(s^4, s^3t, st^3, t^4)$, where $k$ is a field. Let $q=(s^4, t^4)$, if we consider $R$ as a module, then we can consider the length of modules of the form $q^nR/q^{n+1}R$.

Using the definition above, length $R/qR$ should be $2$. But I get stuck computing the general forms. Anyone who can compute cases $n=1,2$ will be helpful. Thanks!

Yuyi Zhang
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  • Hilbert functions, polynomials, and series will provide a general framework to compute the numbers you are interested in. (If you are interested in only the case of $n = 1, 2$, Grobner basis can be also helpful.) There are a few reductions. 1) The length of those modules in quesiotn is finite and equal to the $k$-vector space dimension. 2) The length can be computed in the ring $k[s^4, s^3t, st^3, t^4]$ which is isomorphic to the ring $A=k[x,y,z,w]/(xz-y^2,yw-z^2,xw-yz)$. Hence $q$ corresponds to the ideal $(x,w)A$. I believe that the length of $R/q$ is 3 instead of $2$. – Youngsu Dec 21 '20 at 08:04
  • Why is 3? Isn't it true that the square of the maximal ideal is contained in $q$? Also, from what I am reading, it only provides some general theorems about Hilbert polynomials, but no specific ways of computing them. Maybe Grobner basis will help. – Yuyi Zhang Dec 21 '20 at 14:17
  • Just as a caution, the claim in your first paragraph is wrong . It is not in general true that the length is equal to the power of the maximal ideal annihilating the module . – Mohan Dec 21 '20 at 15:19
  • @Mohan If so, how to compute it? – Yuyi Zhang Dec 22 '20 at 01:09

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