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Consider $T=k[x_1,\ldots,x_n]$ ( $k$ alg. closed and of char $k=0$), and consider the ideal $$I=(x_1,x^{a_2}_2,\ldots,x^{a_n}_n)$$ with $2\leq a_2 \leq\ldots\leq a_n$. I want to prove that $$\sum_{i=0}^t HF(T/I,i)=\prod_{i=2}^n a_i.$$ I know that $I$ is a complete intersection ideal, so $T/I$ is Artinian and Gorenstein, so there exists $\tau$ s.t. $\dim_k [T/I]_{\tau}=1$ (and of course I can cut this sum after the $\tau$ index), but I can't see why I can pass from this summatory to a product of esponents. T tried some combinatorial proofs without much success.

This question is taken by these two paper (resp. page 8 and 2, both at the bottom), which I link you:

https://www.sciencedirect.com/science/article/pii/S0021869312003730

https://arxiv.org/abs/1110.0745

Can anyone help me? Thanks in advance.

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  • $\begingroup$ Identify a basis for the quotient $T/I$. Hint, use a basis consisting of monomials. $\endgroup$
    – Zach Teitler
    Dec 6, 2018 at 14:20
  • $\begingroup$ Or work with the free resolution (which is Koszul). Or identify $T/I$ with the tensor product of algebras $k[x_i]/x_i^{a_i}$, so the dimension of the tensor product is the product of the dimensions of the factors. $\endgroup$
    – Zach Teitler
    Dec 6, 2018 at 14:24

1 Answer 1

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The sum of values of the Hilbert function is equal to the vector space dimension of the algebra: $$ \sum_{i=0}^{\infty} HF(T/I,i) = \dim(T/I). $$ In this case, the vector space $T/I$ has a basis consisting of monomials $x_1^{b_1} \dotsm x_n^{b_n}$ with $0 \leq b_i \leq a_i-1$ for each $i$ (where I am taking $a_1=1$). So the dimension is $\prod_{i=1}^n a_i$. Or, this follows from general facts about complete intersections.

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