# Relation between Hilbert function and complete intersection ideals

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.

• Identify a basis for the quotient $T/I$. Hint, use a basis consisting of monomials. Dec 6 '18 at 14:20
• 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. Dec 6 '18 at 14:24

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.