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My question: Let $R = \Bbbk[x_{1}, \dots, x_{n}]$, $m = (x_{1}, \dots, x_{n}) \in Spec_{m}(R)$, $I = m^{i}$. I want to compute the multiplicity $ e(I, R)$ related to the Hilbert-Samuel polynomial. Essentially, this involves calculating the leading coefficient and degree of $\dim_{\mathbb{k}} R / I^j $.
My attempt: $R / I^{j} = Span_{\Bbbk}\{ x_{1}^{r_{1}}\dots x_{n}^{r_{n}} | 0 \leq r_{1} + \dots + r_{n} \leq ij - 1 \}$, so $\dim_{\mathbb{k}} R / I^j = \sum_{r = 0}^{ij - 1}\binom{r + n - 1}{n - 1}$. But how can the explicit form of this binomial summation be computed? Any hints would be greatly appreciated.

jhzg
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You can show (using induction) that $$ \sum_{r=0}^m \binom{r + k}{k} = \binom{m + k + 1}{k + 1} $$

  1. Base Case ($ m = 0 $): When $ m = 0 $, the summation has only one term: $$ \sum_{r=0}^0 \binom{r + k}{k} = \binom{0 + k}{k} = \binom{k}{k} = 1. $$

The right-hand side is: $$ \binom{m + k + 1}{k + 1} = \binom{0 + k + 1}{k + 1} = \binom{k + 1}{k + 1} = 1, $$

thus, the base case holds.

  1. Inductive step: Assume the identity holds for $ m $, we want to show that it holds for $ m + 1 $. You can split the summation: $$ \sum_{r=0}^{m+1} \binom{r + k}{k} = \sum_{r=0}^m \binom{r + k}{k} + \binom{(m + 1) + k}{k}. $$

By the induction hypothesis: $$ \sum_{r=0}^m \binom{r + k}{k} = \binom{m + k + 1}{k + 1}. $$

Thus: $$ \sum_{r=0}^{m+1} \binom{r + k}{k} = \binom{m + k + 1}{k + 1} + \binom{m + 1 + k}{k} = \binom{(m + 1) + k + 1}{k + 1} $$

Using this identity, your summation becomes: $$ \sum_{r=0}^{ij-1} \binom{r + n - 1}{n - 1} = \binom{(ij - 1) + (n - 1) + 1}{(n - 1) + 1}. $$

Simplifying: $$ \sum_{r=0}^{ij-1} \binom{r + n - 1}{n - 1} = \binom{ij + n - 1}{n}. $$

eta.beta
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  • Does this equation have any practical interpretation? After all, $\binom{r + n - 1}{n - 1}$ can be viewed as distributing $r$ balls into $n$ distinct boxes. – jhzg Dec 15 '24 at 10:03
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    You can look at the sum $\sum_{r=0}^m \binom{r + n - 1}{n - 1}$ as the number of paths in a grid with $r$ horizontal and $n - 1$ vertical steps, starting from $(0, 0)$. Each path consists of $r + n - 1$ steps, with exactly $r$ horizontal and $n - 1$ vertical steps. The binomial coefficient $\binom{r + n - 1}{n - 1} = \binom{r + n - 1}{r}$ counts the ways to arrange these steps.

    Thus, the total number of paths to the point $(r, n)$ is $\binom{r + n - 1}{n}$.

    I'm not sure this is exactly what you need, but I've always found this interpretation useful.

     – eta.beta Dec 15 '24 at 10:22
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