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Let $A$ be a regular local ring of dimension $n$ with maximal ideal $\mathcal m$.

Then one can consider the $A-$module $A/\mathcal m^i$ for all natural numbers $i$.

My question is simply what length this module has. For $i=0$ it is clearly of length $1$. But what about $i=2$ and bigger? Can one write down a formula for the length depending just on $n$?

And do you know a reference for this problem?

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up vote 6 down vote accepted


For regular local rings $A$ of dimension $n$ one has the formula:

$$\mathrm{length}(A/\m^{i+1}) = \binom{i+n}{n}.$$

More generally, let $(A,\m)$ be any Noetherian local ring. Associated to $A$ is the graded $k = A/\m$-algebra

$$\mathrm{gr}(A) = \bigoplus_{j=0}^{\infty} \mathrm{gr}^j(A),$$

where $\mathrm{gr}^j(A):=\m^j/\m^{j+1}$. It is a fact that $\mathrm{gr}(A)$ is a local $k$-algebra of dimension equal to $n = \dim(A)$. The length of $A/\m^{i+1}$ is equal to the sums of the lengths of $\m^j/\m^{j+1}$ for $j \le i$. Since these latter modules are vector spaces, this is the same as saying that:

$$\chi_A(i) := \mathrm{length}(A/\m^{i+1}) = \sum_{j=0}^{i} \dim_k \mathrm{gr}^j(A).$$

The main fact to know is that $\chi_A(i)$ is a polynomial for sufficiently large $i$. This polynomial is known as the Hilbert-Samuel polynomial; it has degree $n$ and leading coefficient $1/n!$.

If $A$ is regular, however, then $\dim \m/\m^2 = n$, and so by Nakayama's lemma, $\mathrm{gr}(A)$ is a quotient of $k[x_1,x_2, \ldots, x_n]$. Since any non-trivial quotient of this has dimension less than $n$, it follows that

$$\mathrm{gr}(A) \simeq k[x_1,x_2, \ldots, x_n].$$

From this it follows that the Hilbert-Samuel polnomial is exactly:


and moreover that $\chi_A(i)$ is given by this formula for all $i \ge 0$.

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