Hilbert–Samuel function

Remarks

The length is finite because $M/I^n M$ is a finitely generated module over the Artinian ring $R/I^n$. There is a polynomial $P_M^I$ with rational coefficients, the Hilbert–Samuel polynomial, such that $\chi_M^I(n) = P_M^I(n)$ for all large $n$. If $M \neq 0$, its degree is the Krull dimension of $M$, so it does not depend on $I$. Example: for $R = k[x, y]_{(x, y)}$, with $k$ a field, $M = R$ and $I = \mathfrak m$, the module $R/\mathfrak m^n$ has the $k$-basis of monomials $x^a y^b$ with $a + b < n$, so $\chi(n) = n(n+1)/2$. Non-example: for the same ring and $I = (x)$, which is not $\mathfrak m$-primary, the module $R/(x)^n$ has infinite length, so the formula does not define a function to $\N$. Some authors use $\ell(M/I^{n+1} M)$ instead, which shifts the variable by one.