Suppose $I$ has $s$ generators. Its <associated graded ring>
$$
\operatorname{gr}_I(R)=\bigoplus_{n\ge0}I^n/I^{n+1}
$$
is generated in degree one by their initial forms, so there is a graded surjection
$$
(R/I)[X_1,\ldots,X_s]\twoheadrightarrow\operatorname{gr}_I(R).
$$
Because $I$ is $\mathfrak m$-primary, $R/I$ has finite <length of a module>. The degree-$n$ piece on the left has length
$$
\operatorname{length}(R/I)\binom{n+s-1}{s-1},
$$
so $\operatorname{length}(I^n/I^{n+1})$ grows with degree at most $s-1$. Summing these lengths shows that $\operatorname{length}(R/I^n)$ has polynomial degree at most $s$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 5 / a / ii / Solution

{title2=$\operatorname{length}_R(M)$}
{wiki=Length_of_a_module}

The length of a module is the number of simple factors in a composition series. It is additive across short exact sequences.


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 Incoming links: Length of a module