KU Math 996 (Combinatorial Commutative Algebra): Lecture 1/26/2026


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For what follows, we let $S = K[x_1, \ldots, x_n]$.

[remark=%counter%] $H_M(t)$ contains all numerical information including $\dim M$ and Hilbert-Samuel multiplicity $e(M)$. Both can be read off of $H_M(t)$.

Consequently, for $M = R/I$ with $I$ a monomial ideal, these invariants do not depend on the field. [/remark]

For what follows, we will let $M = R/I$.

[theorem=%counter%] If $M$ is a graded f.g. module over $S$, then

\[H_M(t) = \frac{h_M(t)}{(1 - t)^d}\]

where $d$ is the Krull dimension of $M$, $h(1) \neq 0$, and $e(M) = h_M(1)$. [/theorem]

[proof] The key part is to first prove the formula for $H_M(t)$. We proceed by induction on $d$.

Base case. Suppose $d = 0$. Then $\ell(M) = 0$ This tells us

\[\ell(M) = 0 \iff \dim M < \infty \iff M = \bigoplus_{a}^b M_i.\]

So then

\[H_M(t) = \sum_{i=a}^b \dim M_i t^i.\]

Now we just set $h_M(t) = H_M(t)$ and write

\[H_M(t) = \frac{h_M(t)}{(1 - t)^0}.\]

Inductive step. If $d > 0$, we use dimension reduction via hyperplane intersection. Take $\ell$ to be a general linear form, that is

\[\ell = \sum \alpha_i x_i\]

where each $\alpha_i \in K$ (first expand $K$ to make it infinite). Then we avoid finite union of closed sets:

\[\ell \notin \bigcup_{\substack{P \in \operatorname{Ass}(M) \\ ) \neq \mathfrak m}}P.\]

Then we have the sequence

\[0 \to C \to M \xrightarrow{\times\ell} M \to M/\ell M\]

where $C = \ker(\times \ell)$. But since $\times\ell$ is injective on $\Supp(M) - \{\mathfrak m\}$ and $\Supp(C) \subseteq \{\mathfrak m\}$, it follows that

\[\dim(C) = 0 \qquad\text{ and }\qquad \dim M/\ell M = d-1.\]

So the sequence above is short exact and doing the usual generating function argument gives

\[H_M(t)(1 - t) = H_{M/\ell M}(t) - H_C(t).\]

[/proof]

Then we briefly had a detour on the quantity $e(M)$. Let $\mathfrak{m} = (x_1, \ldots, x_n)$. Then it is known that

\[e(M) = e_\mathfrak{m}(M) = \lim_{t\to\infty} \frac{\ell(M/\mathfrak{m}^t M)}{t^d} \cdot d!.\]

[remark=%counter% (TODO)] Source for that claim: these lecture notes. [/remark]

According to Long, simple manipulation gives

\[e(M) = h_M(1)\]

provided that we know $H_M(t) = \frac{h_M(t)}{(1 - t)^d}$.

[example=%counter%] If $M = S/I$, then

\[\ell(M/\mathfrak{m}^tM) = \sum_{i < t} \dim_K M_i.\]

[/example]

Long also made the assertion that $e(M)$ is a lot more. It measures the “spatial size” of $M$ in the following sense.

[example=%counter%] Let $M = R/I$ and $X = \operatorname{Proj}(R/I)$ and $K = \CC$. Then

\[e(M) = \# \text{ points } X \cap H\]

where $\dim X = d - 1$ and $H$ is a general linear subspace of codimension $d - 1$. This is why in algebraic geometry $e(M)$ is called the degree in $\mathbb P^2$. Long gave an example of this with $X$ being a curve and $H$ a hyperplane and mentioned that this has relationships to Intersection Theory and Bezout’s Theorem. [/example]