KU Math 996 (Combinatorial Commutative Algebra): Lecture 1/21/2026

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The rough idea of what we are doing for the first half of the semester is covering the content of Stanley’s Proof of the Upper Bound Conjecture.

The prerequisites for the course are:

  • Basic Hilbert function theory for graded rings and modules,
  • Depth,
  • Cohen-Macaulay.

[remark=%counter%] Putting these together roughly gives Stanley-Reisner theory? [/remark]

Let $K$ be a field and $S \coloneqq K[x_1, \ldots, x_n]$. Let $M$ be a $\ZZ$-graded $S$-module with

\[M = \bigoplus_{i} M_i\]

where $M_i$ is the $i$th graded piece. Scalar multiplication by $x_j$ induces a map $M_i \to M_{i+1}$.

[remark=%counter%] Note that each $i$th graded piece is a $K$-vector space. Clearly if $x, y \in M_i$, then $x + y \in M_i$ and since $K$ is the $0$th-graded piece of $S$, it follows that scalar multiplication stays in $M_i$.
[/remark]

[definition=%counter%] With the notation as above,

\[H(i, M) = \dim_{K}(M_i)\]

is the $i$th Hilbert function and

\[H_M(t) = \sum_{i\in \ZZ} t^i \dim_{K}(M_i)\]

is the Hilbert series of $M$. [/definition]

[example=%counter%] Let $S = K$. Then $S$ is trivially graded as an $S$-module by simply setting $S_0 = S$. So then

\[H_S(t) = 1.\]

[/example]

[example=%counter%] Let $S = K[x_1]$. Then $S$ is a graded $S$-module via degree:

\[S = \bigoplus_{i\ge 0} S_i\]

where $S_i \coloneqq K\langle x_1^i\rangle$. So then $\dim S_i = 1$ for all $i$ which implies that

\[H_S(t) = \sum_{i \ge 0} t^i = \frac{1}{1-t}.\]

[/example]

[example=%counter%] Let $S = K[x_1, x_2]$. We define

\[S_i = K\langle x_1^{a_1} x_2^{a_2} \mid a_1 + a_2 = i\rangle.\]

Then $S$ is graded as an $S$-module by

\[S = \bigoplus_{i\ge 0} S_i\]

and we note that $\dim_K S_i = i+1$. Then it follows that

\[H_S(t) = \sum_{i\ge 0} (i + 1)t^i = \frac{1}{(1 - t)^2}.\]

[/example]

[proof=Proof of Formula for $H_S(t)$ in Characteristic 0]

\[\begin{align*} H_S(t) &= \dv{t}\sum_{i \ge 0} t^{i+1} \\ &= \dv{t}\sum_{i \ge 0} t^{i} \\ &= \dv{t}\frac{1}{1 - t} \\ &= \frac{1}{(1-t)^2}. \end{align*}\]

[/proof]

[theorem] Let $S = K[x_1, \ldots, x_n]$. Then

\[H_S(t) = \frac{1}{(1 - t)^n}.\]

[/theorem]

[proof=Combinatorial + GF Proof in Characteristic 0] The grading of $S$ is given by letting

\[S_i = K\langle x_1^{a_1} \cdots x_n^{a_n} \mid a_1 + \cdots + a_n = i\rangle.\]

By the sticks and stones 1-1 correspondence, we know that the size of this basis is given by the rearrangements of $n-1$ sticks and $i$ stones. This gives

\[\dim_{K} S_i = {n + i - 1 \choose n - 1}.\]

So then

\[\begin{align*} H_S(t) &= \sum_{i\ge 0} {n + i - 1 \choose n - 1} t^i \\ &= \sum_{i \ge 0} \frac{(i+n-1)!}{(n-1)!i!} t^i \\ &= \frac{1}{(n-1)!} \sum_{i\ge 0} (i+n-1) \cdots (i+1)t^i \\ &= \frac{1}{(n-1)!}\dv[n-1]{t}\sum_{i \ge 0} t^{i+n-1} \\ &= \frac{1}{(n-1)!}\dv[n-1]{t}\sum_{i\ge -n+1} t^{i+n-1} \\ &= \frac{1}{(n-1)!}\dv[n-1]{t} \sum_{i\ge 0} t^i \\ &= \frac{1}{(n-1)!} \frac{(n-1)!}{(1 - t)^n} \\ &= \frac{1}{(1 - t)^n}. \end{align*}\]

[/proof]

[proof=Commutative Algebraic Proof] Since $S$ is a domain, it follows that the map

\[S \xrightarrow{\cdot x_n} S\]

is injective. So then we have the SES

\[0 \to S \xrightarrow{\cdot x_n} S \to S/x_nS \to 0.\]

Now we consider what happens on the graded components. By construction, $x_n$ induces a map $S_i \to S_{i+1}$ given by multiplication with injectivity inherited from the above sequence. We claim that the sequence

\[0 \to S_i \xrightarrow{\cdot x_n} S_{i+1} \to (S/x_nS)_{i+1} \to 0\]

is short exact. Note that $(S/x_nS)_{i+1}$ actually makes sense because $x_nS$ is a homogeneous ideal. Indeed, note that

\[x_1^{a_1} \cdots x_{n-1}^{a_{n-1}}x_n^{a_n} \in x_nS\]

for all $a_1, \ldots, a_{n-1}, a_n \ge 0$. So then $x_nS$ contains all of the homogeneous monomials with at least one factor of $x_n$. This implies that the homogeneous components of any element of $x_nS$ must be in $x_nS$. So then $S/x_nS$ is graded with its grading inherited from $S$. Since

\[(S/x_nS)_{i+1} = S_{i+1}/(x_nS)_{i+1},\]

it suffices to show that

\[0 \to S_i \xrightarrow{\cdot x_n} S_{i+1} \to S_{i+1}/(x_nS)_{i+1} \to 0\]

is short exact. Surjectivity is immediate so it remains to show that we have exactness at $S_{i+1}$. That is, we want to show that $x_nS_i = (x_nS)_{i+1}$. But this is immediate since nonzero elements (of either side) are necessarily of the form $x_n p(x_1, \ldots, x_n)$ with the degree of $p$ being $i$. Thus, as vector spaces, it follows that

\[\dim_{K}S_i - \dim_{K} S_{i+1} + \dim_{K} \left(S/x_nS\right)_{i+1} = 0.\]

So then

\[\sum_{i\ge 0}\dim_{K}S_it^{i+1} - \sum_{i\ge 0}\dim_{K} S_{i+1}t^{i+1} + \sum_{i\ge 0}\dim_{K} \left(S/x_nS\right)_{i+1}t^{i+1} = 0.\]

This simplifies to

\[tH_S(t) - H_S(t) + H_{K[x_1, \ldots, x_{n-1}]} = 0.\]

Note that $K[x_1, \ldots, x_{n-1}] \cong S/x_nS$. Thus,

\[H_S(t) = \frac{1}{1-t}\cdot H_{K[x_1, \ldots, x_{n-1}]}(t).\]

By we proceed by induction on $H_{K[x_1, \ldots, x_{n-1}]}$ and so we obtain the desired claim. [/proof]