\magnification=1200

\input AMSSYM.DEF
\def\Q{\mathord{\Bbb Q}}
\def\R{\mathord{\Bbb R}}
\def\N{\mathord{{\Bbb N}}}
\def\Z{\mathord{{\Bbb Z}}}

\def\monoid{*monoid}
\def\supp{\mathop{{\rm supp}}}
\def\tmo{\mathrel{\preceq}}
\def\stmo{\mathrel{\prec}}

\def\graver{\mathop{\rm Gr}}
\def\mb{\mathop{\rm MinBal}}


\noindent {\bf Lecture 4. }

\bigskip
\noindent This lecture is based on Chapter I of [Miller-Sturmfels].  We describe Stanley-Reisner ideals and consider some combinatorial data associated with them.

\smallskip
\noindent  Recall:
\item{$\circ$} $I\subseteq \N^n$ is an ideal if $I+\N^n\subseteq I$.  
\item{$\circ$} If $X\subseteq \N^n$, $\langle X\rangle :=X+\N = \{\,\alpha\in \N^n\mid \hbox{$\xi\leq \alpha$ and $\xi\in X$}\,\}$.
\item{$\circ$} By Dickson's Lemma, every ideal in $\N^n$ has a unique set of minimal generators.

\smallskip
\noindent {\bf 4.1.~Fact.}  Suppose $I, J\subseteq \N^n$ are ideals.  Then $I\cap J$ is generated by $\alpha\vee \beta$, where $\alpha$ is a minimal generator of $I$ and $\beta$ is a minimal generator of $J$.   {\it Proof\/}.  $\gamma\in I\cap J$ iff $\gamma\geq \alpha$ and $\gamma\geq \beta$ for some minimal generators $\alpha \in I$ and $\beta\in J$.\hfill/////


\bigskip
\noindent{\it Stanley-Reisner ideals in $\N^n$} 

\medskip
\noindent Associated to any subset $s \subseteq \{\,1,2,\ldots,n\,\}$, we have its characteristic function $$\chi_s: = \sum_{i\in s}\delta^i\in\N^n.$$  In general, we can view an arbitrary element of $\N^n$ as a subset of $\{\,1,2,\ldots,n\,\}$ with ``multiplicities'' attached to its members.  An element all of whose multiplicities are either $0$ or $1$ is the characteristic function of a set.  We call such an elemet {\it setlike\/}.  For $\alpha\in \N^n$, we define $\supp\alpha:= 1\wedge\alpha$, this being the setlike element of $\N^n$ that vanishes at the same indices as $\alpha$.  Occasionally, we write $i\in \alpha$ to mean $\alpha_i\geq 1$.  

\medskip
\noindent {\bf 4.2.~Definition.}  A {\it  simplicial complex\/} $\Delta$ is a collection of subsets of $\{\,1,2,\ldots,n\,\}$ such that $\tau\subseteq \sigma\in \Delta$ implies $\tau \in \Delta$.   (See [MS], p.4, for related terminology.)

\medskip
\noindent An ideal of $\N^n$ is ``upwards closed'', so the complement of an ideal is ``downwards closed''.  (We say $\alpha$ is upwards from $\beta$ if $\alpha=\beta+\gamma$ for some $\gamma\in \N^n$.) Thus, if $I\subseteq \N^n$ is an ideal, then the following collection of subsets of $\{\,1,2,\ldots,n\,\}$ is a simplicial complex: $$\Delta_I:= \{\,\sigma\in\{0,1\}^n\mid\sigma\not\in I\,\}.$$
Note that if $I\subseteq J$, then $\Delta_J\subseteq \Delta_I$.

\medskip
\noindent {\bf 4.3.~Definition.} Suppose $\Delta$ simplicial complex.  The {\it Stanley-Reisner ideal of $\Delta$\/} is 
$$I_\Delta:=\langle\,\sigma \mid\sigma \not\in \Delta\,\rangle = \{\,\alpha\in \N^n\mid \supp\alpha\not\in\Delta\,\}.$$  

\medskip
\noindent {\it 4.3.1~Remark.}  $\alpha\not\in I_\Delta$ iff $\supp\alpha\in \Delta$. 

\smallskip
\noindent {\it Remark.}  If $\Delta\subseteq\Delta'$, then $I_{\Delta'}\subseteq I_\Delta$.  For all $\Delta$, $\Delta=\Delta_{I_\Delta}$. For all $I$, $I\subseteq I_{\Delta_I}$.

\medskip
\noindent {\bf 4.4.~Definition.}  $I$ is said to be {\it prime\/} if its complement $\N^n\setminus I$ is closed under $+$.  

\medskip
\noindent {\bf 4.5.~Fact.}  Ideal $I\subseteq \N^n$ is prime if and only if its minimal elements are among the generators of $\N^n$.  {\it Proof\/}.  If a minimal element of $I$ is not a generator of $\N^n$, it can be expressed as a sum of two elements that are strictly smaller, and hence are not in $I$.\hfill/////

\medskip
\noindent It follows that there is a bijection:
$$\matrix{\hbox{subsets of $\{\,1,2,\ldots,n\,\}$} &\longleftrightarrow &\hbox{prime ideals of $\N^n$}\cr
          \sigma                                   &\longleftrightarrow &\langle\,\delta^i\mid i\in \sigma\,\rangle&.\cr}$$
Note that $\langle\,\delta^i\mid i\in \sigma\,\rangle = \{\,\alpha\in \N^n\mid \sigma\wedge\alpha \not=0\,\}$.

\proclaim 4.6.~Lemma.  $I_\Delta$ is an intersection of prime ideals.  Indeed,
$$I_\Delta = \bigcap_{\sigma\in\Delta}\langle\,\delta^i\mid i\not\in\sigma\rangle.$$

\noindent{\it Proof\/}.  An element $\alpha$ of $\N^n$ belongs to the intersection $\Leftrightarrow$ for each $\sigma\in\Delta$, $\alpha\in \langle\,\delta^i\mid i\not\in\sigma\rangle$ $\Leftrightarrow$  for each $\sigma\in\Delta$, there is $i\in\supp\alpha$ that is not in $\sigma$ $\Leftrightarrow$ $\supp\alpha\not\in\Delta$ $\Leftrightarrow$ $\alpha\in I_\Delta$.\hfill/////  


\bigskip
\noindent {\it Hilbert series}

\medskip\noindent In this section, our point of view changes.  We think in terms of the monomial ideal in $S:=k[x_1, \ldots, x_n] = k[\N^n]$ rather than in terms of monoid ideals in $\N^n$.  One can easily translate back and forth between these two kinds of object.  In this subsection $I_\Delta$ will denote the ideal in $S$ generated by the monomials $x^\sigma$, where $\sigma\in \N^n$ is setlike and $\sigma\not\in\Delta$.

\smallskip\noindent An $S$-module $M$ is $\N^n$-graded if $M=\bigoplus_{\beta\in \N^n}M_\beta$ and $x^\alpha M_\beta\subseteq M_{\alpha+\beta}$.  If $\dim_k(M_\alpha)<\infty$ for all $\alpha$, the $\N^n$-graded Hilbert series is:
$$H(M;x): = H(M;x_1,x_2,\ldots,x_n):=\sum_{\alpha\in \N^n} \dim_k(M_\alpha)x^\alpha.$$

\medskip
\noindent {\bf Facts.}
\item{1.} $\displaystyle{H(S;x) =\prod_{i=1}^n(1+x_i+x_i^2+\cdots)= \prod_{i=1}^n{1\over1-x_i}}$
\item{2.} $\displaystyle{H(\langle x^\alpha\rangle;x) =x^\alpha H(S;x)}$
\item{3.} Observe that $\sum\{x^\alpha\mid\supp\alpha=\sigma\,\}=\prod_{i\in\sigma}{x_i\over 1-x_i}$.  Thus, 
$$\eqalign{
H(S/I_\Delta;x) &=\sum\{\,x^\alpha\mid x^\alpha\not\in I_\Delta\,\}\cr
&=\sum \{\,x^\alpha\mid \supp\alpha\in \Delta\,\}\quad\hbox{(see 4.3.1)} \cr
&=\sum_{\sigma\in\Delta}\prod_{i\in\sigma}{x_i\over 1-x_i}\cr
&=H(S;x)\sum_{\sigma\in\Delta}\bigl(\prod_{i\in\sigma}x_i\cdot\prod_{i\not\in\sigma}(1-x_i) \bigr).   \cr}$$

\noindent The {\it K-polynomial\/} of $M$ is ${\cal K}(M;x):= H(M;x)/H(S;x)$.  Thus, $${\cal K}(S/I_\Delta;x)= \sum_{\sigma\in\Delta}\bigl(\prod_{i\in\sigma}x_i\cdot\prod_{i\not\in\sigma}(1-x_i) \bigr).$$


\end