\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\K{\mathord{{\Bbb K}}}

\def\monoid{*monoid}
\def\supp{\mathop{{\rm supp}}}
\def\card{\mathop{{\rm card}}}
\def\Tor{\mathord{{\rm Tor}}}
\def\im{\mathop{{\rm im}}}
\def\tmo{\mathrel{\preceq}}
\def\stmo{\mathrel{\prec}}

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


\noindent {\bf Lecture 7.  Betti Numbers}
\bigskip
\noindent  In this talk, $S=k[\N^n]$.  All modules will be $(S, \N^n)$-graded-modules.  Let 
$$0\longleftarrow M\longleftarrow F_0 \buildrel\phi_1\over\longleftarrow F_1\buildrel\phi_2\over\longleftarrow\cdots\buildrel\phi_\ell\over\longleftarrow F_\ell\longleftarrow0$$
be an augmented free resolution of $M$ in which the matrices $a_{ji}$ defining the maps $\phi$ have no nonzero entries in degree $0$.  In other words, we are assuming that the resolution is {\it minimal\/}; see [Miller-Sturmfels]=[MS], page 12.  (The minimal free resolution of $M$ is unique up to isomorphism; see the references in [MS] or [Bruns-Herzog], page 36.)  Now suppose
$$F_i = \bigoplus_{\alpha\in\N^n}S(-\alpha)^{\beta_{i\alpha}},$$
with $\beta_{i\alpha}\in \N$ being the number of times that $S(-\alpha)$ occurs as a summand of $F_i$.   These numbers, which are completely determined by $M$, are called the {\it graded Betti numbers of $M$}.  (Graded Betti numbers may be defined in a similar manner for graded rings other than $S$, but there are restrictions on the ring that must be met, without which uniqueness may fail; see [Bruns-Herzog], page 37.)  The goal of this talk is to demonstrate how the $\Tor$ functor can be used to compute the $\beta_{i\alpha}$.  

\bigskip
\noindent  We suggest that the reader review Lecture 5 for information on tensor products $M\otimes_A N$.  (Some parts have been revised since the first version was posted, and comments intended to clarify the following discussion have been included.)

\bigskip
\noindent Let $M$ and $N$ be $A$-modules.  Recall that $\Tor^A(M, N)$ is the homology of 
$F_\bullet\otimes_A N$, where $F_\bullet$ is a free resolution of $M$:
$$\Tor^A_i(M, N) = \ker(\phi_i\otimes 1)/\im(\phi_{i+1}\otimes 1).$$  
When the free resolution is graded, then so is $\Tor^A_i(M, N)$, as the reader should check.
\proclaim Lemma.  Suppose $M$ is a $(S, \N^n)$-graded $k$-algebra.  Then $\beta_{i\alpha} = \dim_k\Tor^S_i(M, k)_\alpha$.

\noindent {\it Proof.} Note that $k$ itself is an $(S, \N^n)$-graded-module, concentrated in degree $0$.  Thus, $S(-\alpha)\otimes_S k = k(-\alpha)$.  Also, note that if $\phi:S(-\alpha)\to S(-\gamma)$ is morphism of degree $0$, then $\phi\otimes 1_k: S(-\alpha)\otimes k\to S(-\gamma)\otimes k$ is the zero map, unless $\alpha = \gamma$.  Thus, if $F_\bullet$ is a minimal resolution, then 
$$\phi_i\otimes 1_k = 0$$
for $i=0,1,\ldots$, since $F_\bullet$ being minimal,  every $a_{ji}$ has degree $>0$.  Accordingly, $$\Tor^A_i(M, k) = \bigoplus_{\alpha\in\N^n}k(-\alpha)^{\beta_{i\alpha}},$$
and in degree $\alpha$, we have just $k(-\alpha)^{\beta_{i\alpha}}$.  This proves the claim.\hfill/////

\bigskip
\noindent Recall that $\Tor^A(M, N) \cong \Tor^A(N, M)$.  This enables us to compute $\Tor^A(M, N)$ from a resolution of $N$ rather than $M$.  In particular, we can compute $\Tor^A(M, k)$ by using the Koszul resolution $\K_\bullet$ of $k$ to compute $\Tor^A(k, M)$.  This works out particularly nicely when $M$ is a monomial ideal of $S$.  In fact, the Betti numbers of $I$ turn out to be the Betti numbers of certain simplicial complexes associated with $I$.

\smallskip
\noindent {\bf Definition.}  Let $I\subseteq k[\N^n]$ be a monomial ideal.  (We can also view $I$ simply as a monoid ideal of $\N^n$.)  Let $\gamma\in \N^n$.   Then
$${\bf K}^\alpha(I):=\bigl\{\,\tau\in\{0,1\}^n\mid \alpha-\tau \in I\,\bigr\}.$$

\proclaim Theorem. {\rm ([MS], 1.34).}  $\beta_{i\alpha}(I) = \dim_k\widetilde H_{i-1}({\bf K}^\alpha(I), k)$.

\noindent{\it Proof.}  Recall that the Koszul resolution $\K_\bullet$ is:
$$0\buildrel\phi_0\over \longleftarrow S\buildrel\phi_1\over \longleftarrow\bigoplus_{|\tau|=1} S(-\tau)\buildrel\phi_2\over \longleftarrow\bigoplus_{|\tau|=2} S(-\tau)\buildrel\phi_3\over \longleftarrow\cdots.$$
Here, we are assuming that $\tau\in\{0,1\}^n$. Observe that $\im \phi_1 = \langle\,x_1, \ldots, x_n\,\rangle$.  Now,
 $$\eqalign{\beta_{i\alpha}(I)&=\dim_k \Tor^S_i(k, I)_\alpha\cr
                        &=\dim_k (\hbox{degree-$\alpha$ part of the $i^{\rm th}$ homology of $\K_\bullet\otimes_S I$}).\cr}$$
Because $I\subseteq S$ and $\K_\bullet$ is a sum of free modules, $\K_\bullet\otimes_SI\subseteq \K_\bullet\otimes_S S = \K_\bullet$, so $\left(\K_\bullet\otimes_SI\right)_\alpha\subseteq \left(\K_\bullet\right)_\alpha$.  We saw in Lecture 6 that $\left(\K_\bullet\right)_\alpha$ is the reduced chain complex of $\supp \alpha$ (shifted up in homological degree).  We shall show that
$\left(\K_\bullet\otimes_SI\right)_\alpha$ is the reduced chain complex of ${\bf K}^\alpha(I)\subseteq \supp \alpha$.  Indeed, $$\eqalign{
	\left(\K_i\otimes_S I\right)_\alpha 
		&=\left(\bigoplus_{|\tau| = i}S(-\tau)\otimes I\right)_\alpha\cr
		&=\bigoplus_{|\tau| = i}I(-\tau)_\alpha\cr
		&\cong k^{{\bf K}^\alpha_i(I)},\cr}$$
since $$I(-\tau)_\alpha =\cases{k, & if $\alpha-\tau\in I$;\cr0, &otherwise.\cr}$$
Thus, $\left(\K_\bullet\otimes_S I\right)_\alpha$ is the reduced chain complex of ${\bf K}^\alpha(I)$, re-numbered so as to start in degree $0$ rather than $-1$.\hfill/////   

 

 

\end  