\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\card{\mathop{{\rm card}}}
\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 6.  Graded morphisms}

\bigskip
\noindent{\it  Part A.  ``Monomial matrices''}  

\medskip\noindent
In the first part of this lecture, we describe the material presented in Miller-Sturmfels, pages 11--13. The goal is to develop an efficient way to record the data that determine a homogeneous degree 0 morphism between free $(A,\Gamma)$-graded-modules
$$\phi:\bigoplus_iA(-\gamma_i)\to \bigoplus_j A(-\delta_j),$$
where $A$ is a $\Gamma$-graded $k$-algebra.  Such morphisms occur in a free resolution of a $(A, \Gamma)$-graded-module, which explains our interest in this question. 

\smallskip To create the description, let $\epsilon_i$ (respectively, $\epsilon_j$) be the generator of $A(-\gamma_i)$  (respectively, $A(-\delta_j)$). Its degree is $\gamma_i$ (respectively, $\delta_j$).  Then $$\phi(\epsilon_j)=\sum_ja_{ji}\epsilon_j,$$ where $\delta_j+\deg(a_{ji})=\gamma_i$, and so,
$$\eqalign{\phi\left(\sum_ib_i\epsilon_i\right) &=\sum_ib_i\left(\sum_ja_{ji}\epsilon_j\right)\cr
&=\sum_j\left(\sum_ia_{ji}b_i\right)\epsilon_j\;.\cr}$$  If we let $b$ denote the column vector $(b_i)^T$ and interpret it as an element of $\bigoplus_iA(-\gamma_i)$ by identifying the column with $1$ in the $i^{th}$ place and $0$ elsewhere with $\epsilon_i$, and if furthermore we view $a=(a_{ji})$ as a matrix with entries from $A$, then $$\phi(b) = ab\;.$$

\smallskip
Additional compression of notation is possible under additional assumptions on $A$.  Suppose $k$ is a field, $A_\gamma$ is one-dimensional for all $\gamma\in \Gamma$ (as is the case when $A$ is the monoid algebra $A=k[\Gamma]$), and suppose further that $\Gamma$ is cancellative.  Then the degrees $\gamma_i$ and $\delta_j$ determine the degree of $a_{ji}$ uniquely: $\deg(a_{ji})=\gamma_i-\delta_j$ (provided $\delta_j+\xi=\gamma_i$ has a solution $\xi=\gamma_i-\delta_j$).  Once a ($k$-vector-space) generator for each $A_\gamma$ is chosen, the the coefficient $a_{ji}$ will be determined completely by an element $\lambda_{ji}\in k$.  Thus, the mapping $\phi$ can be represented by a matrix with entries in $k$.  For example, if $A=k[\Gamma]$, then:
$$a_{ji} = \cases{\lambda_{ji}X^{\gamma_i-\delta_j}&if $\delta_j \preceq   \gamma_i$;\cr 0&otherwise.\cr}$$
The grade vectors $(\gamma_i)$ and $(\delta_j)$ and the matrix $(\lambda_{ji})$ determine $\phi$ completely. 
\smallskip

\smallskip\noindent The example on page 13 of [Miller-Sturmfels] illustrates the presentation of several morphisms in the manner just described.  (The text asserts that the morphisms on page 13 constitute a minimal free resolution. The reader should verify this.)

\vfill\eject

\noindent {\it Part B. The Koszul complex}

\medskip\noindent   We describe the Koszul resolution of $k$ as a $k[\N^n]$-module.  Throughout this example, $k$ is a field.   

\smallskip\noindent As background, we review the definition of the reduced (or augmented) chain complex of an oriented simplicial complex $\Delta$.  Let $F_i=F_i(\Delta)$ denote the set of $i$-dimensional faces of $\Delta$, {\it i.e.}, elements of $\Delta$ of cardinality $i+1$.  For $\sigma\in F_i$, let $e_\sigma\in k^{F_i}$ be the function that has value $1$ at $\sigma$ and is $0$ otherwise.  Define $\partial_i:k^{F_i}\to k^{F_{i-1}}$ by:
$$\partial_i(e_\sigma)=e_{\sigma\setminus 1}-e_{\sigma\setminus 2}+e_{\sigma\setminus 3}-\cdots +(-1)^i e_{\sigma\setminus {i+1}}\;,\eqno(1)$$
where $\sigma\setminus j$ denotes $\sigma$ with its $j^{\rm th}$ element (in the order determined by the given orientation) dropped.  Then, $\widetilde{\cal C}_{\,\bullet}(\Delta; k)$ is the complex:
$$0\longleftarrow k^{F_{-1}}\buildrel\partial_0\over\longleftarrow k^{F_0}\buildrel\partial_1\over\longleftarrow k^{F_1}\buildrel\partial_2\over\longleftarrow\cdots \buildrel\partial_{d}\over\longleftarrow k^{F_{d}}\longleftarrow 0\;,$$
where $d$ is the largest dimension of any face of $\Delta$.  The reduced homology of $\Delta$ is the sequence of vector spaces:
$$\widetilde H_i(\Delta, k) = \ker(\partial_i)/\im(\partial_i)\;.$$

\smallskip\noindent
Now we modify the above construction by using $S:=k[\N^n]$ in place of $k$ and assigning grades in an appropriate way.  Assume the vertices of $\Delta$ are labeled $1,2,\ldots, n$.  Let 
$${\Bbb K}_{\,i}(\Delta):= \bigoplus_{\sigma \in F_{i-1}}S(-\sigma).$$
Here, we are using $\sigma$ to refer to an element of $\N^n$.  For example, $(1,0,\ldots,0)\in F_0$ is the first vertex in $\Delta$, and $(1,1,0,0,\ldots,0)$ is a one-dimensional simplex (which might or might not be in $F_1$).  Note that the index $i$ on the free $S$-modules ${\Bbb K}_{\,i}(\Delta)$ refers to the number of vertices in each of the faces involved in defining ${\Bbb K}_{\,i}(\Delta)$ rather than the dimension.  In particular, ${\Bbb K}_{\,0}(\Delta) = S$ is the free module with one generator in degree $(0,0,\ldots,0)$, corresponding to the unique empty face $(0,0,\ldots,0)$.  Also, ${\Bbb K}_{\,1}(\Delta) = S^n$.  Thus, the numbering of $\Bbb K_{\,\bullet}(\Delta)$ is shifted from $\widetilde{\cal C}_\bullet(\Delta, k)$.  

\smallskip\noindent
For $\sigma\in F_{i-1}$, we let $\epsilon_\sigma$ denote the generator of $S(-\sigma)\subseteq{\Bbb K}_{\,i}(\Delta)$.  We define
$$\partial_i:{\Bbb K}_{\, i}(\Delta)\to {\Bbb K}_{\,i-1}(\Delta)$$ by:
$$\partial_i(\epsilon_\sigma)=X^{\sigma[1]}\epsilon_{\sigma\setminus 1}-X^{\sigma[2]}\epsilon_{\sigma\setminus 2} + X^{\sigma[3]}\epsilon_{\sigma\setminus 3}-\cdots +(-1)^{i-1} X^{\sigma[i]}\epsilon_{\sigma\setminus {i}}\;,\eqno(2)$$
where $\sigma[j]:=\sigma-(\sigma\setminus j)$.  (The choice of coefficient is precisely what is needed to balance degrees.)   Then the complex ${\Bbb K}_{\,\bullet}(\Delta)$ is the following:
$$0\longleftarrow {\Bbb K}_{\,0}(\Delta)\buildrel\partial_1\over\longleftarrow {\Bbb K}_{\,1}(\Delta)\buildrel\partial_2\over\longleftarrow {\Bbb K}_{\,2}(\Delta)\buildrel\partial_3\over\longleftarrow\cdots \buildrel\partial_{d+1}\over\longleftarrow {\Bbb K}_{\,d+1}(\Delta)\longleftarrow 0\;.\eqno(2.1)$$

\smallskip\noindent
Let ${\Bbb K}_{{\,\bullet}\, \alpha}(\Delta)$ denote the part of ${\Bbb K}_{\,\bullet}(\Delta)$ in degree $\alpha\in \N^n$.  Thus, 
$${\Bbb K}_{\,i\, \alpha}(\Delta)\subseteq \bigoplus\{\,S(-\sigma)_\alpha\mid \sigma\in\Delta\,,\; \card \sigma = i \;\&\; \sigma\leq\alpha\,\}\;,$$
and ${\Bbb K}_{\,i\, \alpha}(\Delta)$ has basis, $\{\,e_{\sigma\,\alpha}:=X^{\alpha-\sigma}\epsilon_\sigma\in S(-\sigma)_\alpha\mid \sigma\in\Delta\,,\;\card \sigma = i \;\&\; \sigma\leq\alpha\,\}$.
Let us write $\partial_{i\alpha}$ for the restriction of $\partial_i$ to ${\Bbb K}_{\,i\,\alpha}(\Delta)$.  Then, $\im(\partial_{i\alpha})\subseteq{\Bbb K}_{\,(i-1)\,\alpha}(\Delta)$.  Multiplying (2) by $X^{\alpha-\sigma}$, we see
$$\partial_{i\alpha}(e_\sigma)=e_{\sigma\setminus 1}-e_{\sigma\setminus 2}+e_{\sigma\setminus 3}-\cdots +(-1)^{i-1}e_{\sigma\setminus {i}}\;.\eqno(3)$$  Thus, ${\Bbb K}_{{\,\bullet}\, \alpha}(\Delta)$ is the same as the complex $\widehat{\cal C}(\Delta\wedge\alpha, k)$ (except it is shifted---${\Bbb K}_{{\,\bullet}\, \alpha}(\Delta)$ is indexed starting from $0$ rather that $-1$), where $\Delta\wedge\alpha:=\{\,\sigma\wedge\alpha\mid\sigma\in\Delta\,\} = \{\,\sigma\in\Delta\mid\sigma\subseteq\supp\alpha\,\}$. 


\proclaim Proposition.  Let $\Delta$ be the $(n-1)$-simplex, {\it i.e.} the set of all subsets of $\{\,1,2,\ldots,n\,\}$.  Then the complex ${\Bbb K}_{\,\bullet}:= {\Bbb K}_{\,\bullet}(\Delta)$ is a minimal free resolution of $S/(x_1, \ldots, x_n)$.

\noindent{\it Proof.}  It is necessary only to show that $(2.1)$ is exact.  Since $\Delta$ is the $(n-1)$-simplex, $\Delta\wedge\alpha$ is itself a simplex and therefore has 0 homology.  Thus, ${\Bbb K}_{\,\bullet\,\alpha}$ is exact for each $\alpha$, and this establishes the result we need.\hfill///// 

\medskip\noindent{\it Question for listeners/readers.}  Fill in the missing portion of the proof concerning minimality.  Refer to Definition 1.24 on page 12 of Miller-Sturmfels.
\end  