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C.3 Syzygies and resolutions

Syzygies

Let $R$ be a quotient of $\hbox{Loc}_< K[\underline{x}]$ and let $I=(g_1, ..., g_s)$ be a submodule of $R^r$. Then the module of syzygies (or 1st syzygy module, module of relations) of $I$, syz($I$), is defined to be the kernel of the map $R^s \rightarrow R^r,\; \sum_{i=1}^s w_ie_i \mapsto \sum_{i=1}^s w_ig_i$.

The k-th syzygy module is defined inductively to be the module of syzygies of the $(k-1)$-stsyzygy module.

Note, that the syzygy modules of $I$ depend on a choice of generators $g_1, ..., g_s$. But one can show that they depend on $I$ uniquely up to direct summands.

Example:
 
  ring R= 0,(u,v,x,y,z),dp;
  ideal i=ux, vx, uy, vy;
  print(syz(i));
==> -y,0, -v,0, 
==> 0, -y,u, 0, 
==> x, 0, 0, -v,
==> 0, x, 0, u  

Free resolutions

Let $I=(g_1,...,g_s)\subseteq R^r$ and $M= R^r/I$. A free resolution of $M$ is a long exact sequence

\begin{displaymath}...\longrightarrow F_2 \buildrel{A_2}\over{\longrightarrow} F...
...1}\over{\longrightarrow} F_0\longrightarrow M\longrightarrow
0,\end{displaymath}


where the columns of the matrix $A_1$generate $I$. Note that resolutions need not to be finite (i.e., of finite length). The Hilbert Syzygy Theorem states that for $R = \hbox{Loc}_< K[\underline{x}]$there exists a ("minimal") resolution of length not exceeding the number of variables.

Example:
 
  ring R= 0,(u,v,x,y,z),dp;
  ideal I = ux, vx, uy, vy;
  resolution resI = mres(I,0); resI;
==>  1      4      4      1      
==> R <--  R <--  R <--  R
==> 
==> 0      1      2      3      
==> 
  // The matrix A_1 is given by
  print(matrix(resI[1]));
==> vy,uy,vx,ux
  // We see that the columns of A_1 generate I.
  // The matrix A_2 is given by
  print(matrix(resI[3]));
==> u, 
==> -v,
==> -x,
==> y  

Betti numbers and regularity

Let $R$ be a graded ring (e.g., $R = \hbox{Loc}_< K[\underline{x}]$) and let $I \subset R^r$ be a graded submodule. Let

\begin{displaymath}
R^r = \bigoplus_a R\cdot e_{a,0} \buildrel A_1 \over \longl...
...ts \longleftarrow
\bigoplus_a R\cdot e_{a,n} \longleftarrow 0
\end{displaymath}

be a minimal free resolution of $R^r/I$ considered with homogeneous maps of degree 0. Then the graded Betti number $b_{i,j}$ of $R^r/I$ is the minimal number of generators $e_{a,j}$ in degree $i+j$ of the $j$-th syzygy module of $R^r/I$ (i.e., the $(j-1)$-st syzygy module of $I$). Note that, by definition, the $0$-th syzygy module of $R^r/I$ is $R^r$ and the 1st syzygy module of $R^r/I$ is $I$.

The regularity of $I$ is the smallest integer $s$ such that

\begin{displaymath}
\hbox{deg}(e_{a,j}) \le s+j-1 \quad \hbox{for all $j$.}
\end{displaymath}

Example:
 
  ring R= 0,(u,v,x,y,z),dp;
  ideal I = ux, vx, uy, vy;
  resolution resI = mres(I,0); resI;
==>  1      4      4      1      
==> R <--  R <--  R <--  R
==> 
==> 0      1      2      3      
==> 
  // the betti number:
  print(betti(resI), "betti");
==>            0     1     2     3
==> ------------------------------
==>     0:     1     -     -     -
==>     1:     -     4     4     1
==> ------------------------------
==> total:     1     4     4     1
==> 
  // the regularity:
  regularity(resI);
==> 2