Let $(R,\mathfrak{m},k)$ a Noetherian local ring and $M$ a finitely generated $R$-module. In this case, given $F_{\bullet}$ a free resolution of $M$, what is the relation between $F_{\bullet}$ and a minimal free resolution $G_{\bullet}$ of $M$?
In this paper
Dugger says that we can decompose $F_{\bullet}$ into $G_{\bullet} \oplus Q_{\bullet}$ where $Q_{\bullet}$ is a split exact resolution of zero. I'm new in this area. How I can make this decomposition?
The proof is based on the following observation: If $A \in \mathrm{M}_{n,m}(R)$ is an $n \times m$ matrix with coefficients $a_{ij}$ such that there exists indices $i,j$ where $a_{ij}$ is a unit, then there exist $B \in \mathrm{GL}_n(R)$ and $C \in \mathrm{GL}_m(R)$ such that $$ B A C = \begin{bmatrix} 1 & 0 & \dots & 0 \\ 0 & * & \dots & * \\ \vdots & \vdots & \ddots & \vdots \\ 0 & * & \dots & * \end{bmatrix}. $$
To find $B$ and $C$ you simply reorder your basis and then perform Gaussian row and column elimination. Now we have a commutative diagram $\require{AMScd}$ \begin{CD} R^m @>A>> R^n\\ @V C^{-1} V V @VV B V\\ R \oplus R^{m - 1} @>>1 \oplus A'> R \oplus R^{n - 1} \end{CD}
where $A'$ is the lower right block in $B A C$.
Recall that a free resolution $$F_i \xrightarrow{A_i} F_{i-1} \to \dots \xrightarrow{A_1} F_0 \xrightarrow{} M$$ is minimal if and only if all the coefficients of the matrices $A_i$ are contained in $\mathfrak{m}$. If the resolution is not minimal, there exists some matrix coefficient which is not in $\mathfrak{m}$. Since $R$ is local, this coefficient will be a unit and we can apply the above to write $$ F_\bullet \cong F_\bullet' \oplus (R[l] \xrightarrow{1} R[l - 1]) $$ for some shift $l$. If $F_\bullet'$ is not minimal we inductively continue until we reach a minimal complex $G_\bullet$. Now $F_\bullet \cong G_\bullet \oplus Q_\bullet$, where $Q_\bullet$ is a direct sum of shifts of $R \xrightarrow{1} R$.
