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


<https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-44/issue-3/Betti-numbers-of-almost-complete-intersections/10.1215/ijm/1256060413.full>


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?



Relation between free resolutions and minimal free resolutions
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 $R \xrightarrow{1} R$.


