If a module is second syzygy and has no free summand , then it can be defined via a minimal resolution? Let $M$ be a finitely generated module over a Noetherian local ring $(R,\mathfrak m)$. Assume $M$ has no free summand, and that there exists an exact sequence $0\to M \to R^{\oplus a}\xrightarrow{f} R^{\oplus b}$.
Then, does there exist another exact sequence $0\to M \to R^{\oplus s}\xrightarrow{g} R^{\oplus t}$ such that $Im(g)\subseteq \mathfrak m R^{\oplus t}$ ?
The only thing I can see is that since $M$ has no free summand, so $\ker(f)\subseteq \mathfrak m R^{\oplus a}$. I have no other idea whatsoever.
I have read things like "over a local ring every complex of free modules is a direct sum of a minimal complex and a contractible one..." but I have never seen an argument for that or even a statement which would make it precise.
Please help
 A: I'll show that every complex
$$\dots\rightarrow0\rightarrow R^{\oplus a}\xrightarrow{f} R^{\oplus b}
\rightarrow0\rightarrow\dots$$
is the direct sum of a contractible complex and a complex
$$\dots\rightarrow0\rightarrow R^{\oplus s}\xrightarrow{g} R^{\oplus t}
\rightarrow0\rightarrow\dots$$
with $\operatorname{im}(g)\subseteq\mathfrak{m}R^{\oplus t}$.
Since the contractible summand is acyclic, $\ker{f}=\ker{g}$ and so this answers the  question asked. It is not necessary to assume that $M$ has no free summands.
Proof:
By induction on $a$, it suffices to assume that $\operatorname{im}(f)\not\subseteq\mathfrak{m}R^{\oplus b}$ and prove that there is a (contractible) direct summand of the form
$$\dots\rightarrow0\rightarrow R\xrightarrow{\operatorname{id}_R} R
\rightarrow0\rightarrow\dots.$$
Since  $\operatorname{im}(f)\not\subseteq\mathfrak{m}R^{\oplus b}$, there is at least one summand $R$ of $R^{\oplus b}$ such that $\pi f$ is surjective, where $\pi:R^{\oplus b}\to R$ is the projection map onto this summand.
Then $\pi f$ must split, so there is a map $\varphi:R\to R^{\oplus a}$ such that $\pi f\varphi=\operatorname{id}_R$.
Then there is an embedding of complexes
$$\require{AMScd}$$
\begin{CD}
\dots@>>>0@>>>R@>\operatorname{id}_R>>R@>>>0@>>>\dots\\
@.@.@VV\varphi V@VVf \varphi V\\
\dots@>>>0@>>>R^{\oplus a}@>f>>R^{\oplus b}@>>>0@>>>\dots\\
\end{CD}
which is split by the map
\begin{CD}
\dots@>>>0@>>>R^{\oplus a}@>f>>R^{\oplus b}@>>>0@>>>\dots\\
@.@.@VV\pi f V@VV\pi V\\
\dots@>>>0@>>>R@>\operatorname{id}_R>>R@>>>0@>>>\dots\\
\end{CD}
$\blacksquare$
If we define a complex $(C^\ast, d^\ast)$ of free $R$-modules to be minimal if $\operatorname{im}(d^n)\subseteq\mathfrak{m}C^{n+1}$ for every $n$, then this proof can easily be adapted to show that every bounded complex of finitely generated free $R$-modules is the direct sum of a contractible complex and a minimal complex. With a bit more work, the "bounded" and "finitely generated" assumptions can be removed.
