Any two minimal free resolutions of an R-module are isomorphic Suppose that $(R, m)$ is a local Noetherian ring and $M$ is an R-module
f.g. Show that if $F_*\to M\to 0$ and $G_*\to M\to 0$ are minimal resolutions of $M$ then there is an isomorphism of complexes $\phi_*$ that raises
identity morphism in $M$.
I have thought to do the following:
To prove that there is an isomorphism of complexes of chains $\phi_*$ that raises identity, what I think is that we must build this by induction:
$$\require{AMScd}
\begin{CD}
\dotsb@> f_3 >> F_2 @>f_2>> F_1 @> f_1 >> M @>>>0 \\
@. @V \phi_{2} VV @V \phi_1 VV @V Id VV @. \\
\dotsb@> g_3 >> G_2 @> g_2 >> G_1 @> g_1 >> M @>  >>0
\end{CD}
$$
To define $\phi_1:F_1\to G_1$, we use that $F_1$ is projective and that $g_1:G_1\to M$ is surjective, which immediately gives us that $\phi_1:F_1\to G_1$ exists.
I am stuck in the inductive step: suppose that $n-1>1$ and that the morphism $\phi_{n-1}:F_{n-1}\to G_{n-1}$ has already been built for all $n-1>1$. So how can I define $\phi_{n}:F_{n}\to G_{n}$ such as $\phi_{n-1}f_n=g_n\phi_n$?
To prove that this is an isomorphism of chain complexes I have thought of performing the same process described above but defining $\xi_n: G_n\to F_n$ for all $n>1$ and thus proving that $\phi_n$ and $\xi_n$ are inverse of each other for all $n>1$.
Thanks!
 A: You can construct $\phi_n$ in the same way as $\phi_1$.
Defining $\xi_n$ in this way you only will get that $\xi_\bullet\circ\phi_\bullet$ is homotopic to identity because in general maps extending homomorphisms are not unique. But in your case $\phi$ will be an isomorphism.
Let's check that $\phi_1$ is an isomorphism. The result will follow by induction. As $F$ and $G$ are minimal resolutions, tensoring by $R/m$ that diagram we have
\begin{CD}
F_1/mF_1 @> \bar{f_1} >> M/mM  \\
@V \bar{\phi_1} VV @V Id VV @. \\
G_1/mG_1 @> \bar{g_1} >> M/mM 
\end{CD}
where $\bar{f_1}$ and $\bar{g_1}$ are isomorphisms. Thus $\bar{\phi_1}$ is an isomorphism which implies $coker(\phi_1)/mcoker(\phi_1)=0$. By Nakayama's lemma, $coker(\phi_1)=0$.
Thence we have exact sequence $0\rightarrow\ker\phi_1\rightarrow F_1\xrightarrow{\phi_1}G_1\rightarrow0$ which gives arise to an exact sequence
$$Tor_1(G_1,R/m)\rightarrow\ker\phi_1/m\ker\phi_1\rightarrow F_1/mF_1\xrightarrow{\bar{\phi_1}} G_1/mG_1\rightarrow0.$$
Being $G_1$ projective that Tor vanishes so $\ker\phi_1/m\ker\phi_1=0$. Nakayama again give us $\ker\phi_1=0$.
Note that the hyphothesis that $R$ is local is very important in this prove!
