TITLE: $k[x]/(x^n)$ module with finite free resolution is free
QUESTION [9 upvotes]: How to show a $k[x]/(x^n)$ module with finite free resolution is free? 

Suppose we have a exact sequence $k[x]/(x^n)^{\oplus n_1}\to k[x]/(x^n)^{\oplus n_{0}}\to M\to 0$, how do we get information to show $M$ is free?

REPLY [2 votes]: You can use elementary tools to show this. In here elementary methods are used assuming the  modules involved are finitely generated.
Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:

Let $M$ be free a $R$-module. If $B\subseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.

As $M$ is free, there is a set $I$ such that
$$ M\simeq\bigoplus_IR,$$
so we consider $M$ as this direct sum. Define the function
$$\psi:M\rightarrow\bigoplus_I k,$$
by letting for each $i\in I$ and each $a\in M$ $\text{ }\psi(a)_i$ to be the constant term of $a_i$. As for each $a\in M$ we have $x^{n-1}a=x^{n-1}\psi(a)(1)$, from the linear independence of $B$ over $R$ we get
that the set $\{\psi(x):x\in B\}$ is a linear independent subset of the $k$-vector space $\bigoplus_I k$.
Thus there is a basis $C$ of the $k$-vector space $\bigoplus_I k$ such that $\{\psi(x):x\in B\}\subseteq C$. We claim the set
$$B'=(C\setminus \{\psi(x):x\in B\})\cup B$$
is a basis of $M$ over $R$; notice $\psi$ is $1-1$ on $B'$ and $\psi(B')=C$.
Let us see by reverse induction on $0\leq j\leq n-1$ that for each $a\in M$ with $deg(a_i)\geq j$ or $a_i=0$ for each $i\in I$ we have $a\in$Span$_R(B')$
From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.
If for some $0<j\leq n-1$ the property holds, pick any $a\in M$ with $\deg(a_i)\geq j-1$ or $a_i=0$ for all $i\in I$, then by the inductive hypothesis we may assume that for each $i\in I$, $a_i=b_ix^{j-1}$ for some $b_i\in k$. Now consider the set $\{x^{j-1}b:b\in B'\}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}\psi(b)$, hence as $\psi(B')=C$ and $\psi$ is $1-1$ on $B'$ we get that $a\in$Span$_R(B')$.
Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,\ldots, c_m\in B'$  are distinct and $p_1,\ldots p_m\in R$ are such that
$$p_1c_1+\cdots+p_mc_m=0.$$
Each $p_i$ can be written as $p_i=\sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0\leq j\leq n-1$ that $b_{i,j}=0$ for all $1\leq i\leq m.$
We have
$$0=x^{n-1}p_1c_1+\cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}\psi(c_1)+\cdots+x^{n-1}a_{1,m}\psi(c_m),$$
thus as $\psi$ is $1-1$ on $B'$ and $\psi(B')=C$ we obtain $a_{1,i}=0$ for all $1\leq i\leq m.$
If for some $0\leq j<n-1$ we have that $a_{j',i}=0$ for all $0\leq j'\leq j$ for all $i$, then
$$0=x^{n-2-j}p_1c_1+\cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}\psi(c_1)+\cdots+x^{n-2-j}a_{j+1,m}\psi(c_m);$$
since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=\cdots=p_m=0$.
Therefore $B'$ is a $R$-basis of $M$ with $B'\supseteq B$.
Now suppose $M$ is an $R$-module that has a free finite resolution
$$0\rightarrow M_m\rightarrow M_{m-1}\cdots\rightarrow M_1\rightarrow M\rightarrow 0,$$
where the map $M_j\rightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.
Let us see by reverse induction on $1\leq j\leq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.
So suppose $Im(f_j)$ is free for some $1<j\leq m$. Then by what we prove above we know
$$Im(f_{j-1})\simeq M_{j-1}/\ker(f_{j-1}),$$
is free since $M_{j-1}$ is free and $\ker(f_{j-1})=Im(f_j)$ is free.
Therefore $M$ is free.