# Linear Algebra over a Ring/Chain complexes of finitely generated free modules

Proposition (every chain complex of finitely generated free modules over a Bézout domain is the direct sum of some of its subcomplexes with at most two nonzero terms):

Let

${\displaystyle \cdots {\overset {\partial }{\longrightarrow }}C_{n+1}{\overset {\partial }{\longrightarrow }}C_{n}{\overset {\partial }{\longrightarrow }}C_{n-1}{\overset {\partial }{\longrightarrow }}\cdots }$

be a chain complex whose objects are finitely generated free modules over a Bézout domain ${\displaystyle R}$. This chain complex is then the countable direct sum of chain complexes of the form

${\displaystyle \cdots \longrightarrow 0\longrightarrow 0\longrightarrow L_{n+1}\longrightarrow R_{n}\longrightarrow 0\longrightarrow 0\longrightarrow \cdots }$,

where ${\displaystyle L_{n+1}\leq C_{n+1}}$ and ${\displaystyle R_{n}\leq C_{n}}$.

(On the condition of the countable choice.)

Proof: We shall construct a direct sum decomposition

${\displaystyle C_{n}=L_{n}\oplus R_{n}}$,

where ${\displaystyle R_{n}=\ker \partial }$. Once this is accomplished, we have in fact obtained a direct sum decomposition of the initial chain complex, because elements of ${\displaystyle R_{n}}$ are mapped to zero by ${\displaystyle \partial }$, and elements of ${\displaystyle L_{n}}$ are mapped to ${\displaystyle R_{n-1}}$ due to the chain complex condition ${\displaystyle \partial \circ \partial =0}$.

In order to achieve this decomposition, we invoke Dedekind's theorem for Bézout domains, which tells us that ${\displaystyle \operatorname {im} \partial \leq C_{n-1}}$ is finitely generated and free; indeed, it is finitely generated, since a generating set is given by the image (via ${\displaystyle \partial }$) of a generating set of ${\displaystyle C_{n}}$. Let thus ${\displaystyle b_{1},\ldots ,b_{n}}$ be a basis of ${\displaystyle \operatorname {im} \partial \leq C_{n-1}}$. For each ${\displaystyle j\in \{1,\ldots ,n\}}$, we choose an arbitrary, but fixed ${\displaystyle a_{j}\in \partial ^{-1}(b_{j})}$, and then we define ${\displaystyle L_{n}:=\langle a_{1},\ldots ,a_{n}\rangle }$. This yields the desired direct sum decomposition. Indeed, ${\displaystyle L_{n}\cap R_{n}=\{0\}}$, since whenever

${\displaystyle r_{1}a_{1}+\cdots +r_{n}a_{n}=0}$

for some elements ${\displaystyle r_{1},\ldots ,r_{n}}$ of ${\displaystyle R}$, applying ${\displaystyle \partial }$ to both sides of this equation and using its ${\displaystyle R}$-linearity yields

${\displaystyle r_{1}b_{1}+\cdots +r_{n}b_{n}=0}$,

which implies ${\displaystyle r_{1}=r_{2}=\cdots =r_{n}=0}$. Moreover, ${\displaystyle L_{n}+R_{n}=C_{n}}$, since if ${\displaystyle c\in C_{n}}$ is arbitrary, we may select ${\displaystyle r_{1},\ldots ,r_{n}\in R}$ such that

${\displaystyle \partial (c)=r_{1}b_{1}+\cdots +r_{n}b_{n}}$,

from which we may easily deduce that

${\displaystyle c-r_{1}a_{1}-\ldots -r_{n}a_{n}\in \ker \partial }$. ${\displaystyle \Box }$

Proposition (every chain complex of finitely generated free modules over a Bézout domain splits as the direct sum of two types of elementary chain complexes):

Let

${\displaystyle \cdots {\overset {\partial }{\longrightarrow }}C_{n+1}{\overset {\partial }{\longrightarrow }}C_{n}{\overset {\partial }{\longrightarrow }}C_{n-1}{\overset {\partial }{\longrightarrow }}\cdots }$

be a chain complex whose objects are finitely generated free modules over a Bézout domain ${\displaystyle R}$. This chain complex is the countable direct sum of copies of the following two chain complexes:

1. ${\displaystyle 0\longrightarrow R\longrightarrow 0}$
2. ${\displaystyle 0\longrightarrow R{\overset {r}{\longrightarrow }}R\longrightarrow 0}$ for an element ${\displaystyle r\in R}$, ie. the arrow represents the function which is given by multiplication by ${\displaystyle r}$
(On the condition of the countable choice.)

Proof: Using the notation of the last theorem, we have ${\displaystyle C_{n}=L_{n}\oplus R_{n}}$, where ${\displaystyle L_{n}}$ is finitely generated and ${\displaystyle R_{n}}$ is sent to zero by ${\displaystyle \partial }$. ${\displaystyle \Box }$