# Linear Algebra over a Ring/Modules over Bézout domains

Proposition (extending a single element of a free, finite-dimensional module over a Bézout domain to a basis):

Let ${\displaystyle R}$ be a Bézout domain, and let ${\displaystyle R^{n}}$ be the ${\displaystyle n}$-dimensional free module over ${\displaystyle R}$. Let ${\displaystyle v=(v_{1},\ldots ,v_{n})\in R^{n}}$ be such that ${\displaystyle \gcd(v_{1},\ldots ,v_{n})=1}$. Then there exist elements ${\displaystyle w_{2},\ldots ,w_{n}\in R^{n}}$ such that ${\displaystyle v,w_{2},\ldots ,w_{n}}$ is a basis of ${\displaystyle R^{n}}$.

Proof: We proceed by induction on ${\displaystyle n}$. If ${\displaystyle n=1}$, the statement becomes trivial, so that the induction base is handled. Let now a general ${\displaystyle n}$ be given; we reduce the claim to the same statement for ${\displaystyle n-1}$.

Indeed, if ${\displaystyle v=(v_{1},\ldots ,v_{n})}$ is given, we may set

${\displaystyle d:=\gcd(v_{2},\ldots ,v_{n})}$,

and then ${\displaystyle \gcd(v_{1},d)=\gcd(v_{1},\ldots ,v_{n})=1}$. Hence, there are elements ${\displaystyle r,s\in R}$ such that

${\displaystyle rv_{1}+sd=1}$.

We define

${\displaystyle w:=(s,-rv_{2}/d,\ldots ,-rv_{n}/d)}$,

so that

${\displaystyle rv+dw=e_{1}}$ and ${\displaystyle sv-v_{1}w=(0,v_{2}/d,\ldots ,v_{n}/d)}$,

where for ${\displaystyle j\in \mathbb {N} }$, as usual, ${\displaystyle e_{j}}$ denotes the vector whose every component is zero except the one in the ${\displaystyle j}$-th place, which is one. From these equations it is evident that

${\displaystyle \langle v,w\rangle =\langle e_{1},(0,v_{2}/d,\ldots ,v_{n}/d)\rangle }$.

Also, ${\displaystyle v}$ and ${\displaystyle w}$ are linearly independent, because if there existed ${\displaystyle b,c\in R}$ such that ${\displaystyle bv+cw=0}$, then ${\displaystyle b(0,v_{2}/d,\ldots ,v_{n}/d)=(v_{1}-sc)w}$ and in particular ${\displaystyle 0=(v_{1}-sc)s}$, so that either ${\displaystyle s=0}$ and ${\displaystyle v}$ and ${\displaystyle w}$ are linearly independent (almost) by definition, or ${\displaystyle v_{1}=sc}$, so that ${\displaystyle b=1}$ whenever ${\displaystyle v_{1}\neq 0}$ (once again, if ${\displaystyle v_{1}=0}$, then the vectors are automatically linearly independent). But this implies that ${\displaystyle c|v_{j}}$ for all ${\displaystyle j\in \{1,\ldots ,n\}}$, so that contrary to the assumption ${\displaystyle \gcd(v_{1},\ldots ,v_{n})\neq 1}$.

It now suffices to note that

${\displaystyle R^{n}/\langle v,w\rangle =R^{n}/\langle e_{1},(0,v_{2}/d,\ldots ,v_{n}/d)\rangle \cong R^{n-1}/\langle (v_{2}/d,\ldots ,v_{n}/d)\rangle }$

by the first Noether isomorphism theorem applied to the homomorphism that forgets the first component, which reduces the dimension of the problem by one as we have ${\displaystyle \gcd(v_{2}/d,\ldots ,v_{n}/d)=1}$. ${\displaystyle \Box }$

Theorem (a minimum cardinality generating set of a finitely generated torsion-free module over a Bézout domain is a basis):

Let ${\displaystyle R}$ be a Bézout domain, and let ${\displaystyle M}$ be a finitely generated torsion-free module over ${\displaystyle R}$. Then every generating set of ${\displaystyle M}$ whose cardinality is minimal is a basis of ${\displaystyle M}$.

Proof: We proceed by induction on ${\displaystyle n}$, where ${\displaystyle n}$ is the cardinality of a minimum cardinality generating set of ${\displaystyle M}$. Let ${\displaystyle x_{1},\ldots ,x_{n}}$ be a minimum cardinality generating set of ${\displaystyle M}$. Then we have a homomorphism

${\displaystyle \varphi :R^{n}\to M}$

such that ${\displaystyle \varphi (e_{k})=x_{k}}$ for ${\displaystyle k\in \{1,\ldots ,n\}}$, where ${\displaystyle e_{k}}$ is the element of ${\displaystyle R^{n}}$ whose every entry is zero, except for the ${\displaystyle k}$-th entry, which is one. Since ${\displaystyle \varphi }$ is surjective, the first isomorphism theorem implies that

${\displaystyle M\cong R^{n}/K}$,

where ${\displaystyle K=\ker \varphi }$. If ${\displaystyle K=\{0\}}$, then the claim is demonstrated. We lead ${\displaystyle K\neq \{0\}}$ to a contradiction. Indeed, in this case there is a nonzero vector ${\displaystyle 0\neq v\in K}$. We denote ${\displaystyle v:=(v_{1},\ldots ,v_{n})}$ and set ${\displaystyle d:=\gcd(v_{1},\ldots ,v_{n})}$. Then ${\displaystyle v/d\in K}$, for otherwise the image of ${\displaystyle v/d}$ in ${\displaystyle R^{n}/K}$ via the canonical projection would be a torsion element. Since we have ${\displaystyle \gcd(v_{1}/d,\ldots ,v_{n}/d)=1}$, we may extend ${\displaystyle v}$ to a basis ${\displaystyle \{v,w_{2},\ldots ,w_{n}\}}$ of ${\displaystyle R^{n}}$, and then ${\displaystyle R^{n}/K}$ is generated by the images of ${\displaystyle w_{2},\ldots ,w_{n}}$ via the canonical projection. Yet this contradicts the minimality of ${\displaystyle n}$. ${\displaystyle \Box }$

Theorem (Dedekind's theorem):

Let ${\displaystyle R}$ be a Bézout domain, and let ${\displaystyle M}$ be a torsion-free module over ${\displaystyle R}$. Then every finitely generated submodule ${\displaystyle N\leq M}$ is free.

Proof: If ${\displaystyle M}$ is torsion-free, then every submodule of ${\displaystyle M}$ is torsion-free as well. Hence, ${\displaystyle N}$ as in the theorem's statement is torsion-free and finitely generated. Hence, there exists a basis of ${\displaystyle N}$. ${\displaystyle \Box }$

Proposition (basis extension over Bézout domains):

Let ${\displaystyle R}$ be a Bézout domain, and let ${\displaystyle M}$ be a torsion-free module over ${\displaystyle R}$. If ${\displaystyle v_{1},\ldots ,v_{k}}$ are linearly independent vectors in ${\displaystyle M}$ such that ${\displaystyle M/\langle v_{1},\ldots ,v_{k}\rangle }$ is torsion-free and finitely generated, then we find an ${\displaystyle n\in \mathbb {N} }$ and vectors ${\displaystyle v_{k+1},\ldots ,v_{n}\in M}$ such that the vectors ${\displaystyle v_{1},\ldots ,v_{k},v_{k+1},\ldots ,v_{n}}$ constitute a basis of ${\displaystyle M}$.

Proof: Since ${\displaystyle M/\langle v_{1},\ldots ,v_{k}\rangle }$ is torsion-free and finitely generated, upon choosing a generating set of minimal cardinality we obtain a basis of that module. We shall denote this basis by ${\displaystyle g_{1},\ldots ,g_{m}}$. If moreover

${\displaystyle \pi :M\to M/\langle v_{1},\ldots ,v_{k}\rangle }$

is the canonical projection, we choose a ${\displaystyle w_{j}\in \pi ^{-1}(g_{j})}$ for each ${\displaystyle j\in \{1,\ldots ,m\}}$. We claim that ${\displaystyle \{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{m}\}}$ is a basis. First, we prove that these vectors are linearly independent. Indeed, suppose that ${\displaystyle a_{1},\ldots ,a_{m},b_{1},\ldots ,b_{k}\in R}$ are such that

${\displaystyle 0=a_{1}w_{1}+\cdots +a_{m}w_{m}+b_{1}v_{1}+\cdots +b_{k}v_{k}}$.

Since ${\displaystyle \pi }$ is a homomorphism,

${\displaystyle 0=\pi (0)=\pi (a_{1}w_{1}+\cdots +a_{m}w_{m}+b_{1}v_{1}+\cdots +b_{k}v_{k})=a_{1}g_{1}+\cdots +a_{m}g_{m}}$,

so that ${\displaystyle a_{1}=a_{2}=\cdots =a_{m}=0}$. Thus

${\displaystyle b_{1}v_{1}+\cdots +b_{k}v_{k}=0}$,

which by virtue of the linear independence of ${\displaystyle v_{1},\ldots ,v_{k}}$ implies ${\displaystyle b_{1}=b_{2}=\cdots =b_{k}=0}$.

It remains to show that ${\displaystyle \{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{m}\}}$ is a generating set of ${\displaystyle M}$. In order to do so, suppose that ${\displaystyle u\in M}$ is arbitrary. We choose ${\displaystyle a_{1},\ldots ,a_{m}\in R}$ such that

${\displaystyle \pi (u)=a_{1}g_{1}+\cdots +a_{m}g_{m}}$.

We further define ${\displaystyle u':=u-a_{1}w_{1}-\cdots -a_{m}w_{m}}$, so that ${\displaystyle \pi (u')=0}$. But by the definition of ${\displaystyle \pi }$, this means that ${\displaystyle u'\in \langle v_{1},\ldots ,v_{k}\rangle }$, so that there are ${\displaystyle b_{1},\ldots ,b_{k}\in R}$ such that

${\displaystyle u'=b_{1}v_{1}+\cdots +b_{k}v_{k}}$, hence ${\displaystyle u=u'+a_{1}w_{1}+\cdots +a_{m}w_{m}=b_{1}v_{1}+\cdots +b_{k}v_{k}+a_{1}w_{1}+\cdots +a_{m}w_{m}}$,

which shows that ${\displaystyle \{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{m}\}}$ is a generating set of ${\displaystyle M}$ since ${\displaystyle u}$ was arbitrary. ${\displaystyle \Box }$

Theorem (dimension formula):