Content

Consider a finite free module $F$$F$ over a PID $A$$A$, let $x_1,...,x_n$$x_1,...,x_n$ be the basis of $F$$F$.

Let $x\in F$$x\in F$, define the content of $x=c_1{x}_1+...+c_n{x}_n$$x=c_1x_1+...+c_nx_n$ to be $\mathrm{cont}(x):=gcd(c_1,...,c_n)$$\mathrm{cont}(x):=\gcd(c_1,...,c_n)$. Note that it is a class up to a unit.

We would like to define that $\mathrm{cont}(x)$$\mathrm{cont}(x)$ is not depends on the choice of basis. i.e. it is the property of $F$$F$, not $F$$F$ with basis.

Consider the $A$$A$-Module $F^{\ast }:={\mathrm{Hom}}_{A-\mathrm{Mod}}(F,A)$$F^*:=\mathrm{Hom}_{A-\mathrm{Mod}}(F,A)$, i.e. the dual space of $F$$F$.

It is not hard to see that set $\{\varphi \in F^{\ast }|\varphi (x)\}$$\{\phi\in F^*|\phi(x)\}$ form an ideal in $A$$A$. Indeed, a principal ideal $(c)$$(c)$ since $R$$R$ is a PID.

We aim to prove that $c=\mathrm{cont}(x)$$c=\mathrm{cont}(x)$. Since $c$$c$ is not dependent on basis.

Firstly, we need to prove that $\mathrm{cont}(x)\in (c)$$\mathrm{cont}(x)\in(c)$.

Since $\mathrm{cont}(x)=gcd(c_1,...,c_n)$$\mathrm{cont}(x)=\gcd(c_1,...,c_n)$, by Bezout Theorem, $\mathrm{cont}(x)=a_1{c}_1+...+a_n{c}_n$$\mathrm{cont}(x)=a_1c_1+...+a_nc_n$.

Let ${\phi }_1,...,{\phi }_n$$\varphi_1,...,\varphi_n$ be the dual basis of $x_1,...,x_n$$x_1,...,x_n$.

Then $\mathrm{cont}(x)=\phi (x)=a_1{\phi }_1(x)+...+a_n{\phi }_n(x)=a_1{c}_1{\phi }_1(x_1)+...+a_n{c}_n\phi (x_n)=a_1{c}_1+...+a_n{c}_n$$\mathrm{cont}(x)=\varphi(x)=a_1\varphi_1(x)+...+a_n\varphi_n(x)=a_1c_1\varphi_1(x_1)+...+a_nc_n\varphi(x_n)=a_1c_1+...+a_nc_n$.

For any $\varphi \in F^{\ast },$$\phi\in F^*,$ we have $\varphi (x)=c_1\varphi (x)+...+c_n\varphi (x_n)$$\phi(x)=c_1\phi(x)+...+c_n\phi(x_n)$.

Hence $\mathrm{cont}(x)=gcd(c_1,...,c_n)|\varphi (x)$$\mathrm{cont}(x)=\gcd(c_1,...,c_n)|\phi(x)$, thus $\mathrm{cont}(x)=c$$\mathrm{cont}(x)=c$, up to a unit.