Suppose $R$ be a commutative ring with unity and $N$ is a free $R$-module. m is a maximal ideal of $R$ and $m\in$ m. Then the set of elements $\{ x_1,x_2,...,x_{n-1}, m x_n \}$ which spans $N$ is not linearly independent.
My definition of free module is module with basis.
In case of integral domain, I proved it.
$$x_n=\alpha_1 x_1+...+\alpha_nmx_n$$ $$mx_n=m\alpha_1 x_1+...+m\alpha_nmx_n$$ Subtracting $mx_n$ gives, $m(\alpha_nm-1)=0$. If $R$ is integral domain it forces either $m$ is unit or $0$. unit is not possible . So $m$ has to be 0. Showing that the last vector is redundant
But in general case i couldn't do. Please give a hint
This is trivial.
Suppose it is linearly independent, then
$$x_n=\alpha_1 x_1+\dots+\alpha_nmx_n$$
Which shows $\alpha_nm-1=0$. So $m$ is unit. A contradiction to maximality of m