Let $A$ be a commutative ring with $1.$ Consider the free $A$ module $A^n$ and $\{x_1,\cdots,x_n\}$ set of elements in $A^n$ which is linearly independent. May I conclude that $\{x_1,\cdots,x_n\}$ form a basis for $A^n$ ?
I think it is not true. But I fail to give counter example. Help me.
Hint: Try $A=\mathbb Z$, $n=1$, and $x_1=2$.
In general, $\{x_1,\ldots,x_n\}$ is a basis iff the matrix expressing them in the canonical basis is invertible in $A$, which happens iff its determinant is a unit in $A$.