Prove $\{(2,0)\}$ is not extendable to a basis for $\mathbb{Z}^2$

Question

I want to show that not every independent set in a free module is extendable to a basis.

Let $R=\mathbb{Z}$ and consider the $R$-module $M=R^2=\mathbb{Z}^2$. Then, $M$ is free of rank $2$ and $S=\{(2,0)\}$ is clearly $R$-independent, since $\mathbb{Z}$ is a domain.

How can I prove that $S$ is not extendable to a basis for $M$?

Thanks.

Answer 1

Hint: let $(r, s)$ be arbitrary integers. What does it take to express $(1, 0)$ as an integral combination of $(2, 0)$ and $(r, s)$?

Answer 2

Consider the matrix $$ \begin{pmatrix} 2 & 0 \\ a & b \end{pmatrix} $$ Since its determinant is $2b$, the matrix is never invertible over $\mathbb Z$. Therefore, $\{ (2,0),(a,b) \}$ is never a basis for $\mathbb{Z}^2$.

In general, an integer vector is part of a basis iff its entries are coprime.