## A module which has rank 1 but is not free

Let $R = \mathbb{Z}[x]$ and let $M = (2,x)$ be considered as an $R$-submodule of $R$. Show that $\{2,x\}$ is not a basis of $M$. Show that $M$ has rank 1, but is not free with free rank 1.

Note that $2 \cdot x + (-x) \cdot 2 = 0$, so that $x$ and $2$ are not $R$-linearly independent. Thus $\{2,x\}$ has no hope of being a basis of $M$.

Now suppose $\alpha, \beta \in M$ are nonzero. Similarly, $\beta \cdot \alpha + (-\alpha) \cdot \beta = 0$, so that $\{\alpha, \beta\}$ is not $R$-linearly independent. In particular, the rank of $M$ as an $R$-module is at most 1. On the other hand, since $R$ is an integral domain, every singleton set is linearly independent. So the rank of $M$ over $R$ is exactly 1.

Suppose now that $M$ is free with free rank 1. That is, $M = R\alpha$ for some $\alpha$. Now $2 = r\alpha$ for some $r \in R$, so that $\alpha$ has degree 0. That is, $\alpha \in \mathbb{Z}$, and we have $\alpha \in \{\pm 1, \pm 2\}$. If $\alpha = \pm 1$, then $M = R$. This is a contradiction since (for example) $1 \notin (2,x)$. So, without loss of generality, $\alpha = 2$. But then $M = 2\mathbb{Z}[x]$, and so $x \notin M$, a contradiction. Thus $M$ is not free of free rank 1 as an $R$-module.