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.

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