Finite rank implies finitely generated?

Question

Let $M$ be an $R$-module. If $M$ has finite rank, does it have to be finitely generated?

Answer 1

By your definition, is $\Bbb Q$ a rank one $\Bbb Z$-module?

It certainly is not finitely generated.

Answer 2

Every abelian group is a $\mathbb{Z}$-module, where $\mathbb{Z}$ acts by $z\cdot g = \sum\limits_\text{n times}g$. Therefore $(\mathbb{Z},+)$ and $(\mathbb{Z/2Z},+)$ are both $\mathbb{Z}$-modules.

Now let $$ M = \mathbb{Z}\oplus \mathbb{Z/2Z} \oplus \cdots$$ where we have a countably many copies of $\mathbb{Z/2Z}$

Now, we see that this module is of rank $1$:

We know that given $m = (1,a_1,\dots,a_j, \dots)$, for any $r \in \mathbb{Z}$ we get $r\cdot m = 0 $ iff $r = 0$.

But, given any other element $m' \in M$, where the first entry is $m_1$, we know $$\begin{align} 2\cdot(-m_1 \cdot m + m') &= -2m_1 \cdot m + 2\cdot m' \\ &=(-2m_1, -2m_1\cdot a_1, \cdots) + (2m_1, 2\cdot a_1, \cdots) \\ 2\cdot a_1 = 0 \mod 2\Rightarrow&=(-2m_1, 0, \cdots) + (2m_1, 0, \cdots) \\ &= (0,\cdots) \\ &= \mathbf{0} \end{align}$$

Therefore, there are no elements linearly independent to $m$, meaning the rank of $M$ is $1$. However, clearly $M$ is not finitely generated (since there are countably many combinations of $0,1$s in the infinite copies of $\mathbb{Z/2Z}$.