What's going on here?

The crux of the matter is that without division of scalars, the "spanning power" of a vector is greatly curtailed.

For example, $span \{2\} = 2 \mathbb{Z}$ = the even integers.

Because the scalars do not form a field, the span of $\{2\}$ is not able to reach the odd numbers, and hence, although $\{2\}$ is maximal independent set, it is not able to span the whole space.

To get the full span, we need to add something else, say $\{3\}$. But then our spanning set $\{2,3\}$ contains more elements than the dimension of the space, and is linearly dependent.

Synoposis: The spanning power of the vectors is curtailed. So a maximal independent say may not be spanning. To span whole the space, we may need the combined effects of extra vectors -- but this may cause the set to be linearly dependent.

In contrast, if we take the rationals $\mathbb{Q}$ as a $\mathbb{Q}$-module over itself, we have a field of scalars and hence a vector space, and $\{2\}$ is indeed a spanning set for $\mathbb{Q}$. And $\{2,3\}$ is not a minimal spanning set.

The crux of the matter is that without division of scalars, the "spanning power" of a vector is greatly curtailed.

For example, $span \{2\} = 2 \mathbb{Z}$ = the even integers.

Because the scalars do not form a field, the span of $\{2\}$ is not able to reach the odd numbers, and hence, although $\{2\}$ is maximal independent set, it is not able to span the whole space.

To get the full span, we need to add something else, say $\{3\}$. But then our spanning set $\{2,3\}$ contains more elements than the dimension of the space, and is linearly dependent.

Synoposis: The spanning power of the vectors is curtailed. So a maximal independent say may not be spanning. To span whole the space, we may need the combined effects of extra vectors -- but this may cause the set to be linearly dependent.

In contrast, if we take the rationals $\mathbb{Q}$ as a $\mathbb{Q}$-module over itself, we have a field of scalars and hence a vector space, and $\{2\}$ is indeed a spanning set for $\mathbb{Q}$. And $\{2,3\}$ is not a minimal spanning set.