I have to give examples for modules such that
$1.$ There's a linearly independent set with more elements than a generator set.
$2.$ There's a linearly independent set with the same cardinal than a basis, but is not a basis.
$3.$ Exists a generator with the same cardinal than a basis, but is not a basis.
$4.$ The module is finitely generated and have a submodule that is not finitely generated.
I've made a few:
$(1)$ Seems wrong to me, or at least very counter-intuitive. An idea I got was using a finite dimensional vector space like $\mathbb{R^²}$ and consider a set of vectors $\{v_1,v_2,\dots\}$ where every $v_i$ has only rational entries. I don't know how to proceed formally to do the proof. A second idea: $\mathbb{Z}$ as a $\mathbb{Z}-$module has basis $\{1\}$ while the set {2,3} is linearly independent and with a greater cardinal.
$(2)$ $\mathbb{Z}$ as a $\mathbb{Z}-$module has as basis the set $\{1\}$, and the set $\{2\}$ has the same cardinal as $\{1\}$ but is not a basis because only even elements are generated by it.
$(3)$ My candidate was the vector space $\{f:f:\mathbb{R}\to\mathbb{R}\}$ over $\mathbb{Q}$, but I was unable to prove it.
$(4)$ If $M$ were a cyclic $R-$module where $R$ is the polynomial ring of variables $x_1,x_2,\dots$ and coefficients in a field $\mathbb{K}$. The set $\{x_1,x_2,\dots\}$ generates a submodule, but it can't be generated by a finite set.
Best Answer
For (1), if you think back to linear algebra you'll remember that you can't do this in a vector space. So, it's good that it seems wrong! But it's not. The easiest examples here are noncommutative: for instance, in the ring of linear operators on an infinite-dimensional vector space with basis $e_1,e_2,...$ (as a left module over itself,) the operators $(e_i\mapsto e_{2i})$ and $(e_i\mapsto e_{2i+1})$ are linearly independent even though of course the ring is a cyclic left module for itself. $\{2,3\}$ is not linearly independent over $\mathbb{Z},$ because $3\times 2+(-2)\times 3=0$. You're right on (2). Your massive vector space in (3) will work, as it happens, because the whole space has the same cardinality as a basis, but it's perhaps not the easiest example. Just think of any infinite-dimensional vector space at all. (4) is also correct.
EDIT: My example for (1) is backwards. We need to either look at these two operators as linearly independent for the right module structure, or at the operators $(e_{2i+1}\mapsto e_i, e_{2i}\mapsto 0)$ and $(e_{2i}\mapsto e_i,e_{2i+1}\mapsto 0)$ as left linearly independent. I'll explain the first choice. So let $A$ be the ring of linear operators on a real vector space spanned by $(e_i)_{i\in \mathbb{N}}$. The ring operations here are just like the matrix operations in the finite dimensional case. Let $T_e$ be the operator defined on a basis by $e_i\mapsto e_{2i}$ and $T_o$, similarly, $e_i\mapsto e_{2i+1}$. I claim $T_e$ and $T_o$ are linearly independent for the right $A$-module structure on $A$ given by right multiplication. Indeed suppose $U,V\in A$ and $T_eU+T_oV=0$. If $Ue_i=\sum u_{ij}e_j$ then $T_eU e_i=\sum u_{ij}e_{2j}$ and similarly if $Ve_i=\sum v_{ij} e_j$ then $T_o Ve_i=\sum v_{ij}e_{2j+1}$.
If you think of elements of $A$ as "$\mathbb{N}\times\mathbb{N}$ matrices", then this just says that the even rows of $T_e U$ are the rows of $U$ and the odd rows of $T_e U$ are $0$, and similarly for $T_o V$ (hence the names of $T_e$ and $T_o$.) So, no entries can cancel each other in the sum $T_eU+T_oV$-we've split a single matrix in halves to get all the information from $U$ and of $V$ into it! Thus this sum can only vanish if both $U$ and $V$ do, and $T_e$ and $T_o$ are linearly independent as claimed. Observe that there's nothing special about having chosen $2$, here: I can easily get any finite number of linearly independent elements, or even a countable number, basically just because there's a decomposition of $\mathbb{N}$ into countably many copies of itself. And all this is done in a free module $A$ which can be generated by a single element!