Existence of minimal spanning sets for Modules

Question

It is clear that, for the same reason why Vector Spaces have maximal linearly independent sets, Modules will have maximal linearly independent sets. For Vector Spaces, we can show that these maximal linearly independent sets are also minimal spanning sets, but for modules this is untrue.

What I am asking is if there exists a minimal spanning set for any module. I have tried the following:

Let $\mathcal{T}$ be the (clearly non-empty) set of spanning sets of $V$, ordered in the following way:

For any $S_1,S_2\in \mathcal{T}, S_1\succ S_2$ in case $S_1\subset S_2$

Consider now any (totally ordered) chain $\mathcal{T}'\subseteq \mathcal{T}$ under $\succ$.

I claim if we let $M=\bigcap\limits_{S\in \mathcal{T}'}$, $M$ is a spanning set (and thus in $\mathcal{T}$), and also an upperbound of $\mathcal{T}'$.

That it is an upperbound is clear.

Suppose $M$ is not a spanning set. Then there is some $v\in V$ that is not in the span of $M$. For each $S\in \mathcal{T}'$, let $C_S$ be the set of linear combinations in $S$ that form $v$. It is clear that if $S_1\prec S_2$, $C_{S_1}\supseteq C_{S_2}$, but I can't show that the total intersection $\bigcap\limits_{S\in\mathcal{T}'}C_S$ is non-empty.

Answer 1

Modules need not have minimal spanning sets.

Consider the $\Bbb Z$-module given by the Prüfer p-group $M=\Bbb Z[1/p]/\Bbb Z$. Every finitely generated submodule of $M$ is cyclic, given by $\langle \displaystyle\frac{1}{p^n} \rangle$ for some $n$. In fact, if one looks at the submodule generated by some elements, only the exponent of $p$ in $\displaystyle\frac{a}{p^n}$ (where $p \not \mid a$) matters because $\displaystyle\frac{a}{p^k} \in \langle \displaystyle\frac{b}{p^n} \rangle$ for $p \not \mid a,b$ iff $k \leq n$.
It follows that a subset $S$ of $M$ is a spanning set iff the exponents of $p$ in the denominators of elements in $S$ are unbounded. But if we have such a set, we can easily leave out some elements and still have unbounded exponents of $p$ in the denominators, so $M$ doesn't contain a minimal spanning set.

Edit: there's a different way to come to the same conclusion. In fact, not only every finitely generated submodule is of the form $\langle \displaystyle\frac{1}{p^n} \rangle$, but every proper submodule. In particular, every proper submodule of $M$ is finite. It follows that a subset $S$ is a generating set iff it is infinite. (And we can of course always take some elements away from an infinite subset such that the subset remains infinite.)
This is equivalent to the statement above, because there are only finitely many elements with bounded exponent of $p$ in the denominator.

Answer 2

Unfortunately, there is a mistake in your argument.

You claim that, if $\mathcal{T}'$ is a chain in the family of spanning sets, then $M=\bigcap_{S\in\mathcal{T}'}S$ is a spanning set.

Consider, for a prime $p$, $$ V=\Bigl\{\frac{a}{p^n}:a\in\mathbb{Z},n\ge0\Bigl\} $$ which is clearly a subgroup of $\mathbb{Q}$ (hence a $\mathbb{Z}$-module). Let $$ S_m=\Bigl\{\frac{1}{p^n}:n\ge m\Bigr\} $$ Then $S_m$ is a spanning set for $V$ and $\{S_m:m\ge0\}$ is a chain of spanning sets, with empty intersection.