I am confused by the following exercise from Grillet, Abstract Algebra, VIII.4 Q2:
Show that every left $R$-module $M$ has a maximal linearly independent subset, and that a maximal linearly independent subset of $M$ generates a submodule $S$ that is essential in $M$ ($S\cap A \neq 0$ for every submodule $A \neq 0$ of $M$).
In this context, rings are assumed to have unity, and modules to be unital. I'll follow the author in writing $0$ instead of $\{0\}$.
I am new to modules, and not at all sure what I am doing, so I will try to spell out my reasoning. The proof that every left $R$-module $M$ has a maximal linearly independent subset seems to be a routine application of Zorn's lemma, so I will omit that. But then my confusion starts.
First, I've convinced myself that a maximal linearly independent subset of a nonzero module can be empty. For example, let $M$ be $\mathbb{Z}_2$ as a $\mathbb{Z}$-module. Then $1\cdot 0=0$ and $2\cdot 1=0$, so all singleton subsets of $M$ are linearly dependent, and so $\varnothing$ is a (actually, the unique) maximal linearly independent subset of $M$. But then $S = \mathrm{Span}\,\varnothing = 0$, and the assertion of the exercise is false.
Question 1 Is that correct?
That made me guess there is a typo in the exercise, so I tried to repair it. The counterexample cannot be avoided by requiring $A$ to be a proper submodule. For example, $\mathbb{Z}_2 \times \mathbb{Z}_2$ as a $\mathbb{Z}$-module likewise has $\varnothing$ as its maximal linearly independent subset, but it has proper submodules, and the same problem arises. The best I have come up with is this.
Make the assumption that $M$ is a left $R$-module in which $rx = 0$ implies $r=0$ whenever $x \neq 0$. Then the assertion of the exercise holds.
Proof: Suppose $Y$ is a maximal linearly independent subset of $M$, and let $S = \mathrm{Span} \, Y$. Suppose $A \neq 0$ with $S \cap A = 0$. Pick nonzero $x \in A$. Suppose $$\Big(\sum_{y \in Y} r_y y\Big) + rx = 0, \quad r_y, r \in R, r_y \neq 0 \text{ for only finitely many }y$$ That implies both $\sum_{y\in Y} r_y y=0$ and $rx=0$, so by linear independence, $r_y = 0$ for all $y$, and by the assumption added to the exercise, $r=0$, so $Y \cup \{x\}$ is linearly independent, a contradiction. Thus $S$ is essential.
However, my worry is that the assumption that if $x \neq 0$, then $rx = 0$ implies $r = 0$ is very strong, and slightly artificial. If I understand correctly, it's stronger than the claim that $M$ is torsion-free, which only requires that if $x \neq 0$, then $rx = 0$ implies $r = 0$ for regular $r$ (i.e. for $r$ a non-zero divisor). So my attempted fix of the exercise more or less requires $R$ to be an integral domain and $M$ to be torsion-free. But this can hardly be thought of as repairing a typo any more! So I think I must be quite confused about something.
Question 2 If I'm right in thinking that the exercise is wrong as stated, how should it be repaired?