Proof verification: the simple $\mathbb{Z}$ modules are exactly $\mathbb{Z}/p\mathbb{Z}$ for all primes $p$ I'm reading Noncommutative Algebra by Benson Farb and R. Keith Dennis.
On page 30, it says that the simple $\mathbb{Z}$-modules are $\mathbb{Z}/p\mathbb{Z}$ for all primes $p$.
This is very easy to show using a theorem they prove that $M$ is simple if and only if it is $R/I$ for some maximal left ideal $I$, but I tried proving it without using this theorem as an exercise.
The theorem ended up being very easy to show and it ruled out infinite Abelian groups very quickly, which surprised me. It makes me wonder whether the argument works.
Does this argument work for showing that the simple $\mathbb{Z}$-modules are exactly $\mathbb{Z}/p\mathbb{Z}$?
I'll define a module to be simple if and only if its lattice of submodules has exactly two elements.
Lemma 101: If $G = \langle B \rangle$ where $0 \notin B$ be a nontrivial Abelian group that is not generated by a single element, then $B$ contains two elements that generate distinct proper subgroups of $G$.
By hypothesis, for every $a$ in $B$, $\langle a \rangle$ is a proper subgroup of $B$.
Suppose for contradiction that every pair of elements $a \neq b$ in $B$ generated the same group. Then $G$ would equal $\langle c \rangle$ for every $c$ in $B$. This is clearly not true, so there must exist $a$ and $b$ which generate distinct subgroups.
Note that we did not assume that $B$ was minimal among sets generating $G$.
Theorem: A $\mathbb{Z}$-module $M$ is simple if and only if it is $\mathbb{Z}/p\mathbb{Z}$.
RTL: Fix a prime $p$. Let $M$ is $\mathbb{Z}/p\mathbb{Z}$. Then $0$ generates $\langle 0 \rangle$ and any nonzero element generates $M$, which is distinct from $\langle 0 \rangle$. Therefore $M$ contains exactly two submodules.
LTR: Let $G$ be $M$ construed as an Abelian group. Let $G$ be generated by $B$, where $B$ has size minimal among all sets that generate $G$.
Suppose $B$ is empty, then our group is the trivial group and not a simple module.
Suppose $B$ contains two or more elements. By Lemma 101, there exist $a \neq b$ in $B$ such that $\langle a \rangle \neq \langle b \rangle$. Therefore our subgroup lattice contains at least three elements and we're done.
Suppose $B$ contains exactly one element $a$.
Suppose $a$ has infinite order, then $\langle a + a \rangle$ is a subgroup of $G$ that is distinct from $G$ and $G$ is not simple.
Suppose $a$ has order $n$.
If $n = kl$ where $k$ and $l$ are both not $1$, then $\langle ka \rangle$ is a proper submodule that is not equal to $G$ or $\langle 0 \rangle$.
If $n$ is prime, then $G$ is simple.
This exhausts the possibilities.
Corrections

*

*Changed "empty group" to "trivial group".


*Changed "every $c$ in $B$" to "every nonzero $c$ in $B$" in the proof of Lemma 101.


*Updated the statement of Lemma 101 to rule out $0 \in B$ entirely.
 A: Yes, this argument works, but it's really not any simpler than the proof of the general fact you cite about maximal ideals. Also the use of contradiction in the proof of the lemma is unnecessary, and the lemma itself can be replaced by a simpler and more powerful statement. Here is a streamlined version of the proof, indicating how it needs to be (only slightly) modified for the general case.

Lemma: A module $M$ (over an arbitrary ring) is simple iff $M \neq 0$ and every nonzero element of $M$ generates it. In particular, simple modules are cyclic.

Proof. Let $M$ be a simple module. If $M = 0$ then it cannot be simple. Now let $m \in M$ be any nonzero element. Then the submodule $\langle m \rangle$ generated by $m$ is not zero and so by simplicity it must be all of $M$. Conversely, since every submodule contains a cyclic submodule, if $\langle m \rangle = M$ for all nonzero $m$ then every nonzero submodule is all of $M$, so $M$ is simple. $\Box$
This argument really does not simplify when specialized to the case of $\mathbb{Z}$-modules.

Theorem: The simple $\mathbb{Z}$-modules are the quotients $\mathbb{Z}/p \mathbb{Z}$ where $p$ is prime.

Proof. By the lemma, a simple module $M$ must be cyclic. If $M \cong \mathbb{Z}$ then it has a proper submodule $2\mathbb{Z}$ so is not simple. So $M \cong \mathbb{Z}/n\mathbb{Z}$ for some $n$. If $n = 1$ then $M = 0$ is not simple, so $n > 1$. If $m \in M$, regarded as an ordinary integer, is nonzero (so is not divisible by $n$) then $M/\langle m \rangle \cong \mathbb{Z}/\gcd(n, m)\mathbb{Z}$, so by the lemma again, $M$ is simple iff we always have $\langle m \rangle = M$ for all $m \neq 0$ iff $\gcd(n, m) = 1$ for all $m$ such that $n \nmid m$ iff $n$ is prime. $\Box$
The more general argument for a simple module $M$ over an arbitrary ring $R$ does not require this division into the infinite and finite cases. We just observe that by the lemma $M \cong R/I$ for some left ideal $I$ and then observe that if $m \in M$ is nonzero then $M/\langle m \rangle \cong R/(I + r)$, where $r$ is any lift of $m$ to $R$, so as above $M$ is simple iff $M = \langle m \rangle$ for all $m \neq 0$ iff $M/\langle m \rangle = 0$ for all $m \neq 0$ iff $I + r = R$ for all $r \not \in I$ iff $I$ is maximal.
