Let \(R\) be a ring, and let \(M\) be an \(R\)-module. An \(R\)-submodule \(N \subset M\) is called maximal if \(N \ne M\) and there is no \(R\)-submodule \(P\) satisfying \(N \subsetneq P \subsetneq M\).
Show that an \(R\)-submodule \(N\) of \(M\) is maximal if and only if the quotient module \(M/N\) is a simple \(R\)-module (that is, \(M/N \ne \{0\}\) and \(M/N\) has no submodules other than \(\{0\}\) and \(M/N\)).
Let \(M\) be a \(\mathbb{Z}\)-module (an abelian group). Show that a submodule \(N \subseteq M\) is maximal if and only if the quotient \(|M/N| = p\) for some prime number \(p\).
Let \(M = \mu_\infty = \{z \in \mathbb{C}^\times : z^n = 1 \text{ for some } n \in \mathbb{Z}^+\}\) be the multiplicative group of all complex roots of unity, viewed as a \(\mathbb{Z}\)-module. Show that \(M\) has no maximal \(\mathbb{Z}\)-submodules.
Solution
Goal: Characterize maximal submodules via simplicity of quotients in (a), deduce prime-order quotients for abelian groups in (b), and prove that the divisible group of roots of unity has no maximal submodules in (c).
1. Part (a): \(N \subset M\) is maximal if and only if \(M/N\) is simple.
1.1. By the Submodule Correspondence Theorem (Lattice Isomorphism Theorem for modules), the canonical quotient projection \(\pi: M \to M/N\) induces an inclusion-preserving bijection between the set of submodules of \(M\) containing \(N\) and the set of submodules of \(M/N\):\[P \longleftrightarrow P/N = \pi(P).\]
1.2. Under this correspondence, proper intermediate submodules \(N \subsetneq P \subsetneq M\) correspond bijectively to non-zero proper submodules \(\{0\} \subsetneq P/N \subsetneq M/N\).
1.3. \(N\) is maximal if and only if \(N \ne M\) and no such intermediate submodule \(P\) exists.
1.4. This occurs if and only if \(M/N \ne \{0\}\) and \(M/N\) contains no submodules other than \(\{0\}\) and \(M/N\), which is the definition of \(M/N\) being a simple \(R\)-module.
2. Part (b): A \(\mathbb{Z}\)-submodule \(N \subseteq M\) is maximal if and only if \(|M/N| = p\) is prime.
2.1. A \(\mathbb{Z}\)-module is an abelian group, and \(\mathbb{Z}\)-submodules are subgroups.
2.2. By Part (a), \(N\) is maximal if and only if \(A = M/N\) is a simple abelian group.
2.3. (\(\impliedby\)): If \(|A| = p\) is prime, then by Lagrange’s Theorem the only subgroups of \(A\) are \(\{0\}\) and \(A\). Since \(p \ge 2\), \(A \ne \{0\}\), so \(A\) is simple.
2.4. (\(\implies\)): Suppose \(A\) is a simple abelian group. Since \(A \ne \{0\}\), choose a non-zero element \(x \in A \setminus \{0\}\).
2.5. The cyclic subgroup \(\langle x \rangle \le A\) is a non-zero submodule of \(A\). By simplicity, \(\langle x \rangle = A\), so \(A\) is cyclic.
2.6. If \(A \cong \mathbb{Z}\), then the subgroup \(\langle 2x \rangle \cong 2\mathbb{Z}\) is a non-zero proper subgroup of \(A\), contradicting simplicity. Thus \(A\) must be finite.
2.7. Since \(A\) is finite cyclic, \(A \cong \mathbb{Z}/n\mathbb{Z}\) for some integer \(n \ge 2\).
2.8. If \(n\) is composite, let \(p\) be a prime divisor of \(n\) with \(1 < p < n\). The subgroup \(\langle (n/p) x \rangle\) has order \(p\), so \(\{0\} \subsetneq \langle (n/p) x \rangle \subsetneq A\), contradicting simplicity.
2.9. Therefore \(n = p\) must be prime, so \(|M/N| = |A| = p\).
3. Part (c): \(M = \mu_\infty\) has no maximal \(\mathbb{Z}\)-submodules.
3.1. Suppose for contradiction that \(N \subset M\) is a maximal \(\mathbb{Z}\)-submodule (subgroup).
3.2. By Part (b), the quotient group \(M/N\) has prime order \(|M/N| = p\).
3.3. In any finite group of order \(p\), every element raised to the power \(p\) is the identity: for all \(x \in M\), \((x N)^p = x^p N = N\), which implies \(x^p \in N\).
3.4. We show that \(M\) is \(p\)-divisible: let \(x \in M\). By definition of \(\mu_\infty\), \(x^m = 1\) for some integer \(m \ge 1\).
3.5. In \(\mathbb{C}\), the polynomial \(z^p - x = 0\) has \(p\) solutions. Let \(y \in \mathbb{C}^\times\) be any solution, so \(y^p = x\).
3.6. Then \(y^{p m} = (y^p)^m = x^m = 1\), which proves that \(y \in \mu_\infty = M\).
3.7. Thus every element \(x \in M\) can be written as \(x = y^p\) for some \(y \in M\).
3.8. Applying 3 to \(y \in M\), we have \(x = y^p \in N\).
3.9. This implies \(M \subseteq N\), so \(N = M\), which contradicts the definition of a maximal submodule (\(N \subsetneq M\)).
3.10. Thus no maximal \(\mathbb{Z}\)-submodule of \(M\) exists.
4. Conclusion:
Maximal submodules correspond to simple quotients, simple abelian groups are cyclic of prime order, and the divisible group of roots of unity possesses no maximal submodules.