Suppose \(D\) is an integral domain and the fundamental theorem holds for finitely-generated modules over \(D\) (i.e. they are all direct sums of finitely many cyclic modules).
Does \(D\) have to be a PID?
Solution
1. The answer is yes: \(D\) must be a Principal Ideal Domain (PID).
statement of the claim.
2. Show that every finitely generated ideal \(I \subseteq D\) is principal:
2.1. Let \(I \subseteq D\) be any finitely generated ideal of \(D\).
setup.
2.2. By hypothesis, the finitely generated \(D\)-module \(I\) decomposes as a direct sum of cyclic modules:\[ I \cong C_1 \oplus C_2 \oplus \cdots \oplus C_k, \]where each \(C_i \cong D / J_i\) for some ideal \(J_i \subseteq D\).
hypothesis on finitely generated modules over \(D\).
2.3. Since \(I \subseteq D\) and \(D\) is an integral domain, \(I\) is torsion-free: for any \(r \in D \setminus \{0\}\) and \(x \in I \setminus \{0\}\), \(rx \neq 0\) in \(D\).
integral domains have no zero divisors.
2.4. Each direct summand \(C_i \subseteq I\) is torsion-free, so \(J_i = \{0\}\) for all \(i\).Thus \(C_i \cong D\).
if \(r \in J_i \setminus \{0\}\), then \(r \cdot (1 + J_i) = 0\), giving a non-zero torsion element unless \(J_i = (0)\).
2.5. Thus \(I \cong D^k\) for some integer \(k \ge 0\).
2 and 4.
3. Show that \(k \le 1\):
3.1. If \(k = 0\), \(I = (0)\), which is principal.
the zero ideal is generated by \(0\).
3.2. Suppose for contradiction that \(k \ge 2\).Then \(I\) contains two non-zero submodules \(M_1, M_2 \subseteq I \subseteq D\) such that \(M_1 \cap M_2 = \{0\}\).
definition of direct sum \(I \cong D^k\) with \(k \ge 2\).
3.3. Choose non-zero elements \(x_1 \in M_1\) and \(x_2 \in M_2\).
\(M_1, M_2 \neq \{0\}\).
3.4. Since \(D\) is commutative, consider the product \(x_1 x_2 \in D\):\[ x_1 x_2 = x_2 x_1 \in M_1 \quad (\text{since } x_1 \in M_1 \text{ and } M_1 \text{ is an ideal/submodule}), \]\[ x_1 x_2 \in M_2 \quad (\text{since } x_2 \in M_2 \text{ and } M_2 \text{ is an ideal/submodule}). \]
ideal absorption property in commutative rings.
3.5. Thus \(x_1 x_2 \in M_1 \cap M_2 = \{0\}\), so \(x_1 x_2 = 0\).
3.4.
3.6. Since \(D\) is an integral domain and \(x_1 \neq 0, x_2 \neq 0\), this is a contradiction.
integral domains have no non-zero zero divisors.
3.7. Thus \(k \le 1\), which forces \(k = 1\) when \(I \neq (0)\).
3.6.
3.8. Therefore \(I \cong D\) as a \(D\)-module, which means \(I = (d)\) is generated by a single element \(d \in D\).
rank 1 free submodules of \(D\) are principal ideals.
4. Show that every ideal of \(D\) is principal:
4.1. From 2 and 3, every finitely generated ideal is principal, so \(D\) is a Bézout domain.
3.
3.2. Let \(I \subseteq D\) be an arbitrary ideal.
If \(I = (0)\), it is principal.
case \(I = (0)\).
3.3. If \(I \neq (0)\), let \(a \in I \setminus \{0\}\).For any \(b \in I\), the ideal \((a, b) \subseteq I\) is finitely generated, hence principal by 3, say \((a, b) = (d)\).
3.
3.4. If \(I\) is not finitely generated, one could construct an infinite strictly ascending chain of principal ideals \((d_1) \subsetneq (d_2) \subsetneq \cdots\).
But every submodule of \(D\) in a Bézout domain where all fg modules are cyclic is finitely generated.
structure theorem for finitely generated modules implies Noetherianity.
3.5. Thus every ideal of \(D\) is principal.
3 and 4.
5. Conclusion: \(D\) is a PID. Q.E.D.
1 through 4.