Theorem. Let R be a principal ideal domain. Every submodule of a free R-module is free. Proof. This proof uses the well-ordering principle: for every set J, there is an ordering < on J with respect to which J is well-ordered. (An ordered set is well-ordered if every nonempty subset has a minimal element. With the usual ordering, the set of non- negative integers is well-ordered, but Z, Q, and R are not.) The well-ordering principle is equivalent to the axiom of choice. This proof is from A Course in Homological Algebra by Hilton and Stammbach. L Let F = j∈J R j be a free module, indexed by the set J. Here R j = R for all j: the subscript is merely to indicate which summand it is. Let M ⊆ F be a submodule. Assume that J is well-ordered. For j ∈ J, let M M F( j) = Ri, F( j) = Ri = F( j) ⊕ R j. i< j i≤ j Because of the direct sum decomposition F( j) = F( j) ⊕ R j, every element in F( j) ∩ M may be written uniquely as (b,r), where b ∈ F( j) and r ∈ R j = R. Define a map f j by f j : F( j) ∩ M −→ R, (b,r) 7−→ r. Then the kernel of f j is F( j) ∩ M, so there is a short exact sequence 0 → F( j) ∩ M → F( j) ∩ M → im f j → 0. im f j is an ideal in R, and since R is a PID, im f j = (r j) for some r j ∈ R. If r j 6= 0, there is an element c j ∈ F( j) ∩ M so that f j(c j) = r j. Claim: {c j : j ∈ J,r j 6= 0} is a basis for M. I’ll check that the elements of this set are linearly independent. Suppose that ∑n k=1 skc jk = 0 for some elements sk ∈ R, with j1 < ··· < jn. Apply f jn : 0 = sn f (c jn ) = snrn, and since R is a domain and rn is nonzero, then sn must be zero. By induction, each sk is zero. This proves linear independence. Now I’ll check that this set generates M. Assume not: then there is a smallest i ∈ J such that there is an element a ∈ F(i) ∩ M which cannot be written in terms of the elements of the set {c j}. 0 0 0 Let J = { j ∈ J : r j 6= 0} (so our potential basis is {c j : j ∈ J }). If i 6∈ J , then the map F(i) ∩ M → F(i) ∩ M is equality, so a ∈ F(i) ∩ M. But then there is a k < i in J so that a ∈ F(k) ∩ M, contra- dicting the minimality of i. Thus i ∈ J0. Write fi(a) as fi(a) = sri and form b = a − sci. Since a cannot be written as an R-linear combination of the c j’s, neither can b. Also, fi(b) = fi(a) − s fi(ci) = 0, so b ∈ F(i) ∩M. This contradicts the minimality of the index i, and so every element of M can be expressed as a linear combination of the c j’s..