PID Modules, Free Submodules

The Submodule of a Free Module is Free

First consider the case when R is not a pid. Let R act as an R module, free of rank 1. Let H be an ideal of R that is not principal. In other words, multiple generators are required to span H. If H is free it has rank greater than 1. This cannot fit inside the ring R. This is clear when R is an integral domain, since the basis elements of H would be zero divisors; but it is true in general. If R is not a pid, this theorem has no hope of being true.

Let R be a pid and let F be a free module over R with basis b. Let M be a submodule of F.

Well order the basis b, so that we can consider the basis elements b₀ b₁ b₂ b₃ etc in order. Yes, this requires the axiom of choice, unless b is countable, or can otherwise be well ordered.

For any ordinal j, let F_j be the free submodule of F spanned by b₀ through b_j. Let M_j be the intersection of F_j with M. Let P_j be the projection of M_j onto the j^th copy of R.

Note that P_j is isomorphic to an ideal of R. Since R is a pid, let g_j generate P_j. If g_j = 0 there is nothing to do. If g_j is nonzero, let u_j be anything in M_j whose j^th component is equal to g_j. As we shall see, u_j is a basis for M, whence M is free.

Let's demonstrate linear independence first. Suppose a finite linear combination ∑ c_ju_j = 0. Look at the last term, the largest j. Only u_j contributes g_j to the j^th component. Thus c_j has to be zero, which is a contradiction. The vectors u_j are independent by construction. Each one brings in a component that was not there before, like vectors building a lower triangular matrix.

Suppose u does not span all of M. Every x that is not spanned, lies in some M_j, and naturally, x is not spanned by u₀ through u_j. Let j be the least ordinal such that u₀ through u_j does not span some x in M_j.

Write the j^th component of x as sb_j, for some s in R. If s is 0 then x belongs to a smaller M_j, hence s is nonzero. Write s as tg_j. Let y = x-tu_j. Now x is spanned by u₁ through u_j iff y is spanned. Yet y belongs to a lesser submodule M_i, below M_j. Therefore y is spanned, and so is x.

The arbitrary submodule M of F has a basis, and is free. Furthermore, the rank of M is less than or equal to the rank of F.