Give an example of an integral domain \(R\), a free \(R\)-module \(M\), and an \(R\)-submodule \(N\) of \(M\) such that \(N\) is not a free \(R\)-module.
Solution
Let \(R=\ZZ[x]\), an integral domain; let \(M=R\), a free \(R\)-module of rank \(1\) with basis \(\{1\}\); and let \(N=(2,x)=\{2p+xq : p,q\in\ZZ[x]\}\), an ideal of \(R\) and hence an \(R\)-submodule of \(M\).
1. If \(N\) is a free \(R\)-module, then \(N=(f)\) for some nonzero \(f\in R\).
For nonzero \(a,b\in N\), the relation \(b\cdot a-a\cdot b=0\) has nonzero coefficients \(b\) and \(-a\), so any two elements of \(N\) are linearly dependent over \(R\). A basis of \(N\) therefore has at most one element. Since \(N\ne0\), a basis is a single element \(f\ne0\), and \(N=Rf=(f)\).
2. The ideal \(N\) is not principal.
Suppose \(N=(f)\). Since \(2\in N\), \(f\) divides \(2\) in \(\ZZ[x]\); degrees add in the domain \(\ZZ[x]\), so \(f\in\{\pm1,\pm2\}\). If \(f=\pm1\), then \(1\in N\); but every element \(2p+xq\) of \(N\) has even constant term \(2p(0)\), a contradiction. If \(f=\pm2\), then \(N=2\ZZ[x]\); but \(x\in N\) has coefficient \(1\notin2\ZZ\), a contradiction.
3. Q.E.D.
By steps 1 and 2, the submodule \(N=(2,x)\) of the free \(\ZZ[x]\)-module \(M=\ZZ[x]\) is not free.