Let \(D\) be a commutative ring with unit. A submodule \(N\) of the \(D\)-module \(M\) is said to be pure in \(M\) just in case for every \(y \in N\) and \(a \in D\), \(ax = y\) is solvable in \(N\) if solvable in \(M\).
Show that if \(N\) is a direct summand of \(M\) then \(N\) is pure in \(M\).
Show that if \(N\) is pure in \(M\), \(z \in M\), and \(\mathrm{ann}(z+N) = (d)\), then there is \(w \in M\) with \(z+N = w+N\) and \(\mathrm{ann}(w) = (d)\).
Show that if \(N\) is pure in \(M\), \(M\) is a finitely generated torsion module, and \(D\) is a p.i.d., then \(N\) is a direct summand of \(M\).
Solution
For an element \(v\) of a \(D\)-module, write \(\operatorname{ann}_D(v)=\{a\in D:av=0\}\).
1. A direct summand is pure, proving part (a).
Suppose \(M=N\oplus N'\) and let \(\pi:M\to N\) be the projection onto \(N\). It is \(D\)-linear and restricts to the identity on \(N\). If \(y\in N\), \(a\in D\), and \(ax=y\) has a solution \(x\in M\), then \[ a\pi(x)=\pi(ax)=\pi(y)=y. \] Thus \(\pi(x)\in N\) solves the same equation, as required.
2. A coset with principal annihilator has a representativewith that same annihilator, proving part (b).
Since \(d\in\operatorname{ann}_D(z+N)\), the element \(dz\) lies in \(N\). The equation \(dx=dz\) has the solution \(z\in M\). Purity supplies a solution \(n\in N\). Put \(w=z-n\). Then \(w+N=z+N\) and \(dw=dz-dn=0\), so \((d)\subseteq\operatorname{ann}_D(w)\).
Conversely, if \(aw=0\), then \(a(z+N)=a(w+N)=0\) in \(M/N\). Thus \(a\in\operatorname{ann}_D(z+N)=(d)\). This proves the reverse inclusion and hence \(\operatorname{ann}_D(w)=(d)\). No assumption that \(d\) is nonzero or a nonunit is needed.
3. Under the hypotheses in part (c), the quotient map\(q:M\to M/N\) admits a \(D\)-linear section.
The quotient \(Q=M/N\) is finitely generated, by the images of a finite generating set of \(M\). It is torsion: a nonzero scalar annihilating a representative also annihilates its coset. The structure theorem over the PID \(D\) therefore gives \[ Q\cong\bigoplus_{i=1}^r D/(d_i) \] for nonzero nonunits \(d_i\in D\) [@DF04]. The empty sum is allowed when \(Q=0\). Let \(u_i\in Q\) correspond to \(1\) in the \(i\)th cyclic summand, so \(\operatorname{ann}_D(u_i)=(d_i)\). Choose \(z_i\in M\) with \(q(z_i)=u_i\). By step 2, there is \(w_i\in M\) with \(q(w_i)=u_i\) and \(\operatorname{ann}_D(w_i)=(d_i)\).
Define \[ s:Q\longrightarrow M,\qquad s\left(\sum_{i=1}^r a_i u_i\right)=\sum_{i=1}^r a_i w_i. \] In the displayed direct sum, two coefficient lists represent the same element exactly when their \(i\)th coefficients differ by a multiple of \(d_i\) for every \(i\). Since \(d_iw_i=0\), those changes do not change the proposed value of \(s\). Thus \(s\) is well-defined and \(D\)-linear. Moreover \(q(s(u_i))=u_i\) for all generators, so \(q\circ s=\operatorname{id}_Q\). For \(Q=0\), the zero map is the required section.
4. The section yields the direct summand in part (c).
For every \(m\in M\), \[ m=(m-s(q(m)))+s(q(m)), \] where the first term lies in \(\ker q=N\) and the second in \(s(Q)\). If \(s(u)\in N\), applying \(q\) gives \(u=q(s(u))=0\), and hence \(s(u)=0\). Therefore \(N\cap s(Q)=0\) and \(M=N\oplus s(Q)\).