Let \(R\) be a PID with field of fractions \(F\). Let \(V\) be an \(n\)-dimensional vector space over \(F\).
Show that every finitely generated \(R\)-submodule \(M \subseteq V\) is free of rank \(\le n\).
Let \(M\) and \(N\) be free \(R\)-submodules of rank \(n\) in \(V\). Show that there exists a nonzero element \(\alpha \in R\) such that \(\alpha M \subseteq N\). Use this to show that there exists an \(R\)-basis \(\{e_1, \dots, e_n\}\) of \(M\) and nonzero elements \(\beta_1, \dots, \beta_n \in F\) such that \(\{\beta_1 e_1, \dots, \beta_n e_n\}\) is an \(R\)-basis of \(N\) (Invariant Factor Theorem for Lattices).
Solution
1. An \(R\)-linearly independent subset of \(V\) is \(F\)-linearly independent.
Let \(m_1,\ldots,m_k\in V\) be \(R\)-linearly independent and suppose \(\sum_i c_im_i=0\) with \(c_i\in F\). Choose \(d\in R\setminus\{0\}\) with \(dc_i\in R\) for all \(i\). Then \(\sum_i(dc_i)m_i=0\), so \(dc_i=0\) and hence \(c_i=0\) for all \(i\).
2. In part (a), every finitely generated \(R\)-submodule \(M\subseteq V\) is free of rank at most \(n\).
If \(r\in R\setminus\{0\}\) and \(v\in M\) satisfy \(rv=0\), then \(v=r^{-1}(rv)=0\) in the \(F\)-vector space \(V\); so \(M\) is torsion-free. By the structure theorem for finitely generated modules over a PID, \(M\cong R^k\) for some \(k\ge0\) [@DF04]. An \(R\)-basis of \(M\) is \(F\)-linearly independent by step 1, so \(k\le\dim_FV=n\).
For part (b), let \(u_1,\ldots,u_n\) be an \(R\)-basis of \(M\) and \(w_1,\ldots,w_n\) an \(R\)-basis of \(N\).
3. There is \(\alpha\in R\setminus\{0\}\) with \(\alpha M\subseteq N\).
By step 1, \(w_1,\ldots,w_n\) are \(n\) linearly independent vectors in the \(n\)-dimensional space \(V\), so they form an \(F\)-basis. Write \(u_i=\sum_j(a_{ij}/b_{ij})w_j\) with \(a_{ij},b_{ij}\in R\) and \(b_{ij}\ne0\), and put \(\alpha=\prod_{i,j}b_{ij}\ne0\). Then each \(\alpha u_i=\sum_j(\alpha a_{ij}/b_{ij})w_j\) has coefficients in \(R\), so \(\alpha u_i\in N\); since the \(u_i\) generate \(M\), \(\alpha M\subseteq N\).
4. There are an \(R\)-basis \(e_1,\ldots,e_n\) of \(M\) and \(\beta_1,\ldots,\beta_n\in F^\times\) such that \(\beta_1e_1,\ldots,\beta_ne_n\) is an \(R\)-basis of \(N\).
3.By the stacked-bases theorem for submodules of free modules over a PID [@DF04], there are an \(R\)-basis \(w_1',\ldots,w_n'\) of \(N\) and \(d_1,\ldots,d_n\in R\setminus\{0\}\) such that \(d_1w_1',\ldots,d_nw_n'\) is an \(R\)-basis of \(\alpha M\).Then \(e_i\coloneqq(d_i/\alpha)w_i'\) form an \(R\)-basis of \(M=\alpha^{-1}(\alpha M)\), and with \(\beta_i\coloneqq\alpha/d_i\in F^\times\) we get \(\beta_ie_i=w_i'\), an \(R\)-basis of \(N\).
5. Q.E.D.
Step 2 answers part (a), and steps 3 and 4 answer part (b).