Let \(R\) be a PID, and let \(A,B\) be cyclic \(R\)-modules of nonzero orders \(r,s\) respectively, where \(r\) is not relatively prime to \(s\). Show that the invariant factors of \(A \oplus B\) are \(\mathrm{gcd}(r, s)\) and \(\mathrm{lcm}(r, s)\).
Solution
Since \(A\) and \(B\) are cyclic of orders \(r\) and \(s\), \[ A\cong R/(r), \qquad B\cong R/(s). \] Thus \(A\oplus B\) is presented by the diagonal matrix \[ D=\begin{pmatrix}r&0\\0&s\end{pmatrix}. \]
Let \[ d=\gcd(r,s), \qquad m=\operatorname{lcm}(r,s), \] with representatives chosen up to multiplication by units.
1. The first determinantal divisor of \(D\) is associated to \(d\).
The first determinantal divisor is a generator of the ideal generated by all \(1\times1\) minors of \(D\), namely \[ (r,s). \] In a PID this ideal is generated by \(\gcd(r,s)=d\).
2. The second determinantal divisor of \(D\) is associated to \(rs\).
The only \(2\times2\) minor is \[ \det D=rs. \] Hence the second determinantal divisor is associated to \(rs\).
3. If \(d_1,d_2\) are the invariant factors of \(A\oplus B\), then\[ d_1\sim d, \qquad d_1d_2\sim rs. \]
For a presentation matrix over a PID, the Smith normal form theorem identifies the products of the first \(j\) invariant factors with the \(j\)th determinantal divisors. Apply this for \(j=1,2\) and use 1 and 2.
4. The second invariant factor is associated to \(m\).
In a PID, \[ \gcd(r,s)\operatorname{lcm}(r,s)\sim rs. \] Thus \(dm\sim rs\). Combining this with 3 and \(d_1\sim d\) gives \(d_2\sim m\).
5. Therefore the invariant factors of \(A\oplus B\) are\[ \gcd(r,s), \qquad \operatorname{lcm}(r,s). \]
By 3 and 4 the invariant factors are associated to \(d\) and \(m\). Moreover \(d\mid m\), as required for invariant-factor order, and \(d\) is not a unit because \(r\) and \(s\) are not relatively prime, so neither factor is trivial and \(A\oplus B\) has exactly two invariant factors. Invariant factors are defined only up to multiplication by units, so these representatives give the claimed pair.