Let \(R\) be a commutative principal ideal domain. Show that any \(R\)-module \(M\) generated by two elements takes the form \(R/(a)\oplus R/(b)\) for some \(a,b\in R\). What more can you say about \(a\) and \(b\)?
Give a necessary and sufficient condition for two direct sums as in part (a) to be isomorphic as \(R\)-modules.
Solution
1. There is a surjective homomorphism\[ \pi:R^2\twoheadrightarrow M. \]
Choose generators \(m_1,m_2\) of \(M\) and define \[ \pi(r,s)=rm_1+sm_2. \] By the choice of \(m_1,m_2\), this map is surjective.
2. If \(K=\ker\pi\), then\[ M\cong R^2/K, \]and \(K\) is a finitely generated free \(R\)-module of rank at most \(2\).
The first isomorphism theorem gives \(M\cong R^2/K\). Since \(R\) is a PID, every submodule of the free module \(R^2\) is free of rank at most \(2\). In particular \(K\) is finitely generated.
3. There are bases of \(K\) and \(R^2\) for which the inclusion \(K\hookrightarrow R^2\) has Smith normal form\[ \begin{pmatrix} d_1&0\\ 0&d_2 \end{pmatrix}, \qquad d_1\mid d_2, \]where zero diagonal entries are allowed.
Apply Smith normal form over the PID \(R\) to a matrix representing the inclusion \(K\hookrightarrow R^2\). Since the rank of \(K\) is at most \(2\), after adjoining zero columns or diagonal entries as necessary, the cokernel is represented by a diagonal \(2\times2\) matrix with invariant factors \(d_1,d_2\) satisfying \(d_1\mid d_2\). A zero invariant factor records a free summand, using \(R/(0)=R\).
4. Consequently\[ M\cong R/(d_1)\oplus R/(d_2), \qquad d_1\mid d_2. \]
By 2, \(M\) is the cokernel of the inclusion \(K\hookrightarrow R^2\). Replacing a presentation matrix on the left and right by invertible matrices does not change its cokernel up to isomorphism. Thus 3 gives \[ M\cong\operatorname{coker}\!\begin{pmatrix}d_1&0\\0&d_2\end{pmatrix} \cong R/(d_1)\oplus R/(d_2). \]
5. The invariant factors \(d_1,d_2\) in 4 are unique up to multiplication by units of \(R\).
Uniqueness is the uniqueness part of Smith normal form over a PID: once the diagonal entries are arranged so that \[ d_1\mid d_2, \] the principal ideals \((d_1)\) and \((d_2)\) are determined by the cokernel. Equivalently, replacing either \(d_i\) by an associate does not change \(R/(d_i)\), and these are the only ambiguities in invariant-factor form.
6. Hence, if\[ d_1\mid d_2 \qquad\text{and}\qquad e_1\mid e_2, \]then\[ R/(d_1)\oplus R/(d_2) \cong R/(e_1)\oplus R/(e_2) \]if and only if \(d_1\) is associate to \(e_1\) and \(d_2\) is associate to \(e_2\).
If corresponding invariant factors are associates, the corresponding cyclic quotients are isomorphic, so the direct sums are isomorphic. Conversely, if the direct sums are isomorphic, uniqueness in 5 forces their invariant factors to agree up to associates.
7. For arbitrary \(a,b\in R\), the direct sum\[ R/(a)\oplus R/(b) \]has invariant factors obtained from the Smith normal form of \(\operatorname{diag}(a,b)\).More explicitly, one may choose \(d_1,d_2\) with\[ (d_1)=(a,b), \qquad d_1\mid d_2, \qquad d_1d_2\sim ab, \]where \(\sim\) denotes association and the conventions \(d_2=0\) when exactly one of \(a,b\) is zero and \(d_1=d_2=0\) when both are zero are understood.
For a \(2\times2\) diagonal presentation matrix, the first determinantal divisor is a generator of the ideal generated by its \(1\times1\) minors, namely \[ (a,b). \] The product of the two invariant factors is associate to the determinant \(ab\). These two determinantal divisors determine the Smith normal form. The zero cases give the stated free summands directly.
8. Therefore two arbitrary sums\[ R/(a)\oplus R/(b) \qquad\text{and}\qquad R/(c)\oplus R/(d) \]are isomorphic if and only if their two Smith invariant factors, computed as in 7, agree pairwise up to associates.