Let \(R\) be a PID. For each positive integer \(r\) and sequence of nonzero ideals \(I_1 \supset I_2 \supset \cdots \supset I_r\), choose a sequence \(d_i \in R\) such that \((d_i) = I_i\) and \(d_i \mid d_{i+1}\).
For a given pair of positive integers \(n, m\), let \(S\) be the set of all \(n\times m\) matrices of the form \(\left(\begin{array}{ll}{L_{r}} & {0} \\ {0} & {0}\end{array}\right)\) where \(r=1,2,\cdots,\min(m,n)\) and \(L_r\) is a diagonal \(r\times r\) matrix with main diagonal \(d_i\).
Show that \(S\) is a set of canonical forms under equivalence for the set of all \(n\times m\) matrices over \(R\).
Solution
Two \(n\times m\) matrices over \(R\) are equivalent when one is obtained from the other by multiplication on the left and right by invertible matrices. We prove that every equivalence class contains exactly one matrix in \(S\).
1. Every \(n\times m\) matrix over \(R\) is equivalent to an element of\(S\).
By the Smith normal form theorem over a PID, every matrix \(A\) is equivalent to a diagonal matrix \[ D=\operatorname{diag}(a_1,\ldots,a_r,0,\ldots,0), \] where each \(a_i\ne0\) and \[ a_1\mid a_2\mid\cdots\mid a_r. \] Put \(I_i=(a_i)\). Then \[ I_1\supseteq I_2\supseteq\cdots\supseteq I_r. \] By the choice made in the statement, there are generators \(d_i\) of these ideals with \(d_i\mid d_{i+1}\). Since \(a_i\) and \(d_i\) generate the same principal ideal, they differ by a unit. Multiplying suitable rows (or columns) by these units changes \(D\) to the corresponding matrix in \(S\). Hence every equivalence class meets \(S\).
2. For \(1\le j\le\min(m,n)\), let \(\Delta_j(A)\) be the ideal generatedby all \(j\times j\) minors of \(A\). Then \(\Delta_j(A)\) is invariant undermatrix equivalence.
Let \(P\) be any \(n\times n\) matrix over \(R\). By the Cauchy–Binet formula, each \(j\times j\) minor of \(PA\) is an \(R\)-linear combination of \(j\times j\) minors of \(A\), so \[ \Delta_j(PA)\subseteq\Delta_j(A). \] If \(P\) is invertible, applying this to \(P^{-1}\) and \(PA\) gives the reverse inclusion. The same argument applies to right multiplication by an invertible \(m\times m\) matrix \(Q\). Hence \(\Delta_j(PAQ)=\Delta_j(A)\).
3. If\[ D=\operatorname{diag}(d_1,\ldots,d_r,0,\ldots,0) \]with \(d_1\mid\cdots\mid d_r\), then for \(1\le j\le r\),\[ \Delta_j(D)=(d_1d_2\cdots d_j), \]and for \(j>r\), \(\Delta_j(D)=0\).
Every nonzero \(j\times j\) minor of \(D\) is a product \(d_{i_1}\cdots d_{i_j}\) with \(1\le i_1<\cdots<i_j\le r\). Divisibility of the diagonal entries implies \[ d_1\cdots d_j\mid d_{i_1}\cdots d_{i_j}, \] so every such minor lies in \((d_1\cdots d_j)\). The leading \(j\times j\) minor equals \(d_1\cdots d_j\), giving equality. If \(j>r\), every \(j\times j\) minor vanishes.
4. Two matrices in \(S\) that are equivalent are equal.
Let their nonzero diagonal entries be \[ d_1,\ldots,d_r \quad\text{and}\quad e_1,\ldots,e_s. \] By 2 and 3, the largest index for which \(\Delta_j\ne0\) is both \(r\) and \(s\), hence \(r=s\). Moreover, for each \(j\le r\), \[ (d_1\cdots d_j)=(e_1\cdots e_j). \] For \(j=1\) this gives \((d_1)=(e_1)\). Inductively, if \((d_i)=(e_i)\) for \(i<j\), cancellation in the PID gives \((d_j)=(e_j)\). Thus the two Smith forms determine the same chain of ideals \[ (d_1)\supseteq\cdots\supseteq(d_r). \] The statement fixed once and for all the chosen generator \(d_i\) for each such chain, so the representatives in \(S\) have exactly the same diagonal entries. Hence the matrices are equal.
5. Therefore \(S\) is a set of canonical forms for matrix equivalence.
By 1 every equivalence class contains an element of \(S\), and by 4 it contains at most one.