Let \(R\) be a PID and \(A\in M_{m\times n}(R)\). Prove that there exist invertible matrices \(P\in GL_m(R)\) and \(Q\in GL_n(R)\) such that \[ PAQ=\operatorname{diag}(d_1,\ldots,d_r,0,\ldots,0), \qquad d_1\mid d_2\mid\cdots\mid d_r. \] Explain uniqueness up to multiplication of the \(d_i\) by units.
Solution
We prove existence by induction on \(m+n\).
If \(A=0\), there is nothing to prove. Otherwise use row and column swaps to put a nonzero entry in position \((1,1)\); call it \(d\).
1. Improve the pivot until it divides every entry.Suppose first that \(d\) does not divide some entry in the first column, say \(a\). Since \(R\) is a PID,\[ (d,a)=(g) \]for a gcd \(g\), and Bézout gives\[ g=ud+va. \]An invertible \(2\times2\) row operation on rows \(1\) and the row containing \(a\) replaces the first-column pair \((d,a)^t\) by \((g,0)^t\). If \(d\nmid a\), then\[ (d)\subsetneq(g). \]The same argument with column operations applies when \(d\) fails to divide an entry in the first row.
After clearing the rest of the first row and column, suppose some entry \(b\) in the lower-right block is not divisible by \(d\). Add the row containing \(b\) to the first row. Then \(b\) appears in the first row, and the same Bézout column operation replaces \(d\) by \(\gcd(d,b)\), again strictly enlarging the principal ideal generated by the pivot.
Thus every time the current pivot fails to divide some entry of the matrix, elementary operations replace it by a new pivot whose principal ideal strictly contains the old one. A PID is Noetherian, so there is no infinite strictly ascending chain of ideals. Hence after finitely many steps we reach a pivot \(d_1\) dividing every entry of the current matrix.
Using elementary row and column operations, clear the rest of the first column and first row. We obtain \[ \begin{pmatrix} d_1&0\\ 0&B \end{pmatrix}, \] where \(d_1\) divides every entry of \(B\).
2. Induct on the remaining block.Apply the induction hypothesis to \(B\) using operations only on its rows and columns. This gives\[ B\sim\operatorname{diag}(d_2,\ldots,d_r,0,\ldots,0) \]with\[ d_2\mid d_3\mid\cdots\mid d_r. \]Because every entry of \(B\) was divisible by \(d_1\), and elementary operations form \(R\)-linear combinations of entries, every resulting diagonal entry is still divisible by \(d_1\). Hence\[ d_1\mid d_2. \]This proves existence of Smith normal form.
3. Uniqueness.For \(k\ge1\), let \(\Delta_k(A)\) be the ideal generated by the \(k\times k\) minors of \(A\), with \(\Delta_0(A)=R\). Multiplication on the left or right by an invertible matrix does not change these determinantal ideals.
For a Smith form, \[ D=\operatorname{diag}(d_1,\ldots,d_r,0,\ldots), \qquad d_1\mid\cdots\mid d_r, \] one has \[ \Delta_k(D)=(d_1d_2\cdots d_k) \qquad(1\le k\le r). \] Therefore the ideals \((d_1\cdots d_k)\) are determined by \(A\). Successively taking quotients of principal ideals determines each \((d_k)\), hence each \(d_k\) up to multiplication by a unit.