If \(M\) is free abelian, how can I put quotients of M in some standard form? What was crucial about the integers here (abelian groups being modules over \(\ZZ\))? How does the procedure simplify if the ring is a Euclidean domain, not just a PID?
Solution
Let \(M\cong\ZZ^n\) and let \(N\le M\) be generated by the columns of an integer matrix \[ A\in M_{n\times r}(\ZZ). \] Then \[ M/N\cong\operatorname{coker}(A). \]
1. Smith normal form gives invertible integer matrices \(U,V\) such that\[ UAV=\operatorname{diag}(d_1,\dots,d_s,0,\dots,0), \qquad d_1\mid d_2\mid\cdots\mid d_s, \]with each \(d_i>0\).
The ring \(\ZZ\) is a PID, so every matrix admits Smith normal form. Multiplication by unimodular matrices corresponds to changing bases in the codomain and domain and therefore preserves the cokernel up to isomorphism.
2. Therefore\[ M/N \cong \ZZ^{\,n-s} \oplus \ZZ/d_1\ZZ \oplus\cdots\oplus \ZZ/d_s\ZZ. \]
This is the cokernel of the diagonal Smith matrix.
3. What is crucial is not specifically the integers but the PID property.
Over any PID \(R\), a matrix has Smith normal form and the same argument gives \[ R^n/N \cong R^{n-s}\oplus\bigoplus_{i=1}^s R/(d_i). \] This is the structure theorem for finitely generated modules over a PID.
4. Over a Euclidean domain, the existence theorem is the same but the computation is more algorithmic.
The Euclidean algorithm supplies explicit gcd computations and Bézout coefficients. Repeated elementary row and column operations can use Euclidean division to replace entries by smaller remainders until a gcd appears in a pivot position, then clear its row and column. Repeating produces Smith normal form constructively. Over a general PID the same normal form exists, but one does not necessarily have a Euclidean norm giving this direct reduction algorithm.