State the structure theorem for finitely generated modules over a PID.
Find the decomposition of the \(\mathbb{Z}-\)module \(M\) generated by \(w,x,y,z\) satisfying the relations \[\begin{aligned} 3w + 12y + 3x + 6z &=0 \\ 6y &= 0 \\ -3w -3x + 6y &= 0. \end{aligned}\]
Solution
1. Structure theorem over a PID.
If \(R\) is a PID and \(M\) is finitely generated, then \[ M\cong R^r\oplus R/(d_1)\oplus\cdots\oplus R/(d_t), \qquad 0\ne d_1\mid d_2\mid\cdots\mid d_t, \] with the free rank and invariant factors unique up to associates.
2. For the given module, use the relation matrix\[ A= \begin{pmatrix} 3&3&12&6\\ 0&0&6&0\\ -3&-3&6&0 \end{pmatrix} \]with respect to the ordered generators \((w,x,y,z)\).
Its rows are exactly the coefficient vectors of the three displayed relations.
3. The Smith normal form of \(A\) is\[ \operatorname{diag}(3,6,6,0). \]
The determinantal divisors are \[ \Delta_1=3, \qquad \Delta_2=18, \qquad \Delta_3=108. \] Thus the nonzero invariant factors are \[ 3,\quad 18/3=6,\quad108/18=6. \] The matrix has rank \(3\), so one free summand remains.
4. Hence\[ M\cong\ZZ\oplus\ZZ/3\ZZ\oplus\ZZ/6\ZZ\oplus\ZZ/6\ZZ. \]
Passing to Smith normal form preserves the cokernel, and the cokernel of \(\operatorname{diag}(3,6,6,0)\) is exactly the displayed direct sum.