I am trying to reduce the following matrix to Smith normal form
$$A= \begin{pmatrix}
1&0&0\\
1&2&0\\
1&0&3
\end{pmatrix}$$
Whatever row and column operations I try, I end up with the diagonal matrix
$$A= \begin{pmatrix}
1&0&0\\
0&2&0\\
0&0&3
\end{pmatrix}$$
However for Smith normal form I need diagonal entries $a_i$ such that $a_j|a_{j+1}$. But $2$ does not divide $3$ here. How do I rectify this?
Best Answer
Newman says that Smith form can be accomplished with elementary row and column operations as long as the coefficient ring is Euclidean, as here. Since we are not going to change the determinant, this means diagonal $(1,1,6).$ Newman's assurance means that we can accomplish this for the two by two square with entries $(2,3).$
Take distinct positive integers $p,q$ such that $\gcd(p,q) = 1$ and $$ px + qy = 1. $$ Then the sequence of resulting matrices is $$ \left( \begin{array}{rr} p & 0 \\ 0 & q \end{array} \right), $$ $$ \left( \begin{array}{rr} p & 0 \\ px & q \end{array} \right), $$ $$ \left( \begin{array}{rr} p & 0 \\ 1 & q \end{array} \right), $$ $$ \left( \begin{array}{rr} 1 & (1-p)q \\ 1 & q \end{array} \right) = \left( \begin{array}{rr} 1 & q-pq \\ 1 & q \end{array} \right) $$ $$ \left( \begin{array}{rr} 1 & 0 \\ 1 & pq \end{array} \right) $$ $$ \left( \begin{array}{rr} 1 & 0 \\ 0 & pq \end{array} \right) $$