Smith normal form

For every matrix $A$ over a principal ideal domain, there are invertible matrices $U,V$ such thatThe nonzero diagonal entries are unique up to multiplication by units. Over the integers, $U$ and $V$ are unimodular and the $d_i$ may be chosen positive; their products $d_1\cdots d_k$ are the greatest common divisors of the $k$-rowed minors.

Let $M$ be an $m$-by-$n$ matrix over a principal ideal domain $R$. Choose Smith normal form $PMQ=D$ with exactly $r$ nonzero diagonal entries. Since $P$ and $Q$ are invertible matrices, $Mx=0$ is equivalent to $D(Q^{-1}x)=0$. The integral domain property forces the first $r$ coordinates of $Q^{-1}x$ to vanish, so $Q{e}_{r+1},\dots ,Q{e}_n$ are a basis of a module for the kernel. These vectors form part of a basis of a module of $R^n$, which is stronger than merely finding a nonzero vector in the kernel over the fraction field.

For an integer matrix $M$, its $j$th determinantal divisor is the nonnegative greatest common divisor of all $j$-by-$j$ matrix minors, with ${\Delta }_0=1$. If all these minors vanish, take ${\Delta }_j=0$. Unimodular row and column operations preserve these divisors. For nonzero invariant factors $d_1\mid \cdots \mid d_r$ in the Smith normal form, ${\Delta }_j=d_1\cdots d_j$ for $j\le r$, so $d_j={\Delta }_j/{\Delta }_{j-1}$.