 Smith normal form supplies a kernel basis

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.