Geometric Moments and Rational Lambert Values: Proofs and Further Results

The lattice after evaluation.

For example, $(1,0),(1,6)$ generate $\mathbb{Z}\oplus 6\mathbb{Z}$. They give two residues modulo $2$. Within that lattice, the vectors whose two coordinates are even form $2\mathbb{Z}\oplus 6\mathbb{Z}$; dividing them by $2$ gives $\mathbb{Z}\oplus 3\mathbb{Z}$, of index $3$. These are two different counts: the number of attainable residues before division, and the index of the lattice left after division. Smith normal form computes both for any rank-two lattice. The first count describes congruences available before division; the second describes the integrality restriction that survives it. Neither count estimates the real remainders of the rows.

Let primitive integer rows $u_j\in {\mathbb{Z}}^2$ span a rank-two lattice $L$, and let $g$ be the gcd of their $2\times 2$ minors. Since at least one row is primitive, the Smith invariants are $1,g$ [34]. Hence, for $D\ge 1$,

Indeed an integral unimodular change of coordinates takes $L$ to $\mathbb{Z}\times g\mathbb{Z}$; reduction modulo $D$ and intersection with $D{\mathbb{Z}}^2$ give the two formulas. This is Smith normal form over $\mathbb{Z}$, not over $\mathbb{Z}[p]$. It measures the actual image, rather than the ambient residue space. If two independent divided rows have coordinate height at most $H$ and remainders of absolute value at most $\epsilon$, their nonzero integer determinant gives the necessary inequality $2H\epsilon \ge g/gcd(g,D)$. That is not a necessary condition for producing a single nonzero form. If the specialised rows are not primitive, let $d_1\mid d_2$ be the positive Smith invariants instead. For $D\ge 1$ the corresponding formulas are

They follow by applying the one-dimensional calculation to each summand $d_i\mathbb{Z}$. Polynomial coefficient content alone does not determine either integer Smith invariant after specialisation.