Proposition: Modular original minors

For rational $M$, with all denominators invertible modulo a prime $p$, a nonzero reduced original $r\times r$ minor proves ${\mathrm{rank}}_{\mathbb{Q}}M\ge r$. Finite-field rank alone does not upper-bound rational rank. Taking ${\mathrm{rank}}_{\mathbb{Q}}M\mathrm{mod}p$ is not computing ${\mathrm{rank}}_{{\mathbb{F}}_p}\overline{M}$.

Proof. A zero rational determinant reduces to zero wherever reduction is defined. The matrix $(p)$ has rational rank one and finite-field rank zero, while $1\mathrm{mod}p=1$.