Given two integer matrices $A_1$ and $A_2$. Consider the matrix $M(x_1,x_2) = x_1A_1 + x_2A_2$, where $x_1,x_2 \in \mathbb{C}$. The matrices are chosen such that $M(x_1,x_2)$ has rank at least $d$ (over $\mathbb{C}$) if $x_1 \neq 0$.
Since $x_1 \neq 0$, we can set $t$ to be $x_2/x_1$. Over the polynomial ring $\mathbb{Q}(t)$, the statement says that the Smith normal form $S(t)$ of $\frac{1}{x_1}M(x_1,x_2) = A_1 + tA_2$ has diagonal entries $\text{diag}(1, 1, 1, \dots, 1, p_{d+1}(t), \dots)$, i.e. for all $t$ the rank is at least $d$. Let us denote this equivalence as $$A_1 + tA_2 = P(t) S(t) Q(t).$$ Note that the determinant of $P(t)$ and $Q(t)$ is a rational number, independant of $t$.
Now, let $x_1$ and $x_2$ be integers, with $x_1 \neq 0$. Then we can also consider the Smith normal form of $M(x_1,x_2)$ over $\mathbb{Z}$. I would like to conclude that the product of the $d$ first invariant factors $\alpha_d$, i.e. the gcd of all $d\times d$ minors, divides $\Delta x_1^d$ for some $\Delta$ independant of $x_2$.
I have had two (failed) approaches so far:
(1) Let $M_1$ to $M_s$ denote the $d\times d$ minors of $M(x_1,x_2)$. These are homogeneous of degree $d$. Now $\frac{1}{x_1^d}M_1$ to $\frac{1}{x_1^d}M_s$ are polynomials in $t$ without common zero. By the Nullstellensatz we hence know that there exist polynomials $q_1(t)$ up to $q_s(t)$ such that $\sum q_i(t)\frac{1}{x_1^d}M_i = 1$. Multiplying by $x_1^d$ and by some power $d'$ of $x_1$ such that $x_1^{d'}q_i(t)$ is a polynomial in $x_1$ and $x_2$ gives us that $$\sum (x_1^{d'}q_i(t))M_i = x_1^{d +d'}.$$ Multiplying by the common denominator $\Delta$ gives that $\alpha_d \mid \Delta x_1^{d+d'}$, so the desired result except the power of $x_1$ is too high. (This reasoning cannot be improved as $x_1^2 + x_2^2$ and $x_1x_2$ satisfy the statement, but there is no linear combination reducing them to $x_1^2$.)
(2) As $A_1 + tA_2 = P(t) S(t) Q(t)$, we also have $M(x_1,x_2) = P(t) (x_1S(t)) Q(t)$. It follows that $$\det M(x_1,x_2) = \Delta \det(x_1S(t)),$$ with $\Delta = \det(P(t) Q(t))$ fixed, independent of $x_2$ (as it is independent of $t$). We would like to say that
$$\alpha_d \mid \Delta \cdot \text{ any } d\times d \text{ minor of } x_1S(t).$$ In particular, $\alpha_d \mid \Delta x_1^d$. However, I am not able to prove this claim. I have tried using the formula for a minor of a product: $$x_1S(t) = P^{-1}(t)M(x_1,x_2) Q^{-1}(t) \implies x_1^d = \sum_{r_1,r_2} P^{-1}[(1\dots d),r_1]M[r_1,r_2]Q^{-1}[r_2,(1\dots d)].$$ This shows that $x_1^d$ is a linear combination of the $d\times d$ minors of $M(x_1,x_2)$. However, this combination may include polynomials in $t$, which are not necessarily integers and therefore is useless for estimating $\alpha_d$. (Unless if one uses the same trick as before, but giving a too high power of $x_1$ again.)