While working on a proof, I stumbled on the following fact.
Let $M:=(x_{i,j})_{i,j}\in \mathrm {GL}_{n,d}(\mathbb Z)$. Let's denote by $\alpha_i:=\gcd(x_{1,i},\ldots,x_{n,i})$ for all $i\in \{1,\ldots,d\}$.
Let's also define $N:=\binom nd$ and $\eta_1,\ldots,\eta_N$ be all the minor of $M$ of size $d\times d$.
Then
$$\alpha_1\cdots\alpha_d=\gcd(\eta_1,\ldots,\eta_N).$$
What I tried.
Let $k$ be an integer.
Assume $k$ divides $\alpha_1\cdots\alpha_d$. Let $j\in \{1,\ldots,N\}$. Then $\eta_j$ is a minor of the matrix $M$ of size $d\times d$, extracted by taking only the lines $i_1,\ldots,i_d$. So we have
$$\eta_j=\sum_{\sigma\in \mathfrak S_d}\epsilon(\sigma)\prod_{\ell=1}^d x_{i_\ell,\sigma(\ell)}.$$
But $k\mid \alpha_1\cdots\alpha_d$, so for all $d$-uple $(i_1<\cdots<i_d)$, we have
$$k\mid x_{i_1,1}\cdots x_{i_d,d}= x_{i_1,\sigma(1)}\cdots x_{i_d,\sigma(d)}$$
for all permutation $\sigma\in\mathfrak S_d$. So $k$ divides $\eta_j$ for all $j$, so
$$ k\mid \gcd(\eta_1,\ldots,\eta_N).$$
Reciprocally, I am unable to do it.
Any help would be much appreciated.
This statement seems to be very connected to the theorem in this recent publication:
http://nntdm.net/volume-26-2020/number-3/5-7/ "Equalities between greatest common divisors involving three coprime pairs"
that goes as:
For $a_i$ and $b_i$ positive integers such that $\gcd(a_i, b_i) = 1$ for $i ∈ \{1, 2, 3\}$ and $d_{ij} = |a_ib_j − a_jb_i|$, then $\gcd(d_{32}, d_{31}) = \gcd(d_{32}, d_{21}) = \gcd(d_{31}, d_{21})$.
Basically $a_i$ and $b_i$ can form a 3x2 matrix (your $x$) and $d_{ij}$ are its minors (your $\eta$). I think this can be helpful.