# In which situations $\det(A\mod x) \mod x=\det(A)\mod x$ would help us knowing if $\det(A)=0$?

André Nicolas, in his very neat answer to is the following matrix invertible? uses the fact that the matrix

$$\begin{bmatrix} 1235 &2344 &1234 &1990\\ 2124 & 4123& 1990& 3026 \\ 1230 &1234 &9095 &1230\\ 1262 &2312& 2324 &3907 \end{bmatrix}$$

Modulo 2 is the identity, to show that its determinant is different from zero, something that also requires $$\det(A\mod x)\mod x=\det(A)\mod x$$ Where the modulo in $(A\mod x)$ is applied to the matrix elementwise. This happens because the determinant is a polynomial in the matrix entries and $$(a+bc^d)\mod x=(a\mod x + (b\mod x)(c\mod x)^d)\mod x$$

My question is then whether the successive application of such modular reductions could be a useful method to know if $\det(A)=0$. Perhaps this is so only in some particular situations, it would also be helpful if you show some of them.

It is indeed a useful technique of wide applicability, and not just for determinants: to test whether some difficult-to-compute polynomial identity (over the integers) is true, compute it in the integers mod $p$ where $p$ is some prime. If it's false mod $p$, it's false for the integers as well. If it's true mod $p$, it has some chance of being true. To reduce the number of "false positives", you might use a larger prime $p$: a "random" integer has probability $1/p$ of being $0$ mod $p$. To reduce even more, do this for several different primes. You can also handle rational expressions, if $p$ is chosen so that none of the denominators is $0$; you can handle expressions involving variables by assigning each variable a random number mod $p$.
Yes, if $M$ is an $n \times n$ matrix with entries in $\mathbf{Z}$, then, its determinant is bounded by $n!(\max_{i,j} m_{ij})^n$. Thus, for example, if $\det M \equiv 0 \bmod p$ for all primes $p$ with $p^2 \leqslant n!(\max_{i,j} m_{ij})^n$, then, $\det M$ must be identically $0$.
Edit. As Giuseppe points out, if one starts with Hadamard's inequality for the determinant, then, one gets a better bound. Hadamard's theorem is: $$|\det([v_1, \dots, v_n])| \leqslant \prod_{i=1}^n \|v_i\|$$ where $\|\cdot\|$ is the $2$-norm on $\mathbf{R}^n$.