$\textbf{Question:}$ Is it true that given a matrix $A$ that the $det(A)=0$ iff the column vectors are dependent for $\textbf{ANY}$ field? Below is an example for a particular field to see what I mean (there might be errors). I am not concerned with this example (I am just showing an example how I think one would compute the determinant in a particular field); I want to know if determinants can be generalized over all fields in this way.
Let $\mathbb{Z}_3=\{0, 1, 2\}$ be a Galois Field (i.e. $GF(3)$).
Then, the matrix $A= \begin{bmatrix} 1& 2 & 0\\ 2& 1 & 1\\ 0 & 2 &1 \end{bmatrix}$ has independent vectors $v_1, v_2,$ and $v_3$ where $v_1=(1,2, 0)$, $v_2=(2,1,2)$, and $v_3=(0, 1, 1)$.
$\bullet$ Reasoning as to why $\{v_1, v_2, v_3\}$ are independent (i.e. computing RREF(A)).
$\begin{bmatrix} 1& 2 & 0\\ 2& 1 & 1\\ 0 & 2 &1 \end{bmatrix} \underbrace{\implies}_{R_2\leftarrow 2R_2} \begin{bmatrix} 1& 2 & 0\\ 1& 2 & 2\\ 0 & 2 &1 \end{bmatrix}\underbrace{\implies}_{R_2\leftarrow R_2-R_1} \begin{bmatrix} 1& 2 & 0\\ 1-1& 2-2 & 2-0\\ 0 & 2 &1 \end{bmatrix}=\begin{bmatrix} 1& 2 & 0\\ 0 & 0 & 2\\ 0 & 2 &1 \end{bmatrix} \underbrace{\implies}_{R_2\leftarrow R_2+R_3} \begin{bmatrix} 1& 2 & 0\\ 0 & 2 & 0\\ 0 & 2 &1 \end{bmatrix} \underbrace{\implies}_{R_3\leftarrow R_3-R_2} \begin{bmatrix} 1& 2 & 0\\ 0 & 2 & 0\\ 0 & 0 &1 \end{bmatrix} \underbrace{\implies}_{R_1\leftarrow R_1-R_2} \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix} \underbrace{\implies}_{R_2\leftarrow 2R_2} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$
In $Z_3$, we get \begin{align} det(A) &=1(1*1-1*2)-2(2*1-1*0)+0*(2*2-1*0)\\ &=1(1*1-1*2)-2(2*1-1*0)\\ &=1(1-2)-2(2-0)\\ &=1-2-2(2-0)\\ &=1+1-2(2)\\ &=1+1-1\\ &=1-1+1\\ &=1. \end{align}
So, the $det(A)\neq 0$ which it should in this case.
Yes, the entire discussion on determinants (but also rank, Cramer, Gaussian elimination, Rouché-Capelli...) is valid for $k$-vector spaces of finite dimension, where $k$ may be any field. A small caveat is that some authors, when they work in $\Bbb R$ or $\Bbb C$, define $\det$ as a function such that (amongst other things) $\det(A^1,\cdots, A^i,\cdots, A^j,\cdots, A^n)=-\det(A^1,\cdots, A^j,\cdots, A^i,\cdots, A^n)$ for all $i\ne j$. This requirement must be formulated differently if we want it to make sense for $\operatorname{char}k=2$: $$\det(A)=0\text{ whenever }A\text{ has two identical columns}$$
The two formulations are equivalent when $\operatorname{char}k\ne 2$.
Generally speaking, the gist of it is that everything that involves calculating polynomials and rational functions in the entries of the matrix works the same.