TITLE: Requirements on fields for determinants to bust dependence.
QUESTION [11 upvotes]: Being very much used to working on $\mathbb R^n$ $\mathbb C^n$ I just played around a bit with
$$M = \left[\begin{array}{cc}
2&1\\
1&2
\end{array}\right]$$
If the elements are in "integers mod $3$" field, then $$\det(M) = 2\cdot 2-1\cdot 1=4-1=3\equiv 0 (\mod3)$$
Being so used to real matrices I would at first glance guess they were linearly independent. But if I double first column I actually get $[4,2]^T \equiv [1,2]^T (\text{mod } 3)$. So they actually are dependent anyway?
My real question is: will determinants being $0$ tell us linear dependence for all finite fields in the same way they do if we work over $\mathbb R$ or $\mathbb C$ or is there some special requirement on the field for determinants to bust dependence?

REPLY [5 votes]: The two identities ($ A $ is a square matrix over a commutative ring $ R $)
$$ A \operatorname{adj}(A) = \det(A) I $$
$$ \det(AB) = \det(A) \det(B) $$
respectively imply that if $ \det(A) $ is a unit, then $ \operatorname{adj}(A) (\det(A))^{-1} $ is an inverse for $ A $; and if $ A $ has an inverse, then $ \det(A) $ must be a unit, because $ \det(A^{-1}) $ is then an inverse for $ \det(A) $ in $ R $. In other words, a square matrix with entries in a commutative ring is invertible iff its determinant is invertible in the ground ring.