How to visualize $\operatorname{Lie}(\operatorname{GL}_n)=\mathfrak{gl}_n$ in positive characteristic? I'd like an intuitive explanation as to why the Lie algebra of $\operatorname{GL}_n$ is $\mathfrak{gl}_n$ when working over fields of positive characteristic.  Below I reproduce how I "see" this fact in characteristic $0$, but my intuition breaks down for characteristic $p>0$.
Suppose $k$ has characteristic $0$.  Then one can visualize $I_n$ as a point in $\operatorname{GL}_n(\mathbb{Q})\subset \mathbb{Q}^{n^2}$ which has determinant equal to $1$.  Since $1$ is "far" from $0$ in the usual metric given to $\mathbb{Q}^{n^2}$, and since the determinant is continuous, the point $I_n\in\mathbb{Q}^{n^2}$ is "far" from the zero locus of the determinant inside of $\mathbb{Q}^{n^2}$.  It follows that matrices obtained from $I_n$ by small enough perturbations in any direction are still invertible, and thus the tangent space at $I_n$ is all of $\mathbb{Q}^{n^2},$ which when equipped with the usual bracket gives $\mathfrak{gl}_n$.
When I work through this argument, in my head I'm picturing Euclidean $3$-space inside of which is a one-dimensional curve which is the zero locus of the determinant.  Since the complement of the curve is open, the identity (which lies off of the curve) is contained in a ball which doesn't intersect the curve.  Thus any vector is a tangent vector at the identity (the domain of any curve passing through the identity can be appropriately restricted to completely lie inside of the ball).
Notice how this picture and intuition break down because if $k$ has positive characteristic, I'm not exactly sure what it means for $1$ to be "far" from $0$.  Can anyone share with me how this visualization applies in positive characteristic, or perhaps share with me a different way to easily "see" that the tangent space of $\operatorname{GL}_n$ at $I_n$ is $\mathfrak{gl}_n$.  Note I'm really looking to picture this fact in my mind; I'm not interested in a strictly algebraic proof.
 A: The main question is what we even mean by "the Lie algebra of $\text{GL}_n(k)$" if $k$ is any field other than $\mathbb{R}$ or $\mathbb{C}$, let alone a field of positive characteristic. 
One definition over $\mathbb{R}$ or $\mathbb{C}$ is that it's the tangent space at the identity; but what does that mean for a group that doesn't have the structure of a Lie group? Another definition is that it's the space of one-parameter subgroups; but even in the case of $\text{GL}_n$ where we can define these using matrix exponentiation, the problem is that matrix exponentiation is not well-defined even over $\mathbb{Q}$, let alone over a field of positive characteristic.
The correct definition at this level of generality involves recognizing that $\text{GL}_n$ is an algebraic group, and one of the many wonderful things about algebraic groups is that we can directly make sense of what an "infinitesimal element" of an algebraic group is. That is, we can just directly define $\mathfrak{gl}_n(k)$ to consist of the elements of $\text{GL}_n(k)$ "infinitesimally close to the identity"; these are precisely the elements in the kernel of the natural map
$$\text{GL}_n(k[\varepsilon]/\varepsilon^2) \to \text{GL}_n(k).$$
(This is essentially a computation of the Zariski tangent space at the identity.) Then it's an easy exercise to show that this kernel consists of matrices of the form $I + \varepsilon M$ where $M \in M_n(k)$; the point is that all such matrices are invertible because they have inverse $I - \varepsilon M$. 
