Modular representations of GL_n 
I am interested in the irreducible representations of $G=GL_n(k)$ for a finite field $k$, over $\overline k$.

For complex Lie group representations of $GL_n(\mathbb C)$, the irreducible representations over $\mathbb C$ are classified by Young diagrams, via Young symmetrizers. The same construction should work for our situation. Are these all of the representations?
 A: An extended comment. The question is too wide to be answered comprehensively. Also my competence is limited to only parts of your question.

Your group is finite, so it has only finitely many pairwise non-isomorphic irreducible representations in sharp contrast to the Lie case. Positive characteristic makes a difference even with $G=GL_n(\overline{k})$. For example when $n=2$ the homogeneous bivariate polynomials form an irreducible representation when $k=\Bbb{C}$ (group actiing by linear substitutions). But in characteristic $p$ we have
$(aX+bY)^p=a^p X^p+b^pY^p$ leading to the fact that the monomials $X^p$ and $Y^p$ alone carry a 2-dimensional subrepresentation.

The modular representation theory of groups like $GL_n(\Bbb{F}_q)$ has been studied extensively, but I am definitely the wrong person to give a summary of that (which is why this is just a comment).

But some of the structures do carry over from e.g. $SL_n(\Bbb{C})$ to $SL_n(\overline{k})$. Of the so called algebraic representations (when a generic matrix $A$ of the group $G$ acts via matrices whose entries are rational functions of the entries of $A$) the parametrization of the irreducible reps by the highest weights survives. What happens is that the "characteristic zero irreducibles" you know of, generally don't remain irreducible. The first paragraph gave the simplest mechanism. You can also see it already in the case of $SL_2$. Look at the coefficients the ladder operators act by on the basic weight vectors. Some of those coefficients are divisible by $p$ when the highest weight is large enough. This means that you will not always be able to climb up and down the ladder of weights ($p=0$). Consequently the representation you get by reducing the characteristic zero rep don't stay irreducible when reduced modulo $p$.
That last point is actually kinda nice. In the lucky cases the theory of algebraic group schemes gives us a bridge from characteristic zero to positive characteristic via the route that the groups themselves are defined by a set of polynomials with integer coefficients. When "everything is defined over $\Bbb{Z}$", we can easily extend the scalars to, say $\Bbb{C}$, or first reduce them modulo $p$ and then extend to $\overline{k}$. The theory is quite involved. Humphreys has a classical book on Algebratic groups, and Jantzen wrote a nice summary of what was known about the representation theory of algebraic groups in the 90s.
