Representation theory of the full linear monoid The full linear monoid $M_N(k)$ of a field $k$ is the set of $N \times N$ matrices with entries in $k$, made into a monoid with matrix multiplication.
A representation of $M_N(k)$ on a vector space $V$ over $k$ is a monoid homomorphism
$$   \rho \colon M_N(k) \to \textrm{End}(V). $$
What is the classification of representations of the full linear monoid?
(Note: this is different than asking about representations of $M_N(k)$ viewed as an algebra.)
We get a bunch of irreducible representations of $M_N(k)$ from Young diagrams with $\le N$ rows.  The vector space
$$ (k^N)^{\otimes n}= \underbrace{k^N\otimes \cdots\otimes k^N}_{\mbox{$n$ copies}} $$
is a representation of $M_N(k)$ in an obvious way.   Given a Young diagram with $n$ boxes and $\le N$ rows, we get a minimal central idempotent in $S_n$, and we can use this to project $(k^N)^{\otimes n}$ down to a subspace that is a representation of $M_N(k)$.
I believe these are all the polynomial irreducible representations $\rho$ of $M_N(k)$: that is, those where the matrix entries of $\rho(T)$ are polynomials in the matrix entries of $T \in M_N(k)$.

*

*Is this correct?

We get more irreducible representations using the absolute Galois group of $k$.  Any field automorphism of $k$ gives an automorphism $\alpha$ of the monoid $M_n(k)$, and composing this with a polynomial irreducible representation $\rho \colon M_n(k) \to \mathrm{End}(V)$ we get a new representation $\rho \circ \alpha$ which is still irreducible, but not polynomial unless $\alpha$ is the identity.
But when $N = 1$, at least, there are even more irreducible representations of the full linear monoid!  Then $M_1(k)$ is the multiplicative monoid of $k$, and it has a 1-dimensional irreducible representation sending $x \in k$ to multiplication by $x^n$ for any $n \ge 0$.
The point here is that the multiplicative monoid of $k$ has endomorphisms that don't come from field automorphisms.  There can also be others that combine raising to a power with field automorphisms: e.g. $M_1(\mathbb{C})$ has a 1-dimensional irreducible representation sending $z \in \mathbb{C}$ to multiplication by $z^4 \overline{z}^3$.
Since endomorphisms of the multiplicative monoid of $k$ not arising from field automorphisms don't preserve addition, I don't see how to use them to get extra representations of $M_N(k)$ for $N > 1$.
So:


*Are all the irreducible representations of $M_N(k)$ for $N > 1$ of the form $\rho \circ \alpha$ where $\rho$ comes from a Young diagram with at most $N$ rows and $\alpha$ comes from a field automorphism of $k$, or are there more?


*Do all the irreducible representations of $M_N(k)$ for $N = 1$ come from endomorphisms of the multiplicative monoid of $k$?


*Are all finite-dimensional representations of $M_n(k)$ completely reducible?
 A: As Ben Steinberg has been suggesting in the comments, the representation theory of these have been studied for quite awhile.  His book on this, the Representation Theory of Finite Monoids, is an excellent modern general reference for how to think about these.
The general theme is that the maximal subgroups in $M_n(k)$, namely the groups $GL_m(k)$ for $m=0, \dots, n$, control the representation theory.  For example there is a recollment diagram of the abelian categories:
$$ GL_n(k)\text{- modules}
\begin{array}{c} \stackrel{q}{\longleftarrow} \\[-.08in]
\stackrel{i}{\longrightarrow} \\[-.08in] \stackrel{p}{\longleftarrow}
\end{array}
M_n(k)\text{- modules}
\begin{array}{c} \stackrel{l}{\longleftarrow} \\[-.06in] \stackrel{e}{\longrightarrow} \\[-.08in]
\stackrel{r}{\longleftarrow}
\end{array} M_{n-1}(k)\text{- modules.} $$
From this, one sees that the irreducible modules for $M_n(k)$ come in two flavors: $GL_n(k)$ irreducibles on which singular matrices act trivially, and $M_{n-1}(k)$ irreducible modules `induced' up to $M_n(k)$. Curiously, in the finite field case, at least, these two types are related be tensoring with the determinant representation (perhaps first noticed in a 1980's paper by John Harris and me).
In particular, if $k$ is a finite field of order $q$, the irreducible representations of $M_n(k)$ can be sensibly labelled by $q$--regular Young diagrams with at most $n$ nonzero columns.  If $k$ is infinite, one should be getting all Young diagrams.
Finally, the category of representations of $M_n(k)$ is very much related to the abelian category of functors from finite dimensional $k$--vector spaces to $k$--vector spaces (dubbed the category of `generic representations' of $k$ in some of my papers), and much interesting calculational work has been figured out in this setting.
A: At least over an algebraically field, the polynomial representations are precisely the representations of $M_n(k)$ as an algebraic monoid and I don’t see immediately why that would change over a nonalgebraically closed field, but I am not an expert.
To classify irreducible representations of $M_n(k)$ over any field $F$ is the same as to classify irreducible representations of $GL_r(k)$ for $0\leq r\leq n$ over $F$ (where $GL_0(k)$ is the trivial group).
This follows from old results of Munn and Ponizovskii on representations of finite monoids that were eventually extended to monoids satisfying certain finiteness conditions that $M_n(k)$ satisfies.  I don’t know the classification for $GL_r(k)$ but somebody else might.
The argument is like this.  A principal ideal in a monoid $M$ is a set of the form $MaM$.  The principal ideals of $M_n(k)$ are well known to be of the form $I_r$ where $I_r$ consists of all matrices of rank at most $r$ (with $0\leq r\leq n$.  Let $e_r$ be the diagonal row echelon form rank $r$ idempotent.  Note every rank $r$ idempotent is conjugate to $e_r$.  Note that $e_rM_n(k)e_r\cong M_r(k)$ and so we can identify its group of units $G_r$ with $GL_r(k)$
Let $V$ be an irreducible $M_n(k)$ module over a field $F$.  Then there is a least $r$ with $I_rV\neq 0$.  Then one easily checks that $e_rV$ is an irreducible representation of $e_rM_n(k)e_r\cong M_r(k)$ that is annihilated all elements not belonging to $G_r\cong GL_r(k)$ and so is an irreducible representation of $G_r$.
Conversely, given an irreducible representation $W$ of $Gl_r(k)$ (identified with $G_r$), there is a unique irreducible representation $V$ of $M_n(k)$ with $I_{r-1`}W=0$ and $e_rV\cong W$ as representations of $G_r$.  You can construct $V$ as $(FM_n(k)/FI_{r_1}\otimes_{FG_r}W)/N$ where $N$ is largest submodule annihilated by $e_r$.  This kind of argument is a mixture of Munn-Ponizovksii’s argument for finite semigroups and Green’s condensation technique which coincidentally appears in his book on polynomial representations (he also introduced the study of principal ideals in structural semigroup theory).
The details of how this works can be found in my book on representation theory of finite monoids.  Finiteness can be replaced with taking a monoid $M$ with a finite unrefinable chain of two-sided ideals such that $eMe$is Dededkind finite for all idempotents $e$.  I think this kind of result is in the generality you need in Okninski’s book Semigroups of Matrices.
As to complete reducibility, this I do not know much about for the infinite case.  For $k$ finite and $F$ any field it was first shown by Putcha and Okninski that $FM_n(k)$ is semisimple whenever $F$ is of characteristic $0$.  They showed that this is true for any finite monoid of Lie type (the finite analogues of reductive algebraic monoids).  Kovacs then showed $FM_n(k)$ is Frobenius if and only if the characteristic of $k$ is not the same as that of $F$. You can find his proof in my book.  In fact, Kovacs showed if $R$ is any commutative ring in which the characteristic of $k$ is invertible, then $RM_n(k)$ is isomorphic to a direct product of $M_{n_r}(RGL_r(k))$ with $0\leq r\leq n$ where $n_r$ is the number of $r$-dimensional sub spaces of $k^n$.  In particular, every representation of $M_n(k)$ over $F$ is completely reducible iff the characteristic of $F$ does not divide the order of $GL_n(k)$.  I have a recent paper giving explicitly the Frobenius form for $M_n(k)$ in good characteristic, which is easier than Kovacs’s proof since you can check by hand the form is Frobenius.  The direct product decomposition follows from $M_n(k)$ being von Neumann regular and having a Frobenius algebra by a result you can find in my book
The algebra $kM_n(k)$ is not self-injective if $k$ is finite.
I’d like to hope if $k$ is a field of characteristic $0$, then each finite dimensional representation of $M_n(k)$ is completely reducible but I don’t know.  Let me add Putcha and Okninski showed if $k$ is any field and $F$ is a field of characteristic $0$, then $FM_n(k)$ has trivial Jacobson radical.
