Is $Hom_R(k^2,k^2)=k$? Let $k$ be any field, and $R = Mat_{2 \times2}(k)$.
Is it true that $Hom_R(k^2,k^2)=k$?
I assume the question is asking if the two are isomorphic as $k$-vector spaces. Is this an example of Schur's Lemma? If it is, we would require $k$ to be algebraically closed, wouldn't we?
 A: You do not need Schur's Lemma here. The statement
$$\text{End}_{\text{Mat}_{n\times n}(k)}\left(k^n\right)=\text{Hom}_{\text{Mat}_{n\times n}(k)}\left(k^n,k^n\right)\cong k\tag{*}$$
is true for any positive integer $n$ and for any field $k$.  This is because every matrix $\mathbf{A}\in\text{Mat}_{n\times n}(k)$ that commutes with all matrices $\mathbf{X}$  in $\text{Mat}_{n\times n}(k)$ is a scalar multiple of the identity matrix.  That is, the center of $\text{Mat}_{n\times n}(k)$ is
$$\mathfrak{Z}\big(\text{Mat}_{n\times n}(k)\big)=\big\{\alpha\,\mathbf{I}_{n\times n}\,\big|\,\alpha\in k\big\}\,,$$
where $\mathbf{I}_{n\times n}$ is the $n$-by-$n$ identity matrix.
For a proof, you can check commutativity of $\mathbf{A}$ with $\mathbf{X}$ of the form $\mathbf{E}_{i,j}$ for $i,j=1,2,\ldots,n$ and deduce what $\mathbf{A}$ should look like.  Here, $\mathbf{E}_{i,j}$ is the matrix with $0_k$ at every entry, except $1_k$ at the entry $(i,j)$.  I think $k$ can even be any commutative unital ring for (*) to hold, and the proof remains the same.  The only difference is that $k^n$ is not a simple $\text{Mat}_{n\times n}(k)$-module if $k$ is not a field.  (For an arbitrary, not necessarily commutative, unital ring $k$, the $\text{Mat}_{n\times n}(k)$-module $k^n$ is simple if and only if $k$ is a division ring.)
