Let $k$ be a field, let $D$ be a division ring. Assume the matrix ring $A = M_n(D)$ is endowed with some $k$-algebra structure compatible with its ring structure (namely, for all $\lambda \in k$ and for all $M, N \in A$, $\lambda(MN) = (\lambda M)N = M(\lambda N$)). We are not assuming a priori that there is an embedding of rings $k \to D \subset A$.
Is it possible to show that such an embedding exists and, moreover, the resulting algebra structure on $D$ is naturally the restriction of the algebra structure on $A$? Namely, if $(1, 1, \dots, 1) \in A$ is the identity matrix, then for all $\lambda \in k$, $$ \lambda \cdot (1, 1, \dots, 1) = (\lambda \cdot 1, \lambda \cdot 1, \dots, \lambda \cdot 1). $$
I want to use this fact on a qualifying exam. It may be necessary to assume $k$ is algebraically closed?
I have tried proving this "directly", but have made embarrassingly little progress. Thanks.
A $k$-algebra structure on $A$ can simply be seen as a ring homomorphism $k\rightarrow Z(A)$ into the center of $A$.
Thus if you have a $k$-algebra structure on $M_n(D)$, you have a morphism $k\rightarrow Z(M_n(D))$. But $Z(M_n(D))=Z(D)$ consists of matrices of the form $d I_n$ for $d\in Z(D)$. See here for instance for a proof of this fact.
Thus, the $k$-algebra structure on $M_n(D)$ just comes from a $k$-algebra structure on $D$.