# Vector Dimension of Simple Components of Semisimple $k$-algebra

Suppose that $k$ is an algebraically closed field. Let $R$ be a noncommutative $k$-algebra with $\dim_k R=7$. What are the vector space dimensions of the simple components of $R$? Use this to find the isomorphism classes of simple $R$-modules.

It is obvious that the dimensions of the simple components must be finite. Since maps between simple modules are zero maps or isomorphisms by Schur's Lemma, once one knows the vector space dimension of the simple modules, you can give a description of the simple modules themselves and the isomorphism classes easily based on their dimension. The part I am confused about is how to find these dimensions to begin with. I know this probably makes use of the Artin-Wedderburn Theorem. Since $R$ is semisimple, it is (isomorphic to) a product of matrix rings over division rings. But I do not see exactly what this gets me in relation to the fact that $\dim_k R=7$. How do I proceed with this problem?

"I know this probably makes use of the Artin-Wedderburn Theorem. Since $R$ is semisimple, it is (isomorphic to) a product of matrix rings over division rings. But I do not see exactly what this gets me in relation to the fact that $\dim_k(R)=7$."
The extra piece of information you need is that the division rings involved are also $k$-algebras, and the $k$-algebra structure on a simple component $M_n(D)$ is induced by the $k$-algebra structure on $D$. Thus, $\dim_k(M_n(D)) = n^2\cdot \dim_k(D)$. Now if
$R\cong M_{n_1}(D_1)\times \cdots \times M_{n_r}(D_r)$
and $d_i=\dim_k(D_i)$, then $\dim_k(R) = \sum_{i=1}^r n_i^2\cdot d_i$.
The other two facts that are involved in this problem are that $k$ is algebraically closed (which implies $d_i=1$ for all $i$) and that the ring is noncommutative (so $n_i>1$ for some $i$). This reduces the problem to solving $\sum_{i=1}^r n_i^2 = 7$ in positive integers with $n_i>1$ for some $i$. This has a unique solution up to reordering the $n_i$'s.