Question

Let R be an n2-dimensional algebra over a field k. Show that R ∼ Mn(k) (as k-algebras) iff R is simple and has an element r whose minimal polynomial over k has the form (x − a1) • • • (x − an) where a1, . . . , an ∈ k.

Answer 1

First, suppose $R = \mathbf{M}_n(k)$. Then $R$ is simple, and the matrix with 1's on the line above the diagonal and 0's elsewhere has minimal polynomial $x^n$. Conversely, suppose $R$ is simple and has an element $r$ whose minimal polynomial over $k$ is $(x - a_1) \cdots (x - a_n)$, where $a_1, \ldots, a_n \in k$. Then we have a strictly descending chain of left ideals

$(\#) R \supset R(r - a_1) \supset R(r - a_1)(r - a_2) \supset \cdots \supset R(r - a_1) \cdots (r - a_n) = 0.$

For, if $R(r - a) \cdots (r - a_i) = R(r - a_1) \cdots (r - a_{i+1})$ for some $i$, right multiplication by $(r - a_{i+2}) \cdots (r - a_n)$ would give $(r - a_1) \cdots (r - a_i)(r - a_{i+2}) \cdots (r - a_n) = 0 \in R$, which is impossible. Now, express $R$ in the form $\mathbf{M}_m(D)$ where $D$ is a division $k$-algebra of, say, dimension $d$. Then $n^2 = m^2 d$, and ${}_R R$ has composition length $m$. Appealing to $(\#)$, we have $m^2 \geq n^2 = m^2 d$, so $d = 1$, $n = m$, and $R \sim \mathbf{M}_n(k)$.