# Determine up to isomorphism all semisimple noncommutative rings with order 512

Problem: Determine up to isomorphism all semisimple noncommutative rings of order 512 = $2^9$. (This is problem from an old qualifying exam I am studying from)

So far I have: Let A be a semisimple ring of order 512. A is finite so A must be Artinian. Then Wedderburn-Artin implies A is isomorphic to a direct product of matrix rings over division rings i.e. a direct product of $M_n(\mathbb{F}_{p^k})$ where $\mathbb{F}_{p_k}$ is a finite field of order $p^k$. Now the order of $M_n(\mathbb{F}_{p^k})$ is $p^{kn^2}$. So I'm assuming this becomes a combinatorics problem on how many ways to write $2^9$ as a product of numbers of the form $p^{kn^2}$. When I try to do this I get a very long list: $2^{9*1^2}, 2^{1*3^2}, 2^{8*1^2}*2^{1*1^2}, 2^{2*2^2}*2^{1*1^2}$ and so on. Is this the correct approach? Also, how do I know when A is noncommutative?

• $M_n(\mathbb{F}_{p^k})$ is noncommutative iff $n>1$, to see this just find two matrices that don't commute. Writing $2^9$ as a product of numbers of the form $2^{kn^2}$ is the same as writing 9 as a sum of number of the form $kn^2$, there really aren't that many ways so it's probably easiest to just list them. – Nate Apr 26 '14 at 19:29
• Groups of order $p^1$ and $p^2$ are abelian, so your direct product must have at least one factor $p^{\geq 2}$? This seems to require $2^{1\cdot 3^2}$, $2^{2 \cdot 2^2}$, $2^{1 \cdot 2^2}$, or $2^{\{3,4,5,6,7,8,9\} \cdot 1^2}$ for this "large factor"? – Eric Towers Apr 26 '14 at 19:32
• Are you sure the ring has to be not commutative? Most sources use the adjective t mean "not necessarily commutative (but could be)" – rschwieb Apr 26 '14 at 22:05

• one or two 2 by 2 matrix rings over $F_2$
• one 2 by 2 matrix ring over $F_4$
• one 3 by 3 matrix ring over $F_2$