How can I construct a non commutative ring of order 125?

I think the approach is to take $M_n(F)$, since the ring of matrices over field is always non commutative. But I am not sure how to find $n$ or a specific field $F$.


The set $T$ of upper triangular $2 \times 2$ matrices

$$\begin{pmatrix}a & b\\0 & c\end{pmatrix}$$

with entries in the field $\mathbb{F}_5$ is a noncommutative subring of the ring $M_2(\mathbb{F}_5)$ of $2 \times 2$ matrices.

  • $T$ is a (sub)ring as it is closed for addition and multiplication.

  • its cardinality is $5^3$ because there are 3 independent choices for the entries, each one being in $(\mathbb{F}_5)^3.$

  • the noncommutativity is shown on an example:

$$\text{If} \ A:=\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix} \ \text{and} \ B:=\begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}$$

then: $AB=A$ whereas $BA=0$ (the null matrix).

(thus $A$ and $B$ are "zero-divisors").

Important remark: Let me now cite the Wikipedia article (https://en.wikipedia.org/wiki/Finite_ring):

"If a non-commutative finite ring with 1 has the order of a prime cubed, then the ring is isomorphic to the upper triangular 2 × 2 matrix ring over the Galois field of the prime. The study of rings of order the cube of a prime was further developed in (Raghavendran 1969) and (Gilmer & Mott 1973). Next Flor and Wessenbauer (1975) made improvements on the cube-of-a-prime case".

Final remark: I just found that the problem (and the very same solution I gave !) can be found in a good book "Abstract Algebra Manual. Problems and Solutions" 2nd Edition, Ayman Badawi, Nova ed. (http://www.worldcat.org/title/abstract-algebra-manual-problems-and-solutions/oclc/54674579)

  • $\begingroup$ Thank you. Can you explain why the order is 125? I cannot find any theorem proving the order of $M_n(F) = n^2*|F|$? (I think this is right?) and I can't seem to show it myself. $\endgroup$ – owlvault Apr 23 '17 at 20:36
  • $\begingroup$ I see a second "projective" solution : consider the ring of all $2 \times 2$ matrices with entries in $\mathbb{F_5}$. It has $5^4$ elements. Take the quotient by the ideal composed by the five diagonal matrices $kI_5, k \in \mathbb{F_5}$; the resulting ring should be another solution (to be checked...). $\endgroup$ – Jean Marie Apr 23 '17 at 21:11
  • $\begingroup$ My recommendation to you is to offer as example the upper-triangular $3\times3$ matrices with all $1$’s on the diagonal. $\endgroup$ – Lubin Apr 23 '17 at 21:14
  • $\begingroup$ Based on the counterexample I gave, it is easy to show that this Heisenberg "ring" isn't isomorphic to the ring I gave in my answer ($2 \times 2$ upper triangular matrices with entries in $\mathbb{F}_5$.) $\endgroup$ – Jean Marie Apr 23 '17 at 22:05
  • $\begingroup$ @Lubin In fact,there is a problem (that I had not foreseen) with your proposal: the set of $3 \times 3$ matrices with all $1$s on the diagonal is not closed for addition... $\endgroup$ – Jean Marie May 14 '17 at 20:36

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