non-commutative finite rings - MathOverflow

non-commutative finite rings

2013-01-31T19:01:23Z

This is probably simple. Is there a finite non-commutative ring $R$ with identity in which all of its ideals are two-sided !?

Answer by Torsten Schoeneberg for non-commutative finite rings

2013-01-31T21:23:27Z

For a finite field $k$ with non-trivial automorphism $\sigma$, take the skew polynomial ring $k[X, \sigma]$ (reminder: these are polynomials with coefficients on the left $\sum a_i X^i$ with relation $Xa = \sigma(a)X$) and set $R := k[X, \sigma] / < X^n >$ (i.e. divide out the (two-sided) ideal generated by $X^n$).

The only left as well as right ideals in this ring are the two-sided ones, namely, the ones generated by one of the $X^k$ for $0 \le k \le n$. The key fact to see this is that the units of $R$ are precisely the polynomials $\sum a_i \bar X^i$ with $a_0 \neq 0$ (to see this, use e.g. a geometric series argument). In case $n=2$, it is easily written down explicitly, $R$ can be seen as the set of $a + bX$ with $a, b \in k$ and multiplication given by

$(a+bX)(c+dX) = ac + (b\sigma(c) + ad) X$;
for $a \neq 0$, we have $(a + bX)^{-1} = a^{-1} - a^{-1} \sigma(a^{-1}) b X$.

@ Dimros, concerning your questions to Ralph: A matrix ring $M_n(k)$ over a finite (division) field $k$ (and a fortiori, any product of these) will never meet your conditions, as for $n =1$ it is commutative and for $n >1$ it has left ideals which are not two-sided. So by Artin-Wedderburn, any example will have non-zero radical and its semisimple quotient will be a product of fields. On the other hand, every ring embeds into any matrix ring over itself, so every example is a subring of matrix rings. Maybe you want some special kind of subring.

Answer by Ralph for non-commutative finite rings

2013-02-01T01:04:15Z

A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal.

Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.

Added: The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}_2[Q]$ can be embedded into the matrix ring $M_4(\mathbb{F}_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a faithful representation $Q\hookrightarrow GL(4,2)$ is given by $$x \mapsto \begin{pmatrix}1 & 0 & 1 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 1 \newline 0 & 0 & 0 & 1 \end{pmatrix}\qquad y \mapsto \begin{pmatrix}1 & 1 & 1 & 1 \newline 0 & 1 & 0 & 1 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{pmatrix}$$ Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}_2$.