# Do you know a finite unitary reversible ring that is not isomorphic to its opposite? And the minimal with that property?

Do you know a finite unitary reversible ring that is not isomorphic to its opposite? And the minimal with that property?

The examples of rings not isomorphic to their opposite that I know of are not reversible.

• Would you remind what "reversible" means? And "minimal" is not clearly defined.
– YCor
Commented Mar 9, 2021 at 12:29
• Reversible means $xy=0$ implies $yx=0$ according to Google Commented Mar 9, 2021 at 13:41
• According to math.stackexchange.com/questions/45085/… there are division rings not isomorphic to their opposites and these are reversible. Commented Mar 9, 2021 at 18:56
• @Benjamin Steinberg Okay. Thanks. If you find an example other than division rings, let me know. Commented Mar 10, 2021 at 9:07

Applying the following example with $$q=2$$ gives a finite unital reversible ring $$R$$ with $$2^6=64$$ elements which is not isomorphic to its opposite. I do not know if this is the cardinality-wise smallest possible example.
Fix a prime power $$q$$ and let $$F$$ and $$E$$ denote finite fields with $$q$$ and $$q^3$$ elements, respectively. We assume $$F\subseteq E$$ and let $$\sigma$$ denote a generator of $$\mathrm{Gal}(E/F)\cong C_3$$. Let $$E[X;\sigma]$$ denote the twisted polynomial ring (a.k.a. Ore extension) in the variable $$X$$ over $$E$$. That is, the elements of $$E[X;\sigma]$$ are polynomials $$\sum_{i=0}^t a_i X^i$$ with coefficients in $$E$$, they are added in the usual way, but the product is determined by setting $$X a = \sigma(a)X$$ for all $$a\in E$$. Let $$R=E[X;\sigma]/\langle X^2\rangle,$$ where $$\langle X^2\rangle$$ is the ideal generated by $$X^2$$. Writing $$x$$ for the image of $$X$$ in $$R$$, elements of $$R$$ can be written uniquely as $$a_0+a_1x$$ with $$a_0,a_1\in E$$ and the product in $$R$$ is given by $$(a_0+a_1x)(b_0+b_1x)=(a_0b_0)+(a_0b_1+a_1\sigma(b_0))x$$.
We claim that $$R$$ is reversible and $$R\ncong R^\mathrm{op}$$.
To see that $$R$$ is reversible, note that its Jacobson radical $$J:=Ex$$ satisfies $$J^2=0$$ and $$R/J\cong E$$. That is, $$R$$ is local and it Jacobson radical squares to zero. All such rings are reversible, because if $$f,g\in R$$ satisfy $$fg=0$$, then $$f,g$$ must both lie in the radical and therefore satisfy $$gf=0$$.
Next, for the sake of contradiction, suppose that there is an anti-automorphism $$\tau:R\to R$$. Then $$\tau(J)=J$$, and thus $$\tau$$ induces an automorphism $$\tau_0$$ of $$E\cong R/J$$. Consequently, $$\tau(a)-\tau_0(a)\in J$$ for all $$a\in E$$ (we regard $$E$$ as a subring of $$R$$). Write $$\tau(x)=ux$$ with $$u\in E-\{0\}$$. Then for all $$a\in E$$, we have $$\tau(\sigma(a)x)=\tau(x)\tau(\sigma(a)) =ux\cdot \tau(\sigma(a)) =ux\cdot \tau_0(\sigma(a))=u\cdot \sigma\tau_0\sigma(a)\cdot x$$ where the third equality holds because $$\tau(\sigma(a))-\tau_0(\sigma(a))\in J$$ and $$xJ=0$$. Likewise, one finds that $$\tau(xa)=\tau(a)\tau(x)=\tau_0(a)ux=u\cdot \tau_0(a)\cdot x.$$ Since $$xa=\sigma(a)x$$ and $$u\neq 0$$, it follows that $$\tau_0(a)=\sigma\tau_0\sigma(a)$$ for all $$a\in E$$, or rather $$\tau_0=\sigma\tau_0\sigma$$ in $$\mathrm{Gal}(E/F)$$. Since $$\mathrm{Gal}(E/F)\cong C_3$$, this means that $$\sigma^2=1$$, which is absurd by our choice of $$\sigma$$.