# An example of a division ring $D$ that is **not** isomorphic to its opposite ring

I recall reading in an abstract algebra text two years ago (when I had the pleasure to learn this beautiful subject) that there exists a division ring $D$ that is not isomorphic to its opposite ring. However, the author noted that the construction of such a ring "would take us too far afield". My question is: please give an example of a division ring $D$ that is not isomorphic to its opposite ring.

Let me recall that a division ring is a ring $A$ such that every non-zero element of $A$ is a unit, i.e., has a multiplicative inverse in $A$. However, division rings are not required to be commutative. Let me also recall that if $A$ is a ring, the opposite ring of $A$ is the ring $A^{\text{op}}$ that has the same underlying set as $A$, the same additive structure as $A$, but that the multiplication $*$ on $A^{\text{op}}$ is defined by the rule $a*b=ba$ where "$ba$" denotes the product of $b$ and $a$ with respect to the multiplication in $A$.

It is not hard to find examples of rings which are not isomorphic to their opposite ring and it is trivial to find examples of noncommutative rings that are isomorphic to their opposite rings. (E.g., the $n\times n$ matrix ring over a field $F$ ($n>1$) is isomorphic to its opposite ring via the transpose map $A\to A^{T}$ where $A$ is a matrix and $A^{T}$ denotes its transpose.) Of course, every commutative ring is isomorphic to its opposite ring via the identity map. However, this particular question appears to be difficult.

• There are examples at mathoverflow.net/questions/64370/… . Jun 13, 2011 at 10:49
• I would like to preemptively strike down any attempts at bad puns with "too far afield" here. Jul 1, 2013 at 16:19

Let $F$ be the center of the division ring $D$. Then $D$ represents its class in the Brauer group $Br(F)$. The opposite ring $D^{opp}$ represents the inverse element. The reason why for example the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order 2 in $Br(\mathbf{R})$, and hence equal to its own inverse in the Brauer group.

To get a division algebra that is not isomorphic to its opposite algebra we can use an element of order 3 in the Brauer group. One method for constructing those is to start with a (cyclic) Galois extension of number fields $E/F$ such that $[E:F]=3$. Let $\sigma\in Gal(E/F)$ be the generator. Let $\gamma\in F$ be an element that cannot be written in the form $\gamma=N(x)$, where $N:E\rightarrow F, x\mapsto x\sigma(x)\sigma^2(x)$ is the relative norm map. Consider the set of matrices $$\mathcal{A}(E,F,\sigma,\gamma)=\left\{ \left(\begin{array}{rrr} x_0&\sigma(x_2)&\sigma^2(x_1)\\ \gamma x_1&\sigma(x_0)&\sigma^2(x_2)\\ \gamma x_2&\gamma\sigma(x_1)&\sigma^2(x_0) \end{array}\right)\mid x_0,x_1,x_2\in E\right\}.$$ A theorem of A. Albert tells us that this forms a division algebra with center $F$, and its order in $Br(F)$ is 3, so it will not be isomorphic to its opposite algebra. The theory is described for example in ch. 8 of Jacobson's Basic Algebra II. The buzzword 'cyclic division algebra' should give you some hits.

For a concrete example consider the following. Let $F=\mathbf{Q}(\sqrt{-3})$ and let $E=F(\zeta_9)$, with $\zeta_9=e^{2\pi i/9}$. Then $E/F$ is a cubic extension of cyclotomic fields, $\sigma:\zeta_9\mapsto\zeta_9^4$. I claim that the element $2$ does not belong the image of the norm map. This follows from the fact 2 is totally inert in the extension tower $E/F/\mathbf{Q}$. Basically because $GF(2^6)$ is the smallest finite field of characteristic 2 that contains a primitive ninth root of unity. Now, if $2=N(x)$ for some $x\in E$, then 2 must appear as a factor (with a positive coefficient) in the fractional ideal generated by $x$. But the norm map then multiplies that coefficient by 3, and as there were no other primes above 2, we cannot cancel that. Sorry, if this is too sketchy.

Anyway (see Jacobson again), the product of the $\gamma$ elements modulo $N(E^*)$ is the operation in the Brauer group $Br(E/F)\le Br(F).$ Therefore the opposite algebra should correspond to the choice $\gamma=1/2$, (or to the choice $\gamma=4$, as $2\cdot4=N(2)$). So $$\mathcal{A}(E,F,\sigma,2)^{opp}\cong \mathcal{A}(E,F,\sigma,1/2).$$ and the choices $\gamma=2$ and $\gamma=1/2$ yield non-isomorphic division algebras, as their ratio $=4$ is not in the image of the norm map.

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Edit (added more details here, because another answer links to this answer): In general the cyclic division algebra construction works much the same for any cyclic extension $E/F$. When $[E:F]=n$ we get a set of $n\times n$ matrices with entries in $E$. The number $\gamma$ appears in the lower diagonal part of the matrix. The condition for this to be a division algebra is that $\gamma^k$ should not be a norm for any integer $k, 0<k<n$. Obviously it suffices to check this for (maximal) proper divisors of $n$. In particular, if $n$ is a prime, then it suffices to check that $\gamma$ itself is not a norm.

Edit^2: Matt E's answer here linear algebra over a division ring vs. over a field gives a simpler cyclic division algebra of $3\times3$ matrices with entries in the real subfield of the seventh cyclotomic field.

• Dear Jyrki, this is an excellent answer, needless to say! Thank you very much for your efforts; I really appreciate it. Also, I apologize for taking a long time to accept this answer; I was travelling in the past couple of days. Once again, thanks! Regards, Jun 18, 2011 at 10:21
• After reading Morandi, Patrick J., Bharath Al Sethuraman, and J-P. Tignol. "Division algebras with an anti-automorphism but with no involution." (2005): 485-495. I think there might be a subtle but fatal flaw to the reasoning that "its order in 𝐵𝑟(𝐹) is 3, so it will not be isomorphic to its opposite algebra". Yes, it cannot be both order 3 and isomorphic to its opposite as algebras but it looks like they can still be isomorphic as rings. (cont'd) Dec 30, 2022 at 13:52
• (cont'd) In the article (starting page 4), the authors use the same cyclic algebra construction to give an example of a $3$ dimensional division algebra that does admit an antihomomorphism but has no involution. I may have misread something, though, so I look forward to being proven wrong... Dec 30, 2022 at 13:52
• And when I said but it looks like they can still be isomorphic as rings. I really meant that it might possibly be a loophole, not that I was positive that it was the problem. Dec 30, 2022 at 13:59
• Wow! Thanks for sharing, @rschwieb. I have met Bharath (Al) Sethuraman a few times. I should get in touch with him, and ask him to comment. Dec 30, 2022 at 15:15