## Simplest examples of rings that are not isomorphic to their opposites

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What are the simplest examples of rings that are not isomorphic to their opposite rings? Is there a science to constructing them?

## The only simple example known to me:

In Jacobson's Basic Algebra (vol. 1), Section 2.8, there is an exercise that goes as follows:

Let `$u=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}\in M_3(\mathbf Q)$` and let `$x=\begin{pmatrix} u & 0 \\ 0 & u^2 \end{pmatrix}$`, `$y=\begin{pmatrix}0&1\\0&0\end{pmatrix}$`, where $u$ is as indicated and $0$ and $1$ are zero and unit matrices in $M_3(\mathbf Q)$. Hence `$x,y\in M_6(\mathbf Q)$`. Jacobson gives hints to prove that the subring of $M_6(\mathbf Q)$ generated by $x$ and $y$ is not isomorphic to its opposite.

## Examples seem to be well-known to the operator algebras crowd:

See for example the paper: "A Simple Separable C*-Algebra Not Isomorphic to Its Opposite Algebra" by N. Christopher Phillips, Proceedings of the American Mathematical Society Vol. 132, No. 10 (Oct., 2004), pp. 2997-3005.

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For the same operator algebra people: see also Alain Connes, A factor not anti-isomorphic to itself Ann. Math. (2) 101, 1975, 536-554. – Alain Valette May 9 2011 at 11:23

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Out of curiosity: why do operator people publish papers about these examples? Are they much rarer there? Us mere algebrist don't think much of this, I think (all my algebra undergrad students have had to check that some example or another works) – Mariano Suárez-Alvarez May 9 2011 at 15:21

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@Mariano: It is extraordinarily hard to see whether von Neumann algebras are isomorphic. Results like these are usually hard-won and are accompanied by deep new understanding of the structure of the algebras. (Notice that the above paper of Connes is an Annals paper.) – Jon Bannon May 9 2011 at 15:34
Thanks everyone for the nice answers. I would like to have accepted more than one. – Amritanshu Prasad May 10 2011 at 10:36
@Mariano: My understanding is that there are no analogs of the kinds of examples that the operator algebra people have which mere algebraists could fathom: they are like having semisimple $\mathbf C$-algebras which are not isomorphic to their opposites! – Amritanshu Prasad May 17 2011 at 10:07

## 8 Answers

Here's a factory for making examples. If $\Gamma$ is a quiver, and $k$ a field, then we get a quiver algebra $k\Gamma$. If $\Gamma$ has no oriented cycles, we can recover $\Gamma$ from $k\Gamma$ by taking the Ext-construction. Also, the opposite algebra of a quiver algebra is obtained by reversing all the arrows in the quiver.

Hence you can produce an example by taking the quiver algebra of any quiver with no oriented cycles, which is not isomorphic to its reverse. It's easy to construct lots of quivers with these properties.

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It's clear if you believe that theorem, that you can recover the original quiver! – James Cranch May 9 2011 at 14:16

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Incidentally, the smallest such quiver is the V-shaped quiver with two arrows emanating from one point. This gives a five-dimensional algebra: that's quite neat and tidy! – James Cranch May 9 2011 at 15:09

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First, one finds the vertices of the quiver $\Gamma$, by finding the unique maximal set of orthogonal idempotents $\{e_i\}$; these will correspond to vertices. Then, the number of arrows from $i$ to $j$ will be $Ext^1(\Gamma e_j,\Gamma e_i)$. – Greg Muller May 10 2011 at 15:49

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Alternatively: since the algebra is finite dimensiona, there are a finite number of isomorphism classes of simple modules. Use them as vertices of a graph, and draw $\dim\operatorname{Ext}^1(S,T)$ arrows from the class of $S$ to the class of $T$. – Mariano Suárez-Alvarez May 10 2011 at 16:00

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Whoops, sorry, I confused two different ways of getting the arrows. The $Ext^1$ should be between $k e_j$ and $ke_i$, the simple modules supported at the appropriate vertices. You can also get the arrows as degree 1 graded $Hom$ between projectives $\Gamma e_i$ and $\Gamma e_j$, but the $Ext$ between them certainly vanishes. – Greg Muller May 10 2011 at 16:05
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Hi Amri,

This is a bit late, but it's my favorite class of examples. If $X$ is a smooth affine variety over $\mathbb{C}$ (say), and $\mathcal{D} = \mathcal{D}(X)$ is its algebra of differential operators, then the opposite algebra $\mathcal{D}^{op}$ is isomorphic to $\mathcal{D}(K) = K\otimes \mathcal{D}\otimes K^{-1}$, where $K$ denotes the canonical module of $X$. [This is also true when $X$ is Gorenstein but not necessarily smooth---see work of Yekutieli.]

So one gets answers to your question when $X$ doesn't have trivial canonical bundle. [And of course the story sheafifies for any smooth variety.]

EDIT: I was writing carelessly the first time (thanks to Amri's comment for highlighting this). Note that $\mathcal{D}(K)$ acts on $K$ on the left. Since a left $\mathcal{D}$-module structure on a vector bundle (finitely generated projective module) is the same as a flat connection, one has $\mathcal{D}\cong \mathcal{D}(K)$ if and only if $K$ admits a flat connection. The first chern class of $K$ is an obstruction to the existence of a flat connection. So just pick your favorite such affine variety (see also this MO question for discussion of that). A pretty complete discussion of the (non)triviality of rings of twisted differential operators (TDOs) can be found in Beilinson-Bernstein "A proof of Jantzen conjectures."

This story also illuminates a little bit why differential operators on half-densities, i.e. $\mathcal{D}(K^{1/2}) = K^{1/2}\otimes \mathcal{D}\otimes K^{-1/2}$, plays a special role in the study of rings of differential operators and (twisted) $\mathcal{D}$-modules (it's canonically isomorphic to its opposite algebra).

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Is it always the case that $\mathcal D(K)$ is not isomorphic to $\mathcal D(X)$ if $K$ is not trivial? Why? – Amritanshu Prasad May 11 2011 at 6:11
Thanks Tom! I miss our discussions in the corridor where I learned so many things. – Amritanshu Prasad May 12 2011 at 4:04

A general idea to construct rings which behave different on the left and on the right is the following, which is already contained in Martins's answer: One considers triangular rings ```$$ A=\begin{pmatrix} R & M \\
                   0 & S
   \end{pmatrix}
$$``` where $R$ and $S$ are rings and $M$ is an $R$-$S$-bimodule. The left and right ideals of such a ring can be decribed: for example, the left ideals are isomorphic to $U\oplus J$, where $J$ is a left ideal of $S$, and $U$ an $R$-submodule of $R\oplus M$ with $MJ \subseteq U$. (See Lam's book A First Course in Noncommutative Rings, §1) Suitable choices of $R$, $M$ and $S$ lead to examples with quite different left and right structure. For example, the finite ring ```$$\begin{pmatrix} 
       \mathbb{Z}/4\mathbb{Z} & \mathbb{Z}/2\mathbb{Z} \\
         0                    & \mathbb{Z}/2\mathbb{Z}
   \end{pmatrix}$$``` has 11 left ideals and 12 right ideals, if my counting is right. (This may be the smallest example of a unital ring not isomorphic to its opposite ring, but I'm not sure here.)

Of course, there are lots of examples, since there are many ring theoretic notions which are known to be not left-right symmetric. T. Y. Lam, in his two books (First Course mentioned above and Lectures on Modules and Rings), usually contructs at least one example of a ring being left blah but not right blah, whenever blah is a property which is not left-right symmetric. (Lam's books are generally worth reading, in particular when looking for examples!)

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I don't know whether or not there are others of order 16 (well, aside from its opposite), but there are definitely none of order smaller than 16. – Harry Altman May 9 2011 at 22:03
Thanks for pointing out Lam's book. I looked at it and indeed it is worth reading. Jacobson's example looks suspiciously similar to a triangular ring construction. – Amritanshu Prasad May 10 2011 at 6:15

A particularly simple example of an algebra not isomorphic to its (graded) opposite is the $\mathbb{R}$-algebra $\mathbb{C}$, where $1$ is even and $i$ is odd. This is the ($\mathbb{Z}/2$-graded) real Clifford algebra $Cl(-1) = \langle f \mid f^2 = -1 \rangle$. Its opposite is the Clifford algebra $Cl(1) = \langle e \mid e^2 = 1 \rangle$, whose underlying ungraded algebra is isomorphic to $\mathbb{R} \oplus \mathbb{R}$.

Per the discussion in the other answers, these two algebras represent $1$ and $-1 = 7$ in the graded Brauer group $\mathbb{Z}/8$ of $\mathbb{R}$.

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Here is an explicit example of a central simple algebra over $\mathbb{Q}$ not isomorphic to its opposite (which is merely a detailed example of what Pete explained).

First take a cubic cyclic Galois extension $L/\mathbb{Q}$, for instance $L = \mathbb{Q}[x] / (x^3 + x^2 − 2x − 1)$, and let $\rho$ be a non-trivial element of $\operatorname{Gal}(L/\mathbb{Q})$. Now take an arbitrary element $\gamma \in \mathbb{Q}$ which is not the norm of an element in $L$. Define $$D = L \oplus zL \oplus z^2L,$$ where $z$ is a new "symbol" subject to the relations $z^3 = \gamma$ and $zt = t^\rho z$ for all $t \in L$. Then $D$ is a central simple division algebra of degree $3$ (i.e. of dimension $9$), and since its image in $\operatorname{Br}(\mathbb{Q})$ has order $3$, it is not isomorphic to its opposite.

As you can imagine, this procedure works for any field admitting a cyclic extension (of degree $>2$) for which the norm is non-surjective.

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Here is an easy example. Consider the abelian group $M = \mathbb{Z} \times \mathbb{Q}$. I claim that $R:=\text{End}(M)$ does not have any anti-endomorphism at all. EDIT: My previous proof is flawed. Thanks to Leon Lampret who pointed this out to me. The new proof shows that $R$ has several anti-endomorphisms, but no one is invertible. Thus $R$ is not isomorphic to $R^{\mathrm{op}}$.

Identify $R$ with the matrix ring $\begin{pmatrix} \mathbb{Z} & 0 \\ \mathbb{Q} & \mathbb{Q} \end{pmatrix}$. The endomorphism ring of the underlying abelian group $\mathbb{Z} \times \mathbb{Q} \times \mathbb{Q}$ of $R$ can be identified with the matrix ring $\begin{pmatrix} \mathbb{Z} & 0 & 0 \\ \mathbb{Q} & \mathbb{Q} & \mathbb{Q} \\ \mathbb{Q} & \mathbb{Q} & \mathbb{Q} \end{pmatrix}$.

Assume an anti-endomorphism $\alpha$ of $R$ is given by such a matrix $\begin{pmatrix}a & 0 & 0 \\ b & c & d \\ e & f & g \end{pmatrix}$.

Then $\alpha(1)=1$ yields $a=1, b+d=0, e+g=1$. The determinant is $cg-df$. For all six-tuples $(u,v,w,p,q,r)$ (with $u,p$ integer) we have

$\alpha\left(\begin{pmatrix} u & 0 \\ v & w \end{pmatrix} \begin{pmatrix} p & 0 \\ q & r \end{pmatrix}\right) = \alpha \begin{pmatrix} p & 0 \\ q & r \end{pmatrix} \alpha\begin{pmatrix} u & 0 \\ v & w \end{pmatrix}$

which yields the three equations

1) $a^2 pu = pu$

2) $ap(bu + cv + dw) + (bp + cq + dr)(eu + fv + gw) = bpu + c(qu + rv) + drw$

3) $(ep + fq + gr)(eu + fv + gw) = epu + f(qu + rv) + grw$

If we plug in the three equations we already know from $\alpha(1)=1$, this simplifies of course. Now insert some tuples to get the following equations:

$(0,1,0,0,1,0) \leadsto f^2 = 0 \Rightarrow f = 0$

$(0,1,0,1,0,0) \leadsto c = 0$

This already shows that the determinant of $\alpha$ is zero, thus $\alpha$ cannot be bijective. But we can go even further:

$(1,0,0,1,0,0) \leadsto be=0 \wedge e^2=e \Rightarrow e \in \{0,1\}$

For $e = 0$ we get

$\alpha=\begin{pmatrix}1 & 0 & 0 \\ b & 0 & -b \\ 0 & 0 & 0 \end{pmatrix}$

and for $e=1$ we get

$\alpha=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}$.

Here $b \in \mathbb{Q}$ may be chosen arbitrary. These are all anti-endomorphisms of $R$.

There is a more advanced proof that $R$ is not isomorphic to $R^{\mathrm{op}}$: Observe that $R$ is right noetherian, but not left noetherian.

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Nice answer, Martin! Here's a link to supplement your last line: planetmath.org/encyclopedia/… – Jon Bannon May 9 2011 at 14:46
Thanks Martin. A very striking example from the triangular ring family. – Amritanshu Prasad May 10 2011 at 6:44

To amplify on Bugs Bunny's answer: let $D$ be a finite dimensional central division algebra over a field $K$. Then $D \otimes_K D^{\operatorname{op}} \cong \operatorname{End}_K(D)$. From this it follows that in the Brauer group of $K$, the class of $D^{\operatorname{op}}$ is the inverse of the class of $D$. So a central division algebra over a field is isomorphic to its opposite algebra iff it has order $2$ in the Brauer group, or, in the lingo of that field, period $2$.

So you can get examples by taking any field $K$ with $\operatorname{Br}(K) \neq \operatorname{Br}(K)[2]$. In particular the Brauer group of any non-Archimedean locally compact field is $\mathbb{Q}/\mathbb{Z}$ and the Brauer group of any global field is close to being the direct sum of the Brauer groups of its completions (there is one relation, the so-called reciprocity law, which says that a certain "sum of invariants" map is zero). So for instance a division algebra of dimension $9$ over its center will do and these things can be constructed over the above fields.

If you don't want to restrict to such nice algebras, it becomes much easier to construct examples. (Thanks to L. Moret-Bailly and Todd Trimble for fixing this!) Let $S$ be a set with at least two elements, and consider the left semigroup on $S$: for all $x,y \in S$, $x \cdot y = x$. For any field $K$, let $K[S]$ be the semigroup algebra, and let $A = K[S]^1 = K \oplus K[S]$ be the associated unital algebra. Then, if I am not still mistaken, it should be clear that $A$ is not isomorphic to its opposite algebra.

(Since I know nothing about operator algebras, my opinion is of course going to be that such examples are much more complicated than the ones given above...)

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Unless I miss something, this $L(V)$ is not a ring: we have $x.(y+z)=x$ while $(x.y)+(x.z)=x+x$. – Laurent Moret-Bailly May 9 2011 at 12:17
True, but this can be repaired by passing to a monoid algebra for a monoid $M$ with a similar definition for multiplication (and making due allowance for the identity). – Todd Trimble May 9 2011 at 12:32
Whoops! Good grief, thanks for catching this. I tried to repair it along the lines of Todd's comment, but let me know if it still seems fishy. (I'm off to delete something from my notes...) – Pete L. Clark May 9 2011 at 12:43

Your example is not simple, i.e., it is not a simple algebra! If you want a simple algebra, you need a field whose Brauer group has elements of order more than 2 (the opposite algebra = inversion in Brauer group). If I remember correctly, the p-adic field will do the trick...

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This is so much simpler! – Amritanshu Prasad May 10 2011 at 6:46
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