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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    $\begingroup$ For the same operator algebra people: see also Alain Connes, A factor not anti-isomorphic to itself Ann. Math. (2) 101, 1975, 536-554. $\endgroup$ May 9, 2011 at 11:23
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    $\begingroup$ 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) $\endgroup$ May 9, 2011 at 15:21
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    $\begingroup$ @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.) $\endgroup$
    – Jon Bannon
    May 9, 2011 at 15:34
  • $\begingroup$ Thanks everyone for the nice answers. I would like to have accepted more than one. $\endgroup$ May 10, 2011 at 10:36
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    $\begingroup$ The $*$-operator algebra examples are non-examples for this question: they have a involution, so are rings isomorphic to their opposite rings. The point of the (difficult) examples is that they have no complex-linear anti-automorphism as rings with involution i.e. complex-linear anti-automorphisms that commute with the fixed anti-automorphism $*$ (b.t.w. in $C^*$ algebras all ring [anti]automorphism are real-linear, but not always complex-linear) $\endgroup$
    – user46855
    Feb 15, 2014 at 15:45

9 Answers 9


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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    $\begingroup$ It's clear if you believe that theorem, that you can recover the original quiver! $\endgroup$ May 9, 2011 at 14:16
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    $\begingroup$ 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! $\endgroup$ May 9, 2011 at 15:09
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    $\begingroup$ Ok then could someone indicate what this Ext-construction is which recovers the quiver? $\endgroup$ May 10, 2011 at 15:32
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    $\begingroup$ 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 $Ext^1(\Gamma e_j,\Gamma e_i)$. $\endgroup$ May 10, 2011 at 15:49
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    $\begingroup$ 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 $T$. $\endgroup$ May 10, 2011 at 16:00

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

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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    $\begingroup$ 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. $\endgroup$ May 9, 2011 at 22:03
  • $\begingroup$ 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. $\endgroup$ May 10, 2011 at 6:15

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.

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    $\begingroup$ Unless I miss something, this $L(V)$ is not a ring: we have $x.(y+z)=x$ while $(x.y)+(x.z)=x+x$. $\endgroup$ May 9, 2011 at 12:17
  • $\begingroup$ 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). $\endgroup$
    – Todd Trimble
    May 9, 2011 at 12:32
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    $\begingroup$ Your semigroup example is incorrect. The algebra of a left zero with adjoined zero is isomorphic to its opposite algebra. They are both isomorphic to the path algebra with two vertices and |S|-1 edges from one to the other $\endgroup$ Nov 13, 2020 at 16:21
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    $\begingroup$ More specifically, the elements $1$ and $1-s$ with $s$ in S form a basis and (1-s)(1-t)=1-t for s,t in S so you have a basis which is a right zero semigroup with adjoined identity. $\endgroup$ Nov 13, 2020 at 16:24
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    $\begingroup$ However morally you are correct. The monoid identities satisfies by a left zero semigroup with adjoined identity is axiomatized by the identity $xyx=xy$ and every path algebra of an acyclic quiver is the monoid algebra of such a monoid. So by James Cranch's argument you get lots of examples of this sort. $\endgroup$ Nov 13, 2020 at 16:43

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...

  • $\begingroup$ This is so much simpler! $\endgroup$ May 10, 2011 at 6:46

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.


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).

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

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 $\mathbb{R}$.


Many examples are already given; here is another one, just for its own interest:

Let $V$ be a vector space of infinite countable dimension over a countable field $K$.

Let $E$ be the $K$-algebra of endomorphisms of $V$. I claim that $E$ is not isomorphic to its opposite (even as a ring, i.e., as $\mathbf{Z}$-algebra). Precisely:

  • (1) for every $g\in E-\{0\}$, $gE=\{gf:f\in E\}$ is uncountable [of continuum cardinal]
  • (2) there exists $f\in E-\{0\}$ such that $Ef=\{gf:g\in E\}$ is countable: [namely this holds iff $f$ has finite rank (otherwise it has continuum cardinal)]

Let me justify the non-bracketed assertions. In (2) this holds because if $B$ is a finite subset of $E$ such that $f(B)$ spans $f(E)$, then every element of $Ef$ is determined by its restriction to $B$.

In (1), just fix a line $L$ not in the kernel of $g$, and let $f$ range over the space $Y$ linear maps $V\to L$. Since the dual of $V$ has uncountable dimension, $Y$ has uncountable [continuum] dimension. And $f\mapsto gf$ is injective in restriction to $Y$.

Maybe in this case $E$ and $E^{\mathrm{op}}$ are not elementary equivalent, but this would require another argument.


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