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.

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 ringswith involutioni.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 '14 at 15:452more comments