TITLE: Simplest examples of rings that are not isomorphic to their opposites
QUESTION [36 upvotes]: 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.

REPLY [2 votes]: 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.