Example of an algebra which is not isomorphic to its opposite

Question

I was trying to solve an exercise (marked with a star) that asks to come up with an example of an algebra $A$ which is not isomorphic to $A^{\mathrm{op}}$.

I thought at first that I just need to find an algebra containing elements $a, b$, such that $ab \neq 0, \; ba = 0$ but then I realized that it doesn't give anything so now I don't know what to do.

I googled some examples but they aren't natural for me, for instance, one of them was built out of quiver and I never studied quivers.

Background: most of the Dummit & Foote book.

Answer 1

$\mathbb Z\langle x,y\rangle/(y^2,yx)$ is left Noetherian but not right Noetherian, so it is impossible for it to be isomorphic to its opposite ring.

Answer 2

There are rather explicit examples given in Jacobson's Basic Algebra (vol. 1), Section 2.8, see this MO-post:

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 $x,y\in M_6(\mathbf Q)$. Now the subring of $M_6(\mathbf Q)$ generated by $x$ and $y$ is not isomorphic to its opposite.

Answer 3

This is more of a comment than an answer but if you are willing to consider nonassociative algebras it is really easy to come up with pathological examples like the algebra $\mathbb{R}^n$ with multiplication $xy = \|x\|_1y$ for all $x, y$.