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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.

user26857
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Invincible
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  • You should probably give an idea of your background and level of understanding, so people can try to explain an example that you may understand. – Captain Lama Apr 15 '20 at 10:10
  • @Captain Lama, I read most of the Dummit & Foote book. – Invincible Apr 15 '20 at 10:11
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    See the answers here for example. – Dietrich Burde Apr 15 '20 at 10:23
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    @Dietrich Burde shouldn't examples of $k$-algebras be much simpler that of division rings? – Invincible Apr 15 '20 at 10:24
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    The thing is that your algebra has to be non-commutative, and it is hard to come up with simpler non-commutative algebras than division algebras (or in general central simple algebras). – Captain Lama Apr 15 '20 at 10:26
  • @Captain Lama, I thought that division rings are rather rare so it is hard to even come up with an example of a division ring not speaking about such a rings with given properties (I know only one division ring and couldn't come up with more). – Invincible Apr 15 '20 at 10:34
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    It's a question of point of view, but the fact that they are "rare" as you say means we understand and control them pretty well, so on the contrary it is easy to define one which satisfies the conditions you want. (In this case, when the base field does not have non-trivial involution, it's enough to construct one with odd dimension because a division algebra with an involution which fixes its center always has even dimension over its center). – Captain Lama Apr 15 '20 at 10:38
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    @CaptainLama Just because division rings are structurally very simple does not mean those examples are going to be simple. In fact I find them pretty complicated. There are a lot of simple examples of rings which are “left-not-right something” which are a lot more accessible. – rschwieb Apr 15 '20 at 10:39
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    @rschwieb I agree that your example is very nice, and much simpler to understand than examples based on division algebras. It all depends on our habits; when I saw the question I immediately thought "we just have to take a Brauer class of order not $2$" (but of course it requires knowledge about the Brauer group). There is more machinery involved, but the reasoning in itself is quite simple I think. – Captain Lama Apr 15 '20 at 10:50
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    @CaptainLama, I do think that the division-algebra approach is more conceptual, and more related to other things, than the left-but-not-right Noetherian idea. – paul garrett Apr 15 '20 at 17:57
  • I think this question is at a level, where personal history and specialization correlates better with what one finds more natural. As opposed to, say, general aptitude in algebra. I was lead to study division algebras for their applications in the design of multiantenna radio signals, so I would have gone that way. Others have different backgrounds, and it shows. +1 to y'all. – Jyrki Lahtonen Apr 16 '20 at 11:31

3 Answers3

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

rschwieb
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    That’s a $\mathbb Z$ algebra, but if you need a field I don’t see anything different if you use $\mathbb Q$ in place of $\mathbb Z$ – rschwieb Apr 15 '20 at 10:50
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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.

Dietrich Burde
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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 $x, y$.

Vincent
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