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What is an example of a ring $R$ for which the abelian category of left $R$-modules is not isomorphic to the category of right $R$-modules?

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    $\begingroup$ I'm pretty sure this has been asked before on mathoverflow many times. You need a quiver which is not isomorphic to its opposite quiver and the categories of left and right modules are not equivalent. And I assume you mean equivalent and not isomorphic. $\endgroup$ Nov 3, 2021 at 13:21
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    $\begingroup$ Related question is mathoverflow.net/questions/64370/…. It is not the same but some of the answers there work. $\endgroup$ Nov 3, 2021 at 13:22

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Let $Q$ be a finite acyclic quiver and $K$ a field. Let $Q^{op}$ be the opposite quiver (where all the arrows are reversed). Let $KQ$ be the path algebra. Then the category of left $KQ$-modules is equivalent to the category of right $KQ^{op}$-modules. It is well known that if $Q,Q'$ are acyclic quivers, then the categories of right $KQ$- and $KQ'$-modules are equivalent if and only if $Q$ and $Q'$ are isomorphic as directed graphs. You can recover the quiver from the Ext between simple modules. So if $Q$ is quiver not isomorphic to its opposite then its categories of left and right modules are not equivalent. One can find a suitable orientation of the Dynkin quiver $D_4$ with this property.

Added Nov 4, 2021. In light of @BugsBunny's answer and the comments there, maybe it is worth making things more explicit. Let $K$ be your favorite field and consider the subalgebra of $4\times 4$-upper triangular matrices over $K$: $$A=\begin{bmatrix} K & K & K & K\\ 0 & K & K & K\\ 0 & 0& K&0\\ 0& 0&0&K\end{bmatrix}.$$ This algebra is isomorphic to the path algebra of the Dynkin quiver $D_4$ with a suitable orientation. This algebra has $4$ simple right modules up to isomorphism (since the radical is the ideal of strictly upper triangular matrices in $A$ and the semisimple quotient is isomorphic to the subalgebra of diagonal matrices). Exactly two isomorphism classes of simple modules are projective. The simple right projectives correspond to $$\begin{bmatrix} 0 & 0& K& 0\end{bmatrix}$$ and $$\begin{bmatrix} 0 & 0 & 0 &K\end{bmatrix}$$ acted on by right multiplication. In general the right projective indecomposables are given by the rows "in" $A$. But there is only one simple projective left module $$\begin{bmatrix} K\\ 0 \\ 0\\ 0\end{bmatrix}$$ acted on by left multiplication. In general the left projective indecomposables are given by the columns "in" $A$. So the categories of left and right modules are not equivalent.

Added Nov. 5, 2021. As @JeremyRickard pointed out in the comments, an easier example is the algebra $$B=\begin{bmatrix} K & 0 & K\\ 0 & K & K\\ 0 & 0& K \end{bmatrix}.$$ Arguing as above, this has one right simple projective (coming from the action on vectors of the form of the last row) and two left simple projectives (acting on vectors of the form of the first column and on vectors of the form of the second column).

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  • $\begingroup$ I see - this is intuitive too. $\endgroup$
    – tkf
    Nov 5, 2021 at 5:27
  • $\begingroup$ A slightly simpler example is the $A_3$ quiver with a sink vertex in the middle. $\endgroup$ Nov 5, 2021 at 8:32
  • $\begingroup$ @JeremyRickard, you are right of course. $\endgroup$ Nov 5, 2021 at 12:30
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$\begin{pmatrix} {\mathbb Z} & {\mathbb Q}\\ 0 & {\mathbb Q} \end{pmatrix}$ is a canonical example for such things. Let us list all simple right and left modules: $$R_p=\begin{pmatrix} {\mathbb Z}/(p) & 0\end{pmatrix}, \ R_0 =\begin{pmatrix} 0 & {\mathbb Q} \end{pmatrix}, \ L_p=\begin{pmatrix} {\mathbb Z}/(p) \\ 0 \end{pmatrix}, \ L_0=\begin{pmatrix} 0 \\ {\mathbb Q} \end{pmatrix}$$ for various primes $p$. Now notice that $R_0$ is a projective simple right module. By inspection, there exists no projective simple left module.

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    $\begingroup$ Doc, you are right. What have I been smoking? I will fix it. $\endgroup$
    – Bugs Bunny
    Nov 4, 2021 at 14:59
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    $\begingroup$ Brilliant - anyone learning about modules and categories could naturally ask this question, and this answer would be accessible to them. $\endgroup$
    – tkf
    Nov 4, 2021 at 20:11
  • $\begingroup$ @tkf, I added to my answer to make it more accessible. The example I suggested has a different number of projective simple left and right modules as well, similar to this one. $\endgroup$ Nov 5, 2021 at 3:32

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