Let $A$ be finite dimensional connected algebra. A simple module $S$ is called $n$-regular in case $pd(S)=n$, $Ext_A^i(S,A)=0$ for $i=0,1,...,n-1$ and $Ext_A^n(S,A)$ being a simple $A$-left module. See for example https://www.sciencedirect.com/science/article/pii/S0001870818302809 for the relvance of 2-regular simple modules. We can assume $A$ is a quiver algebra and then being $n$-regular simply means that the injective envelope $I(S)$ of $S$ (having projective dimension $n$) occurs uniquely in the minimal injective coresolution $(I_i)$ of $A$ as a summand of $I_n$.


  1. In case every simple module of projective dimension $n$ (and there exists at least one such simple module) is $n$-regular, does $A$ have global dimension $n$? (In case this is false, is it true when assuming $A$ hsa finite global dimension?)

  2. When every right simple module of projective dimension $n$ is $n$-regular (and there exists at least one such simple module) is the same true for every left simple modules of projective dimension $n$?

edit: Question 1 and 2 were shown to be wrong by the answer of Erik D except the question in the brackets. For clarity I state the remaining open question here again (I also add some bonus questions, that I try to answer myself with the computer).

  1. In case every simple module of projective dimension $n$ (and there exists at least one such simple module) is $n$-regular, does $A$ have global dimension $n$ in case $A$ has finite global dimension? (what if we assume that this holds for simple left and right modules?)

(Bonus question: In case every simple (left and right, or maybe just onesided is enough?) module of projective dimension $n$ (and there exists at least one such simple module) is $n$-regular, does $A$ have finitistic dimension $n$?)

  • 1
    $\begingroup$ Sorry, my favourite piece of pedantry: at least for the first question I presume you want $A$ to be connected? $\endgroup$ Sep 13, 2018 at 17:13
  • $\begingroup$ @JeremyRickard Yes, thanks I added this condition. $\endgroup$
    – Mare
    Sep 13, 2018 at 17:49

1 Answer 1


The answers to both questions are negative. For a simple example, let $k$ be a field, $k[\epsilon]=k[x]/(x^2)$ the algebra of dual numbers, and $A=\begin{pmatrix} k & k[\epsilon] \\ 0& k[\epsilon] \end{pmatrix}$.

(In terms of quivers with relations, $A$ consists of a loop $\epsilon$ and an arrow $\alpha$ pointing at the vertex of $\epsilon$, with the relation $\epsilon^2=0$.)

The indecomposable projective right modules $P_1=\begin{pmatrix} k &k[\epsilon]\end{pmatrix}$ and $P_2=\begin{pmatrix} 0 &k[\epsilon]\end{pmatrix}$ of $A$ have Loewy series $$ P_1:\begin{pmatrix} 1\\2\\2\end{pmatrix} \quad\mbox{respectively}\quad P_2:\begin{pmatrix} 2\\2\end{pmatrix}.$$ The projective dimension of the simple (right) module $S_2$ corresponding to the projective $P_2$ is infinite, and $$ 0\to P_2\to P_1\to S_1\to 0$$ is a projective resolution of $S_1$. So $\mathop{\rm Ext}^1(S_1,A)=\mathop{\rm Ext}^1(S_1,P_2)=\mathop{\rm Hom}(S_1,P_2)$ is $1$-dimensional and thus simple, whilst $\mathop{\rm Ext}^0(S_1,A) = \mathop{\rm Hom}(S_1,A)=0$. Hence $S_1$ is $1$-regular, and $\mathop{\rm gldim}A =\infty \ne 1= \mathop{\rm pd}S_1$.

As for left modules, the algebra $A$ has one simple module of infinite projective dimension, and one simple projective. So there are no $1$-regular simple left modules.

Similarly, let $B=\Lambda/I$, where $$ \Lambda = \begin{pmatrix} k&k&k[\epsilon]\\ 0 & k & k[\epsilon] \\ 0&0& k[\epsilon] \end{pmatrix} \quad\mbox{and}\quad I=\begin{pmatrix} 0&0&k[\epsilon]\\ 0 & 0&0 \\ 0&0&0 \end{pmatrix}.$$ Then the simple right module $S_1$ corresponding to the projective $P_1=\begin{pmatrix} k &k&*\end{pmatrix}$ is $2$-regular, while $\mathop{\rm pd}S_2=1$ and $\mathop{\rm pd}S_3=\infty$. Moreover, $B$ has no $2$-regular simple left modules.

  • $\begingroup$ It may be possible to create a counterexample of finite global dimension by replacing 𝑘[𝜖] by some algebra of finite global dimension for which the set $\{ m\mid m=\mathop{\rm pd}(S),\:S\:\mbox{is simple}\}$ is not an interval, is some smart way. This is only a guess, though. $\endgroup$
    – Erik D
    Feb 23, 2019 at 8:11
  • $\begingroup$ Thanks, I think I mostly checked my question for Nakayama algebras where I found no counterexample (but they are QF-3 algebras and their finitistic dimension should always be equal to the finitistic dimension of their opposite algebra, which makes them very special). I leave the question open in case someone is interested to find an example with finite global dimension. The finitistic dimension of your first algebra is equal to one (=projdim of the simple 1-regular). Do you know whether the finitistic dimension of $B$ is equal to two (=projdim of the simple 2-regular)? $\endgroup$
    – Mare
    Feb 23, 2019 at 11:04

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