# Question on $n$-regular modules

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 $$I_n$$.

Questions:

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 $$n$$?)

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

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.

• 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. Feb 23, 2019 at 8:11
• 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)?
– Mare
Feb 23, 2019 at 11:04