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Let $A$ be a symmetric algebra (finite dimensional and connected) and define $\psi_M:=sup \{ i \geq 1 | Ext^{i}(M,M) \neq 0 \}$ (infinite if this Ext is nonzero infinitely often) for an indecomposable non-projective module $M$. Call $M$ $i$-special in case $\psi_M=i$ is finite. Let $N:=A \oplus X$ with $X$ (having no non-projective summands) containing an $i$-special module as a summand. Then the algebra $B:=End_A(N)$ should contain modules with $Ext^i(D(B),L)=0$ for all $i >0$ (called CMCM-modules) that are not Gorenstein injective. Such CMCM-modules are pretty rare and were found only around 40 years after the definition of Gorenstein projective modules by Auslander and Bridger. I think such algebras $B$ are good candidates to challenge the homological conjectures.

Questions:

1.In a local algebra, can we have $\psi_M$ being an arbitrary finite number?

  1. What are example of modules with $\psi_M=i$ for any $i \geq 2$?

  2. I was able to construct such modules only over algebras over fields containing elements that are not roots of unity. Thus my construction does not work for example for finite fields. Are there modules with finite $\psi_M$ when in the relation of the algebra $A$ there are just the field elements +1 or -1 involved and can one construct such modules over algebras over any field?

  3. Do such algebras $B$ as above satisfy the finitistic dimension conjecture? The analog defined fintistic dominant dimension is infinite for those algebras.

If you have answers to 3. or 4., you would prove something new. Not sure about 1. and 2., have not seen literature on that.

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  • $\begingroup$ For you, what is a symmetric algebra? For a polynomial ring $A=k[x_1,\dots,x_n]$, which many of us consider to be a "symmetric algebra" on $kx_1\oplus \dots \oplus kx_n$, the module $M=A/\langle x_1,\dots,x_n \rangle$ is indecomposable, it is not projective, and $\psi_M$ equals $n$. This continues to be true if you localize $A$ at the maximal ideal $\langle x_1,\dots,x_n \rangle$. $\endgroup$ – Jason Starr Aug 26 '17 at 9:03
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    $\begingroup$ @JasonStarr Im sorry. I forgot to add that my algebras are finite dimensional. So for me symmetric means that it is a Frobenius algebra with symmetric bilinear form. In this situation the algebras always have infinite global dimension expect when they are semisimple. $\endgroup$ – Mare Aug 26 '17 at 9:05

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