Given a ring $R$ with identity $1$, we can define the exterior algebra of order $k$ over $R$ to be the algebra over $R$, generated by elements $x\_1, \dots, x\_k$ satisfying $x\_i^2 = 0$ for each index $i$, and $x\_i x\_j = -x\_j x\_i$ for any distinct indices $i \neq j$.


My question is: **is there a natural analogue of the exterior algebra, generated by some elements whose cubes (instead of squares) equal $0$ (and if so, what is the name of this algebra)?**


In general, I'm interested in getting references to what is known about algebras $A$ over $R$, generated by elements $x\_1, \dots, x\_k$ satisfying $x\_i^3 = 0$ for each index $i$ (and possibly satisfying some additional relations related to the generators may or may not commute). I'm particularly interested in learning about the reprsentation theory of such algebras, or if anything is known about the computational complexity of arithmetic over these algebras.



What is the name for algebras generated by elements, all of whose cubes vanish?
This is a special case of the class of quantum complete intersections when you include the commutativity condition up to a sign, see for example <https://arxiv.org/pdf/0710.2606.pdf> . The representation theory of those algebras will be always wild (also without the commutativity conditions up to a sign the algebras will be wild as they contain the representation theory of the quotients.) and the indecomposable can not be classified when you have at least two variables in the case of cubes.
In characteristic two, those are the group algebras of tensor products of the cyclic group of order 3.


