Let $A$ be a noncommutative Koszul algebra (see here for a definition of Koszul) and let $c \in A$ be a central element. Will the quotient of $A$ by the ideal generated by $c$ again be Koszul. If not what is a counter example and what else could I require to ensure Koszulity?
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asked Dec 16, 2022 at 16:02
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$\begingroup$ Did you look at the book Quadratic algebras? The answer in general is no because Koszul algebras are quadratic, but I recall the book contains some results in the positive direction. $\endgroup$– PedroCommented Dec 16, 2022 at 16:35
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$\begingroup$ Is it easy to see that Koszul algebras are quadratic? $\endgroup$– Didier de MontblazonCommented Dec 16, 2022 at 16:45
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$\begingroup$ Yes. This is probably in the book I mentioned. You compute Tor2. Tor1 informs on the generators and Tor2 on the relations. The diagonal condition gives the algebra is generated in degree 1 (Tor1) with relations in degree 2 (Tor2). $\endgroup$– PedroCommented Dec 16, 2022 at 16:48
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3$\begingroup$ Require $c$ to be a homogeneous element of degree $1$ and a nonzero-divisor. This is certainly enough. $\endgroup$– Leonid PositselskiCommented Dec 16, 2022 at 20:48
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2$\begingroup$ @DidierdeMontblazon What's your definition of a Koszul algebra, and aren't they supposed to be graded, by definition? What is the grading on $A/(c-1)$, if c is a homogeneous element of degree $2$ ? $\endgroup$– Leonid PositselskiCommented Dec 18, 2022 at 2:00
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2 Answers
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With no assumptions, obviously the answer is no. You didn't even require that c is homogeneous.
If $c$ has any components of degree $>2$, then I think the answer is that the quotient is never Koszul: Koszul algebras are quadratic.
If $c$ has degree 0,1 or 2, then certainly there are cases where $A/(c)$ will be Koszul:
- $A=K\langle x,y\rangle$ the free algebra on 2 variables over any field $K$ is Koszul, and $c= xy-yx$ gives $A/c=K[x,y]$ which is again Koszul.
- If $A$ is free or polynomial, the quotient by any degree 1 element will be free or polynomial on one fewer generators.
- If you have a Koszul quotient of the path algebra of a quiver, you can sometimes kill a vertex and get something Koszul. For example, if $x,y$ are the edges of an oriented 2-cycle, the quotient $xy=0$ is 5-dimensional and Koszul, and killing either vertex gives the base field.
I suspect that this is not "typical" behavior (for example, this paper shows that there are quadratic quotients of polynomial rings which are not Koszul: https://arxiv.org/pdf/1605.09145.pdf), but don't have the time/energy to come up with a bunch of counter-examples right now.
answered Dec 16, 2022 at 16:19
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No, take for instance $A = k[x]$ and $c = x^3$. In my opinion, the point is that Koszulity and commutativity/centrality are not related to one other. So I doubt there is any good statement along the lines of what you are looking for.
