Let $A$ be a noncommutative Koszul algebra (see [here](https://en.wikipedia.org/wiki/Koszul_algebra) 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?



Quotients of Koszul algebras
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.


