Quotients of algebras

Question

Let $A$ be a finite dimensional algebra over an algebraically closed field $k$.

Can $A$ admit infinitely many non-isomorphic quotient algebras?

Answer 1

Yes. For instance, let $A=k[x,y]/(x,y)^5$. For any $c\in k\setminus\{0,1\}$, consider the algebra $B_c=A/(xy(x+y)(cx+y))$. Write $V_c=m/m^2$ where $m=(x,y)\subset B_c$ is the unique maximal ideal. We can identify $V_c$ with the set of homogeneous linear polynomials in $x$ and $y$, and thus identify its projectivization with $\mathbb{P}^1_k$. Now note that there is a unique set of four elements of $\mathbb{P}^1_k$ which, if lifted to elements of $V_c$ and then to elements of $B_c$, have a product of $0$, namely the elements corresponding to $x,y,x+y,$ and $cx+y$. Identifying $\mathbb{P}^1_k$ with $k\cup\{\infty\}$ in the usual way, these four points correspond to $\infty,0,1,$ and $c$.

Now, any isomorphism $B_c\to B_d$ would induce a linear isomorphism $V_c\to V_d$; projectivizing this linear isomorphism would give an isomorphism $\mathbb{P}^1_k\to\mathbb{P}^1_k$ which maps $\infty,0,1,$ and $c$ to $\infty,0,1,$ and $d$ in some order. Taking cross-ratios, this means $d$ must be the cross-ratio of $\infty,0,1,$ and $c$ in some order. There are only finitely many such orders, and thus only finitely many values of $d$ such that $B_c$ and $B_d$ are isomorphic. Since there are infinitely many different values $c$ can take, this means the quotients $B_c$ of $A$ must have infinitely many different isomorphism types.