Given an algebraically closed field $F$, for any positive integer $n$, are there always only finitely many non-isomorphic (noncommutative) associative algebras (possibly without identity) with dimension $n$ over $F$?


This questions is motivated by the classification of low dimensional algebras. It seems that at least when $n$ is less than 6, the answer is yes. I'm also guessing that the number of non-isomorphic classes doesn't depend on the choice of algebraically closed fileds--I've convinced myself this is true for low dimensional cases.


So far I have two ideas: 1. To compute the dimension of the variety of associative algebras of dimension n, and then consider the $GL\_n(F)$ action on the variety; 2. Every algebra of dimension $n$ can be embedded as a subalgebra of $M\_{n+1}(F)$. But 1. is also a difficult problem for me and I don't know how to use 2.



Are there only finitely many associative algebras of fixed dimension?
Even for $4$-dimensional algebras with identity it's not true.


For $a\in F$ let $B(a)=F\langle x,y|x^2=y^2=0,xy=ayx\rangle$. Then $B(a)\not\cong B(b)$ unless $a=b$ or $a=b^{-1}$. This is quite easy to see by considering which elements of $B(a)$ square to zero:


If $z=\lambda\_11+\lambda\_xx+\lambda\_yy+\lambda\_{yx}yx$ with $z^2=0$, then clearly $\lambda\_1=0$. So $z^2=\lambda\_x\lambda\_y(a+1)yx$, which is only zero if $\lambda\_x=0$ or $\lambda\_y=0$ (unless $a=-1$, which characterizes $B(-1)$ as the only $B(a)$ with a $3$-dimensional space of square zero elements, so let's assume $a\neq-1$).


So, modulo the ideal $(yx)$, the only square zero elements are scalar multiples of $x$ and $y$, and any two such elements $z$ and $z'$ that generate $B(a)$ satisfy $zz'=az'z$ or $z'z=azz'$. So the isomorphism type of $B(a)$ determines $\{a,a^{-1}\}$.


