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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.

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    $\begingroup$ I would be very surprised if this was true; my guess would be that the classification is wild in sufficiently high dimension. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 17, 2014 at 5:41
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    $\begingroup$ This area is called algebraic geography. Using this term in your favourite search engine might help in finding the relevant articles (Flanigan, Le Bruyn--Reichstein) come to mind. $\endgroup$
    – pbelmans
    Commented Jun 17, 2014 at 8:26
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    $\begingroup$ My guess would be that this is true (but I could be wrong). It is known that every finite-dimensional $\mathbb{C}$-algebra is Morita equivalent to $\mathbb{C}Q/\langle \rho \rangle$, where $(Q, \rho)$ is a quiver with relations. Now the problem reduces to the question whether there are only finitely many quivers with relations giving algebras of a given dimension, and whether there are only finitely many algebras of a given dimension that are Morita equivalent to such an algebra. See, for instance, Richard Vale's lecture notes math.cornell.edu/~rvale/fdalgebras.pdf. $\endgroup$
    – Tom De Medts
    Commented Jun 17, 2014 at 9:25
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    $\begingroup$ @TomDeMedts: Well, the quiver of my example(s) is very simple, just one vertex and two loops $x$ and $y$, with the relations I gave. Given a dimension $d$, there are clearly only finitely many possible quivers, but the relations can involve arbitrary scalars, and in general changing the scalars will change the isomorphism class of the algebra. $\endgroup$
    – Jeremy Rickard
    Commented Jun 17, 2014 at 13:09
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    $\begingroup$ @JeremyRickard: Ah, that's the point -- thanks! Now the picture is complete. $\endgroup$
    – Tom De Medts
    Commented Jun 17, 2014 at 13:26

2 Answers 2

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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 $\{a,a^{-1}\}$.

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  • $\begingroup$ I don't understand the "quite easy to see" part. Could you elaborate? (The elements that square to zero are of the form $\lambda x + \mu xy$ or $\lambda y + \mu xy$, independent of $a$, unless $a=-1$.) $\endgroup$
    – Tom De Medts
    Commented Jun 17, 2014 at 10:17
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    $\begingroup$ You can recover $a^{\pm1}$ as the ratio between $wz$ and $zw$ whenever $w$ and $z$ are square-zero elements such that $wz\neq0$. $\endgroup$
    – Eric Wofsey
    Commented Jun 17, 2014 at 10:33
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    $\begingroup$ @TomDeMedts: I've edited my answer with an explanation. But yeah ... basically what Eric said. $\endgroup$
    – Jeremy Rickard
    Commented Jun 17, 2014 at 10:39
  • $\begingroup$ Yeah, I realized I made a big mistake. This is not true even when dimension is $3$, $B(a)$ without identity contains an infinitely family of algebras, i.e. the algebra with basis ${x,y,z}$ has multiplication table $x^2=y^2=z^2=xz=zx=yz=zy=0$ and $xy=z,yx=az$. $\endgroup$
    – user27976
    Commented Jun 17, 2014 at 16:40
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Even for commutative associative algebras it is not true. The article of Björn Poonen "Isomorphism types of commutative algebras of finite rank over an algebraically closed field" gives a classification in dimension $n\le 6$ for algebraically closed fields of arbitrary characteristic, where there are only finitely many isomorphism classes. Then there are examples given in dimension $7$ of infinitely many different algebras.

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