15
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ 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
    Jun 17, 2014 at 5:41
  • 6
    $\begingroup$ 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
    Jun 17, 2014 at 8:26
  • 1
    $\begingroup$ 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
    Jun 17, 2014 at 9:25
  • 4
    $\begingroup$ @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
    Jun 17, 2014 at 13:09
  • 1
    $\begingroup$ @JeremyRickard: Ah, that's the point -- thanks! Now the picture is complete. $\endgroup$
    – Tom De Medts
    Jun 17, 2014 at 13:26

2 Answers 2

Reset to default
30
$\begingroup$

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

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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
    Jun 17, 2014 at 10:17
  • 3
    $\begingroup$ 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
    Jun 17, 2014 at 10:33
  • 1
    $\begingroup$ @TomDeMedts: I've edited my answer with an explanation. But yeah ... basically what Eric said. $\endgroup$
    – Jeremy Rickard
    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
    Jun 17, 2014 at 16:40
20
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.