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I think this is obvious and I am just missing something but:

Given some finitely presentable algebra, can we always reduce the relations to commutator relations?

And moreover, if yes, then is there a reasonable algorithm?

As I said, I feel like I should know this, so I apologize if it is obvious.


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What exactly is a commutator relation? – Qiaochu Yuan Sep 20 '10 at 23:14
You're probably going to want to put severe restrictions on the type of algebra you're interested in. For example, the relations among the monomials in a group ring are just the relations from the underlying group. – Ryan Budney Sep 20 '10 at 23:36
up vote 4 down vote accepted

(Apologies for putting this in the form of an answer - I haven't got enough rep yet to make it a comment)

What about elements of finite order? It seems like even a simple algebra such as the one generated by $a^m = b^n = e$ couldn't possibly have a commutator presentation, unless I'm missing something immensely obvious...

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sorry for the dumb question. – BBischof Sep 21 '10 at 20:20

No, you cannot even do it in groups. For if you could, consider the identity $x^n=e$ with $n>1$. If this were equivalent to a set of commutator words, then every abelian group would satisfy the relation, but of course this is false: only those abelian groups that are of exponent $n$ satisfy it.

More generally, in groups every identity is equivalent to a pair of identities, one a commutator word and one a power word; equivalently, if $w$ is a word in the free group $F$ on countably many generators $x_1,\ldots,x_n,\ldots$, then there exist words $w_1$ and $w_2$, with $w_1=x_1^a$ for some positive integer $a$, and $w_2\in [F,F]$, such that for all groups $G$, $G$ satisfies $w$ as an identity (every evaluation of $w$ in $G$ is trivial) if and only if $G$ satisfies $w_1$ and $w_2$ as identities. This is due to Bernhard Neumann; write $w=x_1^{n_1}\cdots x_m^{n_m}c$, with $c$ a commutator word, by using the collection process. Then let $a=\gcd(n_1,\ldots,n_m)$, and $w_2=c$.

So at least in groups you also need "power identities".

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