Associator-Commutator Identity

Question

I'm currently working on a hw problem that I've been stuck on for far longer than I should be, and I'm starting to wonder if the problem was written wrong in the first place. I would just like someone to verify whether the identity that I am proving is correct or not so I know if I'm wasting my time or there's just something I'm not seeing.

The identity is as follows:

$a[b,c]-[ab,c]+[a,b]c=[a,c,b]-[a,b,c]-[c,a,b]$

where $[a,b]:=ab-ba$ and $[a,b,c]:=(ab)c-a(bc)$

a,b,c are elements of an arbitrary binary ring.

Thanks in advance

It is clearly written wrong. While the ring might not be commutative, multiplication is certainly associative, which would give that the right hand side is always zero. — rnrstopstraffic, Mar 08 '15 at 05:40
I don' think we can assume that multiplication is associative otherwise it would be an associative binary ring. — AnotherPerson, Mar 08 '15 at 06:11
(R,+,) where (R,+) is an abelian group and distributes over +. — AnotherPerson, Mar 08 '15 at 06:37
What textbook is this from? Traditionally, a ring is understood to be associative unless it is specifically stated to be nonassociative. — rnrstopstraffic, Mar 08 '15 at 06:41
https://math.berkeley.edu/~wodzicki/H113.S15/H113-20150305.pdf — AnotherPerson, Mar 08 '15 at 07:36
In the context, you are correct. Those notes indicate that a ring does not a priori have associativity. In that case, @David Wheeler has a good explanation. — rnrstopstraffic, Mar 08 '15 at 20:54

score 1 · Accepted Answer · answered Mar 08 '15 at 08:54

1

I think it may be written wrong. This is what I get:

$[a,c,b] - [a,b,c] - [c,a,b] = (ac)b - a(cb) - (ab)c + a(bc) - (ca)b + c(ab)$

$= a(bc) - a(cb) + c(ab) - (ab)c + (ac)b - (ca)b$

$= a[b,c] + [c,ab] + [a,c]b = a[b,c] - [ab,c] + [a,c]b$

So I believe the $b$ and $c$ in the third term of the LHS of your identity have been juxtaposed in error.

answered Mar 08 '15 at 08:54

David Wheeler

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That's the exact conclusion I came to, thanks for the confirmation. – AnotherPerson Mar 09 '15 at 00:07

Associator-Commutator Identity

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