Understanding Ring Relationships and the Frobenius Automorphism

[VoleMeister]

I'm having trouble with what seemed like a trivial problem at first, but now I'm rather stuck. If R is a ring with xy=-yx for any x,y from the ring, xyz+xyz=0 must be true for any x,y,z from the ring. I'm trying to show why that is.

Letting y=x yields x^2+x^2=0. Thus then breaking it up into two cases: x=0 (in which case xyz+xyz=0 trivially), and if not - then I get x^2z+x^2z=0. But I'm not sure if I can do this (as this doesn't necessarily show that xyz+xyz=0 but just that x^2z+x^2z=0) just because we can say y=x in some cases it doesn't seem like we can generalize that to all cases. Am I missing something simple or is there some other way of going about it? Any ideas would be much appreciated.

Also, I'm trying to find some information on the small Frobenius (basically the proof about why the map from x -> x^c is an automorphism where c is the char(F) for finite field F). I haven't been able to find anything very succinct or understandable on this.

 

[d_leet]

You know that xy=-yx for every x and y in the ring, so what is zxy? Is there some way you can manipulate xyz into zxy?

 

[Hurkyl]

Have you tried plugging in really simple values for your variables?




For your second question, have you tried just applying the definition of homomorphism to see what you get?

 

[VoleMeister]

Hey guys, thanks for the tips. They really helped out. The one with xy and -yx was indeed really simple once I looked at it that way. Also the way the binomial expansion on the frobenius field problem canceled out was rather nifty.

 

[member 11137]

Hey guys, thanks for the tips. They really helped out. The one with xy and -yx was indeed really simple once I looked at it that way. Also the way the binomial expansion on the frobenius field problem canceled out was rather nifty.

When you are writing a relation like x.y + y. x = 0 on a ring I don't think that you are writing a trivial relation. Now if you can explain me the relation with frobenius, I shall be happy to learn something.