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Example of algebras over GF(2) 
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#1
May811, 08:52 AM

P: 15

Anyone know of an example of an algebra over the field [tex]\mathbb{Z}_2[/tex] with the following properties?
1. commutative; 2. associative; 3. [tex] x^3 = 0 [/tex], for all x; and 4. Exists x and y such that [tex] x^2y \neq 0 [/tex]. Grateful! 


#2
May811, 09:21 AM

Mentor
P: 18,086

What about [tex]\mathbb{Z}_2[X]/(X^3)[/tex]? It satisfies your first three properties, and alse the last one with y=1 and x=X...



#3
May811, 10:55 AM

P: 15

micromass,
Note that condition 3 implies that the algebra can not have unity. Therefore [tex] \mathbb{Z}_2[X]/(X^3) [/tex] is not an example. 


#4
May811, 11:10 AM

Mentor
P: 18,086

Example of algebras over GF(2)
Oh sorry, I forgot to read "for all x" Well, I'll look for another example...



#5
May811, 04:00 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

What goes wrong with the direct way to approach the problem? (e.g. like micromass's, except working with algebras rather than rings)



#6
May811, 06:01 PM

P: 429

Ermm, can't you just take all polynomials over two variables x and y modulo the relation x^3=y^3=0 ?



#7
May911, 12:07 PM

P: 15




#8
May911, 02:12 PM

P: 429

Sorry, I didn't mean it like that, I should describe my algebra a bit better.
Take as our set of elements {0,x,x^2,x^3,y,y^2,y^3} and all multiples and linear combinations of them with the obvious rules of addition and multiplication subject to the condition that x^3=y^3=0. There is no unity here. [Edit:ignore me, this algebra has elements in it which don't cube to zero] 


#9
May911, 02:22 PM

P: 429

Ok, how about my algebra up there but with the relation (x)(y^2)=(x^2)(y).
It seems at a first glance that this works. 


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