# Example of algebras over GF(2)

by Lie
Tags: algebra, algebra help
 P: 15 Anyone know of an example of an algebra over the field $$\mathbb{Z}_2$$ with the following properties? 1. commutative; 2. associative; 3. $$x^3 = 0$$, for all x; and 4. Exists x and y such that $$x^2y \neq 0$$. Grateful!
 Mentor P: 18,240 What about $$\mathbb{Z}_2[X]/(X^3)$$? It satisfies your first three properties, and alse the last one with y=1 and x=X...
 P: 15 micromass, Note that condition 3 implies that the algebra can not have unity. Therefore $$\mathbb{Z}_2[X]/(X^3)$$ is not an example.
 Mentor P: 18,240 Example of algebras over GF(2) Oh sorry, I forgot to read "for all x" Well, I'll look for another example...
 Emeritus Sci Advisor PF Gold P: 16,091 What goes wrong with the direct way to approach the problem? (e.g. like micromass's, except working with algebras rather than rings)
 P: 429 Ermm, can't you just take all polynomials over two variables x and y modulo the relation x^3=y^3=0 ?
P: 15
 Quote by Jamma Ermm, can't you just take all polynomials over two variables x and y modulo the relation x^3=y^3=0 ?
Jamma, same remark:
 Quote by Lie micromass, Note that condition 3 implies that the algebra can not have unity.[...]
 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]
 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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