Polynomials/ galois field question

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[SOLVED] polynomials/ galois field question

Im reading through a section that deals with polynomials Galois fields and ran into something that I am not quite understanding.

Say we have an irreducible polynomial, f(x), which has coefficients from GF(2) and roots
[tex]\beta[/tex], [tex]\beta[/tex][tex]^{2}[/tex], [tex]\beta[/tex][tex]^{4}[/tex], [tex]\beta[/tex][tex]^{8}[/tex], ...[tex]\beta[/tex][tex]^{2}[/tex][tex]^{e}[/tex][tex]-1[/tex] where e is the smallest integer such that [tex]\beta[/tex][tex]^{2}[/tex][tex]^{e}[/tex] = [tex]\beta[/tex]

given by

f(x) = [tex]\prod[/tex][tex]^{e-1}_{i=0}[/tex] ( X + [tex]\beta[/tex][tex]^{2^i}[/tex])

Note: Beta term is Beta^(2^i)

In the section I am reading, they do a test to prove f(x) is irreducible. I will state the test below

Say f(x) = a(x).b(x) where a(x) and b(x) are polynomials with coefficients from GF(2)

if we sub one of the roots of f(x) in, say [tex]\beta[/tex], f([tex]\beta[/tex]) = 0 which means that either a([tex]\beta[/tex]) = 0 or b([tex]\beta[/tex]) = 0, hence a(x) = f(x) or b(x) = f(x). This understanding also says that a(x) or b(x) (depending which one had [tex]\beta[/tex] subbed into it) has all the roots of f(x) (A theory in my textbook says that if f([tex]\beta[/tex]) = 0, f([tex]\beta[/tex][tex]^{2^i}[/tex])=0 for any i)

I get how they arrive at their result, however I am still clueless as to how this proves that
f(x) is irreducible.

insight is appreciated

regards
James
 
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Ok I read over the notes again and think I may have the answer

Since f(x) = a(x) or f(x) = b(x) when the root is substituted in, it cannot be divided into a smaller polynomial with a non-zero degree. Therefore f(x) must be irreducible.

Thoughts, comments, insights ??
 
I'm confused. You start out by saying that f(x) is irreducible, presumably over GF(2). Then you go on to prove that f(x) is irreducible - this time over what?

Also, why is this in the Set Theory, Logic, Probability, Statistics forum? It really should be in the algebra forum.
 
I'm confused. You start out by saying that f(x) is irreducible, presumably over GF(2). Then you go on to prove that f(x) is irreducible - this time over what?

I answered my own question (see f(x)=a(x).b(x) proof), it was just that I didn't read over the notes properly.

Sorry about the confusion

Also, someone pls move this topic to the correct forum. ta
 
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