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JamesGoh
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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
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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