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
\beta, \beta^{2}, \beta^{4}, \beta^{8}, ...\beta^{2}^{e}-1 where e is the smallest integer such that \beta^{2}^{e} = \beta
given by
f(x) = \prod^{e-1}_{i=0} ( X + \beta^{2^i})
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 \beta, f(\beta) = 0 which means that either a(\beta) = 0 or b(\beta) = 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 \beta subbed into it) has all the roots of f(x) (A theory in my textbook says that if f(\beta) = 0, f(\beta^{2^i})=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
\beta, \beta^{2}, \beta^{4}, \beta^{8}, ...\beta^{2}^{e}-1 where e is the smallest integer such that \beta^{2}^{e} = \beta
given by
f(x) = \prod^{e-1}_{i=0} ( X + \beta^{2^i})
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 \beta, f(\beta) = 0 which means that either a(\beta) = 0 or b(\beta) = 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 \beta subbed into it) has all the roots of f(x) (A theory in my textbook says that if f(\beta) = 0, f(\beta^{2^i})=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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