# [SOLVED] polynomials/ galois field question

by JamesGoh
Tags: field, galois, polynomials or, solved
 P: 135 Im reading through a section that deals with polynomials Galois fields and ran into something that Im 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 Im 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 Im still clueless as to how this proves that f(x) is irreducible. insight is appreciated regards James
 P: 135 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 ??
 HW Helper Sci Advisor P: 2,020 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.
P: 135

## [SOLVED] polynomials/ galois field question

 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.