Hi, How do I show that the polynomial f(x)=x^3-9 is irreducible mod 31. I know t

Question

Hi,

How do I show that the polynomial f(x)=x^3-9 is irreducible mod 31.

I know that a polynomial is irreducible if I can't write it as f(x)=g(x)h(x).

Should I write x^3-9=(x^2-a)(x-b) and show that it doesn't work for values of a,b... I'm lost. Please help.

Explanation / Answer

31 is a prime, it is a field, and so no zero divisors, i.e ab = 0 => a=0 or b=0. f is a deg 3 poly so if it can be factorised, it can be factorised into a poly of deg 2 and deg 3. x^3 - 9 = (x^2 + bx + c) (x + d) = x^3 + (b+d) x^2 + (bd+c)x+ cd so b+d = 0 (mod 31) and cd = -9 = 22 (mod 31) and bd + c = 0 (mod 31) b+d = 0 (mod 31) => d = -b (mod 31) -cb = -9 (mod 31) => cb = 9 (mod 31) c=-bd = b^2 (mod 31) => cb = 9 = b^2 . b = b^3 (mod 31) => b^3 = 9 (mod 31) but no b in Z_31 the above equation has a solution. So certainly, x^9 - 3 is irreducible over Z_31.

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