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Math Help - irreducible polynomial

  1. #1
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    irreducible polynomial

    Prove that for every prime p, the polynomial x^p-x-1 is irreducible over Z_p

    Don't even know where to start, can you please help me?
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    Prove that for every prime p, the polynomial x^p-x-1 is irreducible over Z_p

    Don't even know where to start, can you please help me?
    We will prove something stronger that if a\in \mathbb{Z}_p^{\times} then f(x) = x^p - x - a is irreducible over \mathbb{Z}_p. Notice that if \alpha (in some larger field) is a root of f(x) then \alpha+1 is a root. This is because (\alpha + 1)^p - (\alpha + 1) - a = \alpha^p + 1 - \alpha - 1 - a = 0. Consequently, this means \alpha,\alpha+1,...,\alpha + (p-1) are the roots of f(x) (since degree is p). Therefore, \mathbb{Z}_p(\alpha) is a splitting field over f(x) of \mathbb{Z}_p. Now say that f(x) = g_1(x)...g_m(x) where g_i(x) are irreducible over \mathbb{Z}_p. Let \beta_i be any root of g_i(x) and \beta_j be any root of g_j(x). Remember by above \mathbb{Z}_p(\beta_i) and \mathbb{Z}_p(\beta_j) are splitting fields over f(x) of \mathbb{Z}_p. Thus, \mathbb{Z}_p(\beta_i) = \mathbb{Z}_p(\beta_j). This means, obviously, that [\mathbb{Z}_p(\beta_i) : \mathbb{Z}_p] = [\mathbb{Z}_p(\beta_j):\mathbb{Z}_p] \implies \deg (\beta_i, \mathbb{Z}_p) = \deg (\beta_j, \mathbb{Z}_p) and so \deg g_i(x) = \deg g_j(x). In other words, all polynomials in the factorization of f(x) have the same degree. But \deg f(x) = p a prime! Thus, either all g_i(x) are linear or else f(x) is its own irreducible polynomial factorization. It cannot be that all g_i(x) are linear for that would imply that the roots of f(x) lie in \mathbb{Z}_p which is impossible because x_0^p - x_0 - a = -a\not = 0 for all x_0 \in \mathbb{Z}_p by Fermat's theorem. Thus, f(x) is irreducible.
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  3. #3
    Moo
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    Hello,

    I'd like to know if we could use this property :
    \text{In } \mathbb{Z}/p \mathbb{Z},~ \forall x,~ x^p-x= \prod_{k=0}^{n-1} (x-k)
    I cannot get the relationship yet, so if someone can enlighten me... Thanks :P
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  4. #4
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    Quote Originally Posted by Moo View Post
    I cannot get the relationship yet, so if someone can enlighten me... Thanks :P
    Sure

    A non-constant polynomial f(x) \in F[x] (where F is a field, like \mathbb{Z}_p,\mathbb{Q},\mathbb{R}) is irreducible iff we cannot write f(x) = g(x)h(x) where \deg g(x),\deg h(x) < \deg f(x) i.e. factorize f(x) into lower degree polynomials.

    If f(x) is irreducible then f(x) has no zeros in F. But the converse is not true. Consider f(x) = x^4 + 2x^2 + 1. Certainly f(x) has no zeros in \mathbb{Q}. However, f(x) = (x^2+1)(x^2+1). Thus, f(x) is reducible.

    Does that make sense?
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