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Math Help - show that x^p - x + a is irreducible over GF(p) if a is different from 0

  1. #1
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    show that x^p - x + a is irreducible over GF(p) if a is different from 0

    Show that xp - x + a is irreducible over GF(p) if a is different from 0.

    I assumed for contradiction that xp - x + a is reducible. Then it must split into irreducible factors (none of them linear) over GF(p) or just Zp. Say one of the irreducible factors has degree m < p. Then the size of its splitting field must be pm. The splitting field of xp - x + a consists of all the roots of that polynomial. Let b be a root of that polynomial. Then since the multiplicative group of the splitting field is cyclic, all roots of that polynomial can be expressed as some power of b. So I need to find the minimum polynomial of b. If the minimum polynomial of b is just xp - x + a then I am done. So I assume the minimum polynomial of b is the irreducible factor with degree m < p that I mentioned above. Then the splitting field of the irreducible factor of degree m must be the same as the splitting field of xp - x + a since they both contain all the roots of xp - x + a.

    I am having trouble finding a contradiction in this argument. Can someone help me with this proof? Thanks.
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    Re: show that x^p - x + a is irreducible over GF(p) if a is different from 0

    suppose b is a root of x^p - x + a.

    let k be any element of \mathbb{Z}_p.

    then (b+k)^p - (b+k) + a = b^p + k^p - b - k + a = (b^p - b + a) + (k^p - k) = 0 + 0 = 0.

    we then have p distinct roots of x^p - x + a \in \mathbb{Z}_p(b), so this IS the splitting field of x^p - x + a.

    now consider \tau_k \in \text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b)) given by \tau_k(b) = b+k.

    the assignment \tau_k \to k is a group homomorphism \theta:\text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b)) \to \mathbb{Z}_p.

    since \text{ker}(\theta) = \text{id}_{\mathbb{Z}_p(b)}, we know that \text{Im}(\theta) = \{0\} or:

    \text{Im}(\theta) = \mathbb{Z}_p. the former implies that [\mathbb{Z}_p(b):\mathbb{Z}_p] = 1,

    that is, that b is in \mathbb{Z}_p. since a ≠ 0, this cannot be true, so [\mathbb{Z}_p(b):\mathbb{Z}_p] = p,

    which in turn means x^p - x + a is irreducible over \mathbb{Z}_p.
    Last edited by Deveno; March 4th 2012 at 08:32 AM.
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    Re: show that x^p - x + a is irreducible over GF(p) if a is different from 0

    why is  \text{Im}(\theta) = \{0\} or  \text{Im}(\theta) = \mathbb{Z}_p ? also, could you please explain your motivation for defining the group homomorphism from  \text{Aut}(\mathbb{Z}_p(b)) to  \mathbb{Z}_p ? i don't understand the reasoning for doing this.
    Last edited by oblixps; March 4th 2012 at 03:52 PM.
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    Re: show that x^p - x + a is irreducible over GF(p) if a is different from 0

    because \theta is a homomorphsim, so it's image is a subgroup of the additive group of \mathbb{Z}_p. but \mathbb{Z}_p has prime order, so...


    my motivation was to use the galois correspondence: [\mathbb{Z}_p(b):\mathbb{Z}_p] = |\text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b))|.
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    Re: show that x^p - x + a is irreducible over GF(p) if a is different from 0

    thanks for the reply. i understand that the image must be {0} or  \mathbb{Z}_p but what I am still trying to figure out is why does  \text{Im}(\theta) = \{0\} imply that  [\mathbb{Z}_p(b): \mathbb{Z}_p] = 1 ? I am having trouble seeing the connection between the degree of the field extension and the image of the map \theta .

    sorry i haven't learned galois theory yet so i was confused for a bit. but now that you posted the galois correspondence, it all makes sense now. thanks!
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