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

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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 $\displaystyle x^p - x + a$.

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

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

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

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

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

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

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

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

    which in turn means $\displaystyle x^p - x + a$ is irreducible over $\displaystyle \mathbb{Z}_p$.
    Last edited by Deveno; Mar 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 $\displaystyle \text{Im}(\theta) = \{0\} $ or $\displaystyle \text{Im}(\theta) = \mathbb{Z}_p $? also, could you please explain your motivation for defining the group homomorphism from $\displaystyle \text{Aut}(\mathbb{Z}_p(b)) $ to $\displaystyle \mathbb{Z}_p $? i don't understand the reasoning for doing this.
    Last edited by oblixps; Mar 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 $\displaystyle \theta$ is a homomorphsim, so it's image is a subgroup of the additive group of $\displaystyle \mathbb{Z}_p$. but $\displaystyle \mathbb{Z}_p$ has prime order, so...


    my motivation was to use the galois correspondence: $\displaystyle [\mathbb{Z}_p(b):\mathbb{Z}_p] = |\text{Aut}_{\mathbb{Z}_p}(\mathbb{Z}_p(b))|$.
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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 $\displaystyle \mathbb{Z}_p $ but what I am still trying to figure out is why does $\displaystyle \text{Im}(\theta) = \{0\} $ imply that $\displaystyle [\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 $\displaystyle \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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