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Why is x^p - a irreducible over a field of characteristic p?

  1. May 11, 2012 #1
    If [itex]K[/itex] is a field of characteristic [itex]p[/itex], and there exists an element [itex]a \in K[/itex]
    which is not a [itex]p[/itex]th power (i.e. the Frobenius endomorphism is not
    surjective), then I am told we can show [itex]x^p - a[/itex] is an irreducible polynomial
    (and since it is not separable our field is imperfect). I see that
    [itex]x^p - a[/itex] has no roots in [itex]K[/itex], but how do we know that there does not exist
    any factorization of [itex]x^p -a[/itex] into factors of lesser degree?
  2. jcsd
  3. May 11, 2012 #2
    Hint: Let [itex]f(x) = x^p - a[/itex], let F be a splitting field for f and let [itex]\alpha[/itex] be a root of f(x) in F. Can you take it from there?
    Last edited: May 11, 2012
  4. May 11, 2012 #3
    OK, how about this? Over [itex]F[/itex] we have [itex]\alpha[/itex] as the only root of [itex]f(x)[/itex] (with multiplicity [itex]p[/itex]). Let [itex]f(x)=p_1(x)\dotsb p_n(x)[/itex] be a factorization of [itex]f(x)[/itex] in [itex]K[x][/itex] into monic irreducibles. Then each of these must be the minimal polynomial of [itex]\alpha[/itex] over [itex]K[/itex]. So they all must have the same degree. So the degree of [itex]p_1(x)=\dotsb=p_n(x)[/itex] divides [itex]p[/itex], so it is either 1 or [itex]p[/itex]. But it cannot be 1, so it must be [itex]p[/itex], so [itex]f(x)[/itex] itself is irreducible.

    Is that what you would suggest? I don't see an easier way to show [itex](x-\alpha)^m\notin K[x][/itex] for all positive integers [itex]m<p[/itex].
  5. May 11, 2012 #4
    Your argument looks OK. Here's what I had in mind: We have that [itex]f(x) = (x - \alpha)^p = x^p - \alpha^p[/itex] and so [itex]a = \alpha^p[/itex]. Suppose that f(x) = g(x)h(x) is a proper, non-trivial factorization of f(x) over K. Then, by unique factorization, [itex]g(x) = (x - \alpha)^s[/itex], for some 0 < s < p. The constant term of g(x) is [itex]\alpha^s[/itex], so [itex]\alpha^s \in K[/itex]. Since gcd(s, p) = 1, there exist integers m and n such that sm + pn = 1. Therefore, [itex]\alpha = \alpha^{sm+pn} = (\alpha^s)^m (\alpha^p)^n[/itex] is an element of K. But [itex]a = \alpha^p[/itex], contradicting the assumption that a is not a pth power in K. Therefore, no such factorization of f(x) exists.
  6. May 11, 2012 #5
    I like your way, very elegant. Thanks for the help!
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