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Homework Help: Irreducible polynomial in extension field

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]p\in\mathbb{Q}[x][/tex] be an irreducible polynomial. Suppose K is an extension field of [tex]\mathbb{Q}[/tex] that contains a root [tex]\alpha[/tex] of p such that [tex]p(\alpha^2)=0[/tex]. Prove that p splits in K[x].

    3. The attempt at a solution
    I was thinking contradiction, but if p does not split in K, the only logical conclusion I can come to is that there is an extension field L[x] such that p splits and [tex]K[x] \subseteq L[x][/tex].
  2. jcsd
  3. May 9, 2009 #2
    I have some thoughts but they are not complete. By some theorem, Q(alpha) is isomorphic to Q[x]/<p(x)>. Use this to show p(x^2)=p(x)q(x) (assume WLOG that p is monic if you want).

    Now show p(alpha^4)=0. In fact, alpha^{power of 2} is a root of p. All these powers of alpha can't be distinct. Obtain that alpha is a root of unity.

    Here is where my gut tells me you can conclude p is a cyclotomic polynomial, but I don't know enough about them to prove that myself.
  4. May 9, 2009 #3
    Nice work, Billy Bob! You're essentially done. Since f is irreducible in [tex] \mathbb{Q}[x] [/tex], f is the minimal polynomial of [tex] \alpha [/tex]. However, [tex] \alpha [/tex] is a root of unity, so the minimal polynomial is, by definition, some cyclotomic polynomial [tex] \Phi_k [/tex] for some k. Thus, [tex] f = \Phi_k [/tex] (without loss of generality, f is monic), so the roots of f are just [tex] \alpha, \alpha^2, \alpha^3, \ldots [/tex]. In particular, f splits in [tex] \mathbb{Q}(\alpha)[x] \subseteq K[x] [/tex].
  5. May 10, 2009 #4
    I see all of it, except the logic you use to conclude that [tex]\alpha[/tex] is a primitive root of unity. My thought is because p(x) is irreducible and [tex]\alpha, \alpha^2, \alpha^3, \ldots [/tex] are all roots of it, then [tex] \alpha =\alpha^2 =\alpha^3 =\ldots =1[/tex] implying that [tex]\alpha [/tex] is a root of unity.

    Any thoughts?
  6. May 10, 2009 #5
    If [tex] \alpha [/tex] is a primitive [tex] s [/tex]th root of unity, then [tex] \alpha = e^{2 \pi r/s} [/tex] for some [tex] r [/tex]. We can rewrite this as [tex] \alpha = e^{2 \pi t/n} [/tex], where [tex] \textrm{gcd}(t,n) = 1[/tex]. In other words, every [tex] s [/tex]th root of unity is a primitive [tex] n [/tex]th root of unity for some [tex] n [/tex], so its minimal polynomial is cyclotomic.
  7. May 10, 2009 #6


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    An easier way to carry out Billy Bob's plan:

    If E is the splitting field of p, then there is an element of Gal(E/Q) that, among other things, sends alpha to alpha^2. Where does it send alpha^2?
  8. May 10, 2009 #7
    [tex]\alpha, \alpha^2, \alpha^4, \alpha^8, \alpha^{16} \ldots [/tex] are all roots, but maybe not [tex]\alpha^3[/tex] etc. The point is, there can't be infinitely many roots, so [tex]\alpha^i=\alpha^j[/tex] for some i not equal to j. Now an example will give you an idea for how to handle the general case. If [tex]\alpha^8=\alpha^2[/tex], then [tex]\alpha^8-\alpha^2=0[/tex], then [tex]\alpha^6-1=0[/tex], so alpha is a sixth root of unity.

    Thanks VKint and Hurkyl for helping me out.
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