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Math Help - Polynomial irreducible over an extension

  1. #1
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    Polynomial irreducible over an extension

    Can someone verify whether or not this proof works for me? Thanks.

    Suppose that p(x) \in F[x] and E is a finite extension of F. If p(x) is irreducible over F and \mbox{deg}(p(x)) and [E:F] are relatively prime, show that p(x) is irreducible over E.

    Suppose p(x) is reducible over E. Then p(x) has a root in E, call it a (this part feels true but I can't verify it nicely...basically if it's reducible but not by splitting out x-a, the coefficients will be from F, but then it would be reducible over F). Take the subextension F(a). Then since F(a) \cong F[x]/\langle p(x) \rangle we have [F(a):F]=\mbox{deg}(p(x)). Then [E:F]=[E:F(a)][F(a):F] but this implies that [F(a):F] divides [E:F] which it does not since [F(a):F]=\mbox{deg}(p(x)) and \mbox{deg}(p(x)) and [E:F] are relatively prime.
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  2. #2
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    Re: Polynomial irreducible over an extension

    Quote Originally Posted by AlexP View Post
    Can someone verify whether or not this proof works for me? Thanks.

    Suppose that p(x) \in F[x] and E is a finite extension of F. If p(x) is irreducible over F and \mbox{deg}(p(x)) and [E:F] are relatively prime, show that p(x) is irreducible over E.

    Suppose p(x) is reducible over E. Then p(x) has a root in E, call it a (this part feels true but I can't verify it nicely...basically if it's reducible but not by splitting out x-a, the coefficients will be from F, but then it would be reducible over F). Take the subextension F(a). Then since F(a) \cong F[x]/\langle p(x) \rangle we have [F(a):F]=\mbox{deg}(p(x)). Then [E:F]=[E:F(a)][F(a):F] but this implies that [F(a):F] divides [E:F] which it does not since [F(a):F]=\mbox{deg}(p(x)) and \mbox{deg}(p(x)) and [E:F] are relatively prime.
    no, your proof is not complete because p(x) may not have any root in E. you need to pick a root a of p(x) in some extension of F and let q(x) be the minimal polynomial of a over E. clearly

    \deg q(x) \leq \deg p(x).  \ \ \ \ \ \ \ (1)

    we also have

     [E(a):E)][E:F]=[E(a):F(a)][F(a):F]. \ \ \ \ \ \  (2)

    from (2) and the problem hypothesis it follows that \deg p(x) \mid \deg q(x) and so by (1)

    \deg p(x)=\deg q(x). \ \ \ \ \ \ (3)

    let r(x)=q(x)-p(x) \in E[x]. then \deg r(x) < \deg q(x) by (3). we also have r(a)=0 and hence r(x)=0 by the minimality of q(x). hence q(x)=p(x) and we are done.
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  3. #3
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    Re: Polynomial irreducible over an extension

    Thanks. If you don't mind, could you explain your approach/way of thinking about the problem? Often when I don't succeed in finding a proof and see it from someone else, I know I never would have thought of it. I'm really curious about your method.
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  4. #4
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    Re: Polynomial irreducible over an extension

    Quote Originally Posted by AlexP View Post
    Thanks. If you don't mind, could you explain your approach/way of thinking about the problem? Often when I don't succeed in finding a proof and see it from someone else, I know I never would have thought of it. I'm really curious about your method.
    here is how i found the solution:

    i said p(x) is irreducible over E iff p(x) = q(x). so i need to show that  r(x) = q(x) - p(x) = 0. so, considering the facts that r(x), q(x) are in E[x] and r(a)=q(a)=0, i only need to show that \deg r(x) < \deg q(x). but both p(x) and q(x) are monic and so i need to show that \deg p(x)= \deg q(x) in order to prove \deg r(x) < \deg q(x). finally, the identity (2) is just a natural way of relating \deg p(x)=[F(a):F] and \deg q(x)=[E(a):E].
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  5. #5
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    Re: Polynomial irreducible over an extension

    Sorry this reply is late.

    Is p(x) monic because otherwise it would be reducible (every nonzero element in a field is a unit)?

    Thank you for your help.
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  6. #6
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    Re: Polynomial irreducible over an extension

    Quote Originally Posted by AlexP View Post
    Sorry this reply is late.

    Is p(x) monic because otherwise it would be reducible (every nonzero element in a field is a unit)?

    Thank you for your help.
    no, multiplying a polynomial by a non-zero scalar doesn't change its reducibility or irreducibility. so from the beginning you may assume that p(x) is monic. another way is to let r(x)=cq(x)-p(x), where c is the leading coefficient of p(x).
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