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Math Help - Roots and Irreducible polys

  1. #1
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    Roots and Irreducible polys

    Any ideas?
    f(x) belongs to F[x] and alpha is root of f(x) is some field extension of F. Show irreducible poly. g(x) belonging to F[x] has property that h(x) belonging to F[x] also has alpha as a root then g(x)/h(x) in F[x].
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  2. #2
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    Let F be a field and E be its extension field. Let \alpha \in E be algebraic over F (how do we know it is algebraic?). Next define the evaluation homomorphism:
    \phi_{\alpha} : F[x]\mapsto E. We know that the \ker \phi_{\alpha} = \{p(x) \in F[x]| p(\alpha) = 0\} is a principal ideal. So \ker \phi_{\alpha} = \left< q(x) \right> for some q(x) \in F[x]. This polynomial q(x) must be of minimial degree because otherwise if q_1(x) is smaller degree then we require that q_1(x) = q(x)\cdot q_2(x) for some q_2(x) \in F[x] because the ideal is principal. But that is a contradiction because \mbox{deg} q_1 (x) < \mbox{ deg} q(x)q_2(x). Thus, q(x) is of minimal degree. Next we claim that q(x) is irreducible because if not q(x) = r(x)s(x) and so one of the polynomial has the property that (WLOG) r(\alpha)=0 but that cannot be because r(x) has a smaller degree.

    So we see that all the polynomial having this zero, i.e. \ker \phi_{\alpha} are generated by an irreducible polynomial q(x). And so if g(x)\in \ker \phi_{\alpha} then g(x) = q(x)\cdot q_3(x) for some q_3(x). This shows that q(x) divides g(x).
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  3. #3
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    Kernel/Ring homomorphisms

    In our abstract class we have not yet been taught about kernels and ring homomorphisms so this has to be proved without using these. Thanks
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  4. #4
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    Quote Originally Posted by PacManisAlive View Post
    In our abstract class we have not yet been taught about kernels and ring homomorphisms so this has to be proved without using these. Thanks
    1)Okay define N= \{ f(x) \in F[x] | f(\alpha) = 0\}.
    2)Convince yourself that N\triangleleft F[x].
    3)Since F[x] is a PID it follows N=\left< q(x) \right> for some q(x)\in F[x].
    4)Use a similar argument as above.

    Note, this is basically the same approach except kernels eliminate the need to prove #1 which is a little lengthy but straightforward checking of the field axioms.
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