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Math Help - Extension fields / splitting fields proof...

  1. #1
    ElGamal
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    Extension fields / splitting fields proof...

    Hello, put this in the homework help but think its probably best off here, Im trying to prove the following by induction....

    Given a polynomial, f(X) \in F[X], of degree n, there exists
    an extension field K (sub set of) F such that f(X) has n roots
    in K.

    This is what I did first time round...

    First assume that F is irreducible and argue by induction on n.
    Suppose the result holds for irreducible polynomials of degree at
    most n-1. Set E= F[x]/f(x).
    Now, in E, f(x) has at least one root, so we can write f(x)
    = (x - \alpha_{1})(x - \alpha_{2})...(x-\alpha_{r})g(x) with
    \alpha_{1}, \alpha_{2}, ..., \alpha_{r} \in E and g(x) \in
    E[x] irreducible. Since deg g < n, by the inductive assumption,
    the result also applies to g.

    But ive been told this is wrong because there could be more than one irreducible factor of degree > 1, also I shouldnt be trying to take irreduciblity through the proof.

    If someone knows how to prove this it would be much appreciated, its part of a huge project and needs to be done by tomorrow!

    Cheers!

    ElGamal.
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  2. #2
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    Quote Originally Posted by ElGamal View Post
    Hello, put this in the homework help but think its probably best off here, Im trying to prove the following by induction....

    Given a polynomial, f(X) \in F[X], of degree n, there exists
    an extension field K (sub set of) F such that f(X) has n roots
    in K.
    You need to use something call Kronecker's theorem. It is not hard, so we will prove it. Let F be a field and f(x) in F[x] be a non-constant polynomial. Then there exists an extension field K over F and \alpha \in K such that f(\alpha) = 0. By unique factorization we can write f(x) as a product of irreducible polynomials. Thus it is necessary an sufficient that only of those factors p(x) has a zero \alpha \in K. Since p(x) is a non-constant polynomial which is irreducible over F it means F / \left< p(x) \right> is a field. Let us call this field K. We can identify that K contains F (up to isomorphism) by \phi (a) = [a]_{p(x)} (where [a]_{p(x)} is the equivalence class mod p(x)) for a\in F. Next we note that [x]_{p(x)} is a zero of p(x)= a_nx^n+...+a_10+a_0 for p([x]_{p(x)}) = [a_nx^n+...+a_1x+a_0]_{p(x)} = [0]_{p(x)}. Thus K is an extension field that we desired. Now to go into induction you have to note that if f(x) is an arbitrary non-constant polynomial then we factorize it p_1(x)...p_k(x) were all p_i(x) are irreducible. Now apply the argument on every irreducible factor k times. Thus, there exists an extension field K such that f(x) splits (meaning factors into linear factors).

    You can make this result a little stronger: Given field F there is extension field K for a non-constant polynomial such such [K:F]<=n! where n = deg f(x) and f(x) splits over K.
    Last edited by ThePerfectHacker; December 19th 2007 at 09:06 AM.
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