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Math Help - Splitting fields and irreducible polynomials

  1. #1
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    Splitting fields and irreducible polynomials

    Let f(x) be an irreducible cubic polynomial in Q[X]. Show that the degree of its splitting field over Q is 6 if f ' (x) > 0 for all x.

    This seems like quite an interesting question but I have no idea how we are meant to go about solving it - it seems quite strange to have a condition relating to the derivative in "splitting field" type questions.
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    Quote Originally Posted by Amanda1990 View Post
    Let f(x) be an irreducible cubic polynomial in Q[X]. Show that the degree of its splitting field over Q is 6 if f ' (x) > 0 for all x.

    This seems like quite an interesting question but I have no idea how we are meant to go about solving it - it seems quite strange to have a condition relating to the derivative in "splitting field" type questions.
    Think of the graph of f(x). If f'(x) > 0 it means the polynomial is always increasing. Thus, it cross the x-axis precisely at one point. Let \alpha be the real root of f(x). Now if \beta is the other root of f(x) then it has to be complex*. Thus, the final root is \bar \beta. The splitting field is K=\mathbb{Q}(\alpha, \beta, \bar \beta). Note that \sigma : K\to K by complex conjugation is an automorphism. Now there are three possibilities to send \alpha i.e. \alpha \to \alpha, \alpha \to \beta, \alpha \to \bar \beta by isomorphism extension theorem. Call these maps \tau_1,\tau_2,\tau_3. Then we see that S_3\subseteq \left< \sigma,\tau_1,\tau_2,\tau_3 \right> .


    *)You may want to say, but maybe f has only real roots. But then it forces f(x) = (x-a)^3 for some a\in \mathbb{R}. However, f'(a) = 0 is not larger than 0. So this is not possible.
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  3. #3
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    OK thanks for the hints but I really can't see how this helps very much! I haven't come across many actual theorems relating to splitting fields - does your answer rely on some "standard" result?
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  4. #4
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    Quote Originally Posted by Amanda1990 View Post
    OK thanks for the hints but I really can't see how this helps very much! I haven't come across many actual theorems relating to splitting fields - does your answer rely on some "standard" result?
    Define, \sigma : K \to K to be \sigma (x) = \bar x i.e. complex conjugation.
    Define, \tau_1 : K \to K to be \tau_1 (\alpha) = \alpha (i.e. the identity mapping).
    Define, \tau_2 : K \to K to be \tau_2 (\alpha) = \beta.
    Define, \tau_3 : K \to K to be \tau_3 (\alpha) = \bar \beta.
    Notice that \tau_2,\tau_3 exist by the isomorphism extension theorem.
    We know that  \tau_1,\tau_2,\tau_3, \sigma are all distinct automorphism of K.
    But that means, \sigma \tau_2, \sigma \tau_3 are distinct automorphism too.
    Thus, we have shown we have at least 6 automorphism of K.
    Howevever, |\text{Gal}(K/\mathbb{Q})| = [ K : \mathbb{Q} ] \leq 3! = 6.
    Thus, we found all the automorphisms
    This means [K:\mathbb{Q}] = 6.

    Do you have any other questions?
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