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Thread: Splitting Fields and Galois Groups

  1. #1
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    Splitting Fields and Galois Groups

    Hi,

    I've been set a question and think I've managed about half of it but stuck on some other bits..

    Show that $\displaystyle f := x^3 - 3x + 1$ is irreducible over $\displaystyle \mathbb{Q}$. Let $\displaystyle a\in\mathbb{C}$ be given as a root of $\displaystyle f$. Show that $\displaystyle (1-a)^{-1}$ is also a root of $\displaystyle f$. Hence, find a splitting field $\displaystyle L$ for $\displaystyle f$ over $\displaystyle \mathbb{Q}$. Identity the Galois group $\displaystyle \tau(L : \mathbb{Q})$. (You must justify any claim that a particular map is an isomorphism. Be sure to state carefully
    any facts that you assume in doing this.)

    first part is easy just reducing by $\displaystyle \mathbb{Z}_{2}$ to get $\displaystyle \bar{f} := x^3 +x + 1\in\mathbb{Z}$\$\displaystyle \mathbb{Z}_{2}(x)$ which has no zeros in $\displaystyle \mathbb{Z}_{2}$ and thus irreducible over $\displaystyle \mathbb{Z}$ and hence over $\displaystyle \mathbb{Q}$ also.

    I think I can identify the Galois group using Prop 22.4 which gives $\displaystyle \Delta= -27(-1)^2 - 4(-3)^3 = 81$ , a perfect square thus the Galois Group is $\displaystyle \mathbb{A}_{3}\cong C_{3}$

    I can't think how to show that $\displaystyle (1-a)^{-1}$ is also a root of $\displaystyle f$. For the splitting field all I can think to do is say that as $\displaystyle a\in\mathbb{C}$ is a root of $\displaystyle f$then $\displaystyle f$ splits in $\displaystyle \mathbb{C}(a)$ and so $\displaystyle \mathbb{C}(a)$ is the splitting field for $\displaystyle f$ over $\displaystyle \mathbb{Q}$.

    Any help for this last bit or if you can point out any errors I've already made would be greatly appreciated!
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  2. #2
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    Re: Splitting Fields and Galois Groups

    As I posted just thought that to show $\displaystyle (1-a)^{-1}$ is also a root of $\displaystyle f$ you would say something along the lines of $\displaystyle f$ factors into $\displaystyle f:= (x-a)(x^2+bx-\frac{1}{a})$ for some $\displaystyle b\in\mathbb{C}$ and somehow this ends up with $\displaystyle (1-a)^{-1}$ as a root... I know this is obviously not correct but am I along the right lines??
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  3. #3
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    Re: Splitting Fields and Galois Groups

    one can verify $\displaystyle \frac{1}{1-a}$ is a root directly:

    $\displaystyle f(\frac{1}{1-a}) = \left(\frac{1}{1-a}\right)^3 - \frac{3}{1-a} + 1$

    $\displaystyle = \frac{1}{(1-a)^3} - \frac{3(1-a)^2}{(1-a)^3} + \frac{(1-a)^3}{(1-a^3)}$

    $\displaystyle = \frac{1 - 3 + 6a - 3a^2 + 1 - 3a + 3a^2 - a^3}{(1-a)^3}$

    $\displaystyle = \frac{-(1 - 3a + a^3)}{(1-a)^3} = 0$
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  4. #4
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    Re: Splitting Fields and Galois Groups

    Haha, cheers; I'm a fool. Other than that does the rest seem OK?
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