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Math Help - Field Extensions - Dummit and Foote Chapter 13

  1. #1
    Super Member Bernhard's Avatar
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    Field Extensions - Dummit and Foote Chapter 13

    Dummit and Foote Chapter 13, Exercise 2, page 519 reads as follows:

    "Show that  x^3 - 2x - 2 is irreducible over  \mathbb{Q} and let  \theta be a root.

    Compute  (1 + \theta ) ( 1 + \theta + {\theta}^2) and  \frac{(1 + \theta )}{ ( 1 + \theta + {\theta}^2)}  in  \mathbb{Q} (\theta)

    ---------------------------------------------------------------------------------------------------------------------------------

    My attempt at this problem so far is as follows:

     p(x) = x^3 - 2x - 2 is irreducible over  \mathbb{Q} by Eisenstein's Criterion.

    To compute  (1 + \theta ) ( 1 + \theta + {\theta}^2) I adopted the simple (but moderately ineffective) strategy of multiplying out and trying to use the fact that  \theta is a root of p(x) - that is to use the fact that  {\theta}^3 - 2{\theta} - 2 = 0  .

    Proceeding this way one finds the following:

     (1 + \theta ) ( 1 + \theta + {\theta}^2) = 1 + 2{\theta} + 2{\theta}^2 + {\theta}^3

     = ({\theta}^3 - 2{\theta} - 2) + (2{\theta}^2 + 4{\theta} + 3)

     2{\theta}^2 + 4{\theta} + 3

    Well, that does not seem to be going anywhere really! I must be missing something!

    Can someone please help with the above and also help with the second part of the question ...

    Peter
    Last edited by Bernhard; September 26th 2013 at 02:52 PM.
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  2. #2
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    Re: Field Extensions - Dummit and Foote Chapter 13

    Don't just try to simplify immediately. First, find more relationships. Note that you can reduce any term of \theta^n for n>2 to an expression where the highest power of \theta is 2.

    \theta^3 = 2\theta+2
    \theta^4 = 2\theta^2 + 2\theta
    \begin{align*}\theta^5 & = 2\theta^3 + 2\theta^2 \\ & = 2(2\theta+2) + 2\theta^2 \\ & = 2\theta^2 + 4\theta+4\end{align*}

    Also note that if you do this for \theta^6, the expression you get is exactly the one you would get from (\theta^3)^2. So, these expressions become exhaustive fairly quickly.

    Next, remember that you are using the rationals, so you need to figure out inverse powers of theta, too.

    2 = \theta^3-2\theta so
    \theta^{-1} = \dfrac{\theta^2-2}{2}
    \begin{align*}\theta^{-2} & = \dfrac{\theta}{2} - \theta^{-1} \\ & = \dfrac{2 + \theta -\theta^2}{2}\end{align*}

    Again, these expressions become exhaustive fairly quickly.

    Does that help at all?
    Last edited by SlipEternal; September 27th 2013 at 05:42 AM.
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    Re: Field Extensions - Dummit and Foote Chapter 13

    To offer a bit more assistance, 1+\theta = \dfrac{\theta^3}{2} and \theta^2 = \dfrac{\theta^5-2\theta^3}{2}. So, adding those together, we get: 1+\theta+\theta^2 = \dfrac{\theta^3}{2} + \dfrac{\theta^5-2\theta^3}{2} = \dfrac{\theta^5-\theta^3}{2} = \dfrac{\theta^3}{2}(\theta^2-1) = (1+\theta)(\theta^2-1). This will help with the second part of your problem.
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    Re: Field Extensions - Dummit and Foote Chapter 13

    For the second part of the problem, first find the multiplicative inverse of

    \theta ^2+\theta +1 in Q(\theta )

    This can be done by using the division algorithm to find f(x) and g(x) such that

    \left(x^3-2x-2\right)f(x)+\left(x^2+x+1\right)g(x)=1
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  5. #5
    Super Member Bernhard's Avatar
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    Re: Field Extensions - Dummit and Foote Chapter 13

    Thanks SlipEternal, most helpful however I need a bit more help:

    Can you explain how you proceed in the following:

     2 = {\theta}^3 - 2 \theta \ \Longrightarrow  \ {\theta}^{-1} = \frac{{\theta}^2 - 2}{2}

    How does the above follow?

    Peter
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    Re: Field Extensions - Dummit and Foote Chapter 13

    Quote Originally Posted by Bernhard View Post
    Thanks SlipEternal, most helpful however I need a bit more help:

    Can you explain how you proceed in the following:

     2 = {\theta}^3 - 2 \theta \ \Longrightarrow  \ {\theta}^{-1} = \frac{{\theta}^2 - 2}{2}

    How does the above follow?

    Peter
    Divide both sides by \theta then divide both sides by 2.
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