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Thread: t^3-3t+1 splits or is irreducible

  1. #1
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    t^3-3t+1 splits or is irreducible

    Prove that for any field K, $\displaystyle t^3-3t+1$ splits or is irreducible in K.

    Hint: show that any zero is a rational expression in any other zero.

    Using the formula for the sum of roots: $\displaystyle \sum \alpha_i = \frac{-a_{n-1}}{a_n}$ where $\displaystyle \alpha_i$ are the roots, and $\displaystyle a_i$ the coefficients of the i-th power term in the polynomial, we get $\displaystyle \alpha_1+\alpha_2+\alpha_3=0$

    Also using the formula for the product of the roots: $\displaystyle \prod \alpha_i=(-1)^n (\frac{a_0}{a_n}) \$ we get $\displaystyle \alpha_1\alpha_2\alpha_3=-1$

    I only seem to be able to combine these equations, and the equations resulting from the fact that each $\displaystyle \alpha_i$ is a root of our polynomial to get equations of two variables where a power higher than 1 is present in each variable, and quadratic formulas don't seem to resolve those nicely. I think if I'm reading the problem right, we don't want any roots in our relating equation so I guess we need another rational expression (other than $\displaystyle \alpha_1+\alpha_2+\alpha_3=0$) involving the three roots to combine with the previous to get a rational expression involving two roots.

    Ideas?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Assuming this is the right path to go down, you could get a third equation from the coefficient of $\displaystyle t $, namely $\displaystyle \alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3 =-3 $.

    Now you have three equations and three unknowns. See if that does the trick. See here for more info.
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  3. #3
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    Ok, using that information relating coefficients of our polynomial to symmetric polynomials in the roots I have the equations:

    $\displaystyle 0=\alpha_1+\alpha_2+\alpha_3$

    $\displaystyle -3=\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha _3$

    $\displaystyle -1=\alpha_1\alpha_2\alpha_3$

    or solving for $\displaystyle \alpha_1$ so I can directly relate two variables:
    $\displaystyle \alpha_1=-\alpha_2-\alpha_3=-\frac{1}{\alpha_2 \alpha3}=\frac{-3-\alpha_2 \alpha_3}{\alpha_2+\alpha_3}$

    When I manipulate those equations, I just seem to get things like $\displaystyle \alpha_2^n*\alpha_3^m*p(\alpha_2)=0$ where p is the original polynomial and this just confirms that my variables are roots ( I already know they're nonzero).

    It's not enough to simply solve for one root in terms of another. I have to relate them using only add/subtr, mult/div (no radicals).

    Am I perhaps starting off with the wrong approach?
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  4. #4
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    The problem became much easier once I actually solved for the roots and got them in a good looking form, namely $\displaystyle \alpha_1=2cos(2\pi/9),\alpha_2=2sin(\pi/18),\alpha_3=-2cos(\pi/9)$
    I ended up only using one of the prior formulas $\displaystyle \alpha_1+\alpha_2+\alpha_3=0$ and the double angle formulas.
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