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Solve the algebraic equation,Galois group of FF, and prove/disprove.

  1. Apr 24, 2006 #1
    1. x^4+2x^3+3x^2+4x+5=0

    2. is the equation x^2=2 solvable in 7-adic numbers?

    3. find the group Gal(F_p^n^2/ F_p^n), I got n^2-n+1... is it right?
  2. jcsd
  3. Apr 25, 2006 #2

    matt grime

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    1. Isn't a question it's an equation

    3. n^2-n+1 is not necessarily the most universal notation for a group. You mean the cyclic group of that order, right?

    Show your work.
  4. Apr 25, 2006 #3
    1. is to solve the equation, means to get the x, and I have no clue....

    2.I think it's solvable, thus I used hensel lemma and got f'(x)=2x with |f(x)|_7... but haven't got a number wich can satisfy the equation

    3. for n+1, up to n^2 there will be n^2+1, and minus the n one in p^n, thus u got n^2-n+1.
  5. Apr 25, 2006 #4
    Please someone , these questions will be on the final examination and it'll take place tomorrow, none of us had got these questions so far.... :(
  6. Apr 25, 2006 #5


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    It's never good to wait until the last minute! :tongue:

    For (1), surely you've done (or seen) similar problems already, and have an idea about what you should be doing?

    For (2) and (3), you are very unclear...

    For example, I haven't the slightest idea what "with [itex]|f(x)|_7[/itex]" means -- what exactly are the conditions for Hensel's lemma?

    (P.S. don't forget that there is a very easy way to find solutions to an equation modulo 7...)

    For (3) I have absolutely no idea what you're doing at all. Maybe if you could write more clearly what you are trying to do, it would be more clear to me... and more importantly, it would be more clear to you.
  7. Apr 26, 2006 #6

    matt grime

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    If I read this on its own I have no idea that you're talking about Galois Theory or even that you are talking about groups.
  8. Apr 29, 2006 #7


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    This has "ugly" complex roots (2 pairs of complex conjugate roots). You need the general quartic method to solve this one.

    Proving the equation has no real roots is easy. x= 1 is obviously not a solution of the original equation.

    Let the orig. expression in x be represented as f(x)

    [tex]{(x-1)}^2f(x) = {(x-1)}^2(x^4+2x^3+3x^2+4x+5) = x^6-6x+5 = g(x)[/tex] (say)

    All zeros of f(x) comprise a subset of the zeros of g(x).

    The degree 6 polynomial [tex]g(x)[/tex] obviously has a repeated root at x = 1. It would have 4 other (not necessarily distinct) roots.

    But [tex]g'(x) = 6(x^5 - 1)[/tex] has no real zeros other than x = 1 (the others are the complex fifth roots of unity). Hence g(x) has no intersections with the x-axis other than forming a tangent at the repeated root x = 1. And we've observed that x = 1 is obviously not a zero of f(x).

    Hence, f(x) has no real zeros.
    Last edited: Apr 29, 2006
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