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Math Help - Conclusions after finding the Galois Group

  1. #1
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    Conclusions after finding the Galois Group

    A lot of effort is put into finding the Galois group of a polynomial. I'd like to know about what you can say when you go in the other direction.

    1. Suppose I tell you that the Galois Group of a polynomial is trivial (or Z2, or something). What can I conclude from this?

    2. In particular, if I know the Galois Group is trivial, am I supposed to conclude that the polynomial is / is not solvable by radicals?
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  2. #2
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    partial answer: the trivial group, and Z2 are both solvable. therefore a polynomial which has these groups as galois groups is solvable by radicals.

    this is part of the beauty of galois theory, questions about polynomials (which are generally hard), are turned into questions about groups (which are easier).

    so, if you start with the "hard part solved" (you've found, or been given, the galois group of a polynomial), you can use the "easy part" to recover information

    about the polynomial without much effort.

    the classic example is p(x) = x^5 - x - 1. its galois group is S5, which is not solvable, because A5 is simple. we can therefore abandon any hope of solving

    p(x) by radicals (S5/A5 is abelian, but A5/{e} is not, and there is no other normal subgroup of A5).
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  3. #3
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    Quote Originally Posted by Deveno View Post
    partial answer: the trivial group, and Z2 are both solvable. therefore a polynomial which has these groups as galois groups is solvable by radicals.


    In this particular case yes, in general you have to be working in characteristic noit dividing the order of the Galois group (for

    example, characeristic zero always fulfils this, and this is why in undergraduate courses this is the usual condition)

    Tonio



    this is part of the beauty of galois theory, questions about polynomials (which are generally hard), are turned into questions about groups (which are easier).

    so, if you start with the "hard part solved" (you've found, or been given, the galois group of a polynomial), you can use the "easy part" to recover information

    about the polynomial without much effort.

    the classic example is p(x) = x^5 - x - 1. its galois group is S5, which is not solvable, because A5 is simple. we can therefore abandon any hope of solving

    p(x) by radicals (S5/A5 is abelian, but A5/{e} is not, and there is no other normal subgroup of A5).
    .
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  4. #4
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    yeah, weird things happen in extensions of Zp. my answer wasn't intended to be comprehensive. usually the polynomials of interest are in Q[x].
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  5. #5
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    Thank you. Yes, I am concerned right now with polynomials with integral coefficients.

    So I'm concluding from your comments that if the Galois Group is any group not containing a simple subgroup (trivial group, Z2, Z3, Z4, S3, S4...), you can solve it by radicals. Next question, how do I deduce the procedure for solving it by radicals?
    In other words, what is the equivalent of the quadratic formula for getting the roots?
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  6. #6
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    two things:

    1) a group can be simple, and still solvable. Zp, for example is simple, yet solvable, because {{e},Zp} is a composition series with abelian factor groups.

    2) the procedure for finding actual roots by "solving by radicals" can be somewhat complicated. for example, the general quadratic ax^2 + bx + c, can be turned

    into a monic quadratic x^2 + rx + s by setting r = b/a, s = c/a (assuming a is not 0). in a splitting field, this becomes (x - α)(x - β).

    now if α is in F, there are two possibilities: α = 0, or not. if α is not 0, then β = s/α is also in F. if α = 0, then the polynomial is of the form:

    x^2 + rx = x(x + r), which splits in F. so if α is in F, the galois group is Gal(F/F) = 1.

    if α is not in F, we have β = -r - α, so β is in F(α) = F(β), and we have the galois group Gal(F(α)/F) = {1,σ}, where

    σ(α) = β is the automorphism of F(α) with F(β). now F(α) is not obviously an extension of F by radicals, since α^2 is not

    necessarily in F (although it may be, for example if r = 0, and s = 2, and F = Q).

    but α^2 + αr = -s, so 4α^2 + 4αr + r^2 = r^2 - 4s, so (2α + r)^2 is in F, and clearly F(2α + r) = F(α) is an extension by radicals of F.

    (note: if char(F) = 2, this isn't true, since then F(2α + r) = F(0 + r) = F(r) = F. quadratic equations have to be handled

    differently in a field of characteristic 2).

    this means we can obtain a splitting field for x^2 + rx + s by adjoining 2α + r to F or what is the same thing: F(√(r^2 - s)).
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  7. #7
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    Thank you.

    If we look at concrete examples of polynomials with integral coefficients, there are known ways of solving quadratics (quadratic formula), cubics ( Cardan's formula) and quartics ( Ferrari's method). The search for formulas of this type breaks down for 5th degree equations and this was 'explained' by Galois theory (since S5 has subgroup A5).

    I get lost when talking just about theory.

    If I have a polynomial whose roots are {1,2,3,4,5}, (i.e., (x-1)(x-2)(x-3)(x-4)(x-5) - a 5th degree monic polynomial ), I presume the Galois group of this polynomial is the trivial group. Since the group is trivial, this implies (?) there is a way to find the roots. So my concrete question is:

    Suppose someone gave you the equation :

    x^5 - 15 x^4 +85 x^3 -225 x^2 + 274 x -120,

    (which is (x-1)(x-2)(x-3)(x-4)(x-5) all multiplied out, if I did the math right )

    how do I find the roots (which are 1,2,3,4,5) ?
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  8. #8
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    well, if you knew that it did indeed factor over Q[x], and thus factored over Z[x], you would know that the roots are integers. since 120 = 5!, the only possible integral roots are:

    {1,2,3,4,5}. this is a small enough set, that you can test each root in turn:

    p(1) = 1 - 15 + 85 - 225 + 274 = 0, so x - 1 is a factor. dividing by x - 1, you would get a quartic:

    x^4 - 14x^3 + 71x^2 - 154x + 120, which we know is solvable.

    but determining if a quintic is indeed solvable is by no means an easy task. finding the galois group

    of a quintic polynomial can depend on whether the polynomial is irreducible or not.

    here is an actual example: find the galois group of f(x) = x^5 - 4x^2 + 2. by the eisenstein criterion,

    this is irreducible, so the galois group acts transitively on the roots, and so contains a 5-cycle.

    by considering f'(x) = 5x^4 - 8x, we see that f(x) has 2 turning points, since f(0) > 0, we can see that it turns

    downwards at 0 after crossing the x-axis once. f((8/5)^(1/3)) ~ -1.283, so it only turns upwards again

    after it has crossed the x-axis again, and is thereafter increasing (and so crosses the x-axis one last time).

    so f has 3 real roots, and 2 complex roots. complex conjugation is an automorphism of the splitting field for f

    which fixes Q (and also the real roots), so the galois group of f contains a 5-cycle, and a transposition,

    so it must be S5, and f is not solvable by radicals. for an example of a quintic that is solvable,

    determine the galois group of: x^5 - 6x^4 + 9x^3 + 6x^2 - 22x + 12.

    as regards your original polynomial, indeed the galois group is trivial. if someone told you that fact, that would imply the roots were integers.

    so we can move "the hard part of solving" from polynomials to the galois group, but finding the galois group of a given polynomial

    isn't always a trivial task.
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  9. #9
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    Thank you for your response. Your concrete example was quite helpful.

    You asked for the Galois group of:
    f(x) = x^5 - 6x^4 + 9x^3 + 6x^2 - 22x + 12
    (alternatively, just write the coefficients of f = (1,-6,9,6,-22,12) )

    Any integral roots must be from the set { 1,2,3,4,6,12 (all +/-) }.

    By inspection, x=1 is a root. Dividing out (x-1) leaves:

    f1(x) = (1,-5,4,10,-12)

    Testing, we find x=2 is a root. Dividing out (x-2) leaves:

    f2(x) = (1,-3,-2,6)

    Testing, we find x=3 is a root. Dividing out (x-3) leaves:

    f3(x) = (1,0,-2)

    This is x^2 = 2, with roots = +/- sqroot(2) by inspection.

    So the entire set of roots is: (1,2,3,+sqroot(2), -sqroot(2)).
    So I think the Galois Group is Z2.


    I worked through your analysis of:
    f(x) = x^5 - 4x^2 + 2
    Part of your analysis used calculus to find the zeroes of f to be at 0 and
    cube root of (8/5). Then you looked at f' and f'' to graph what f(x)
    looked like. From this, you can see it has 3 real roots. Since it's a
    quintic, there must be 5 total roots, so the remaining 2 roots must be a
    complex pair. This complex pair gives you the transposition in your Galois
    group analysis.
    From Eisenstein (prime=2) you deduced the presence of a 5-cycle.
    Then since there exists a 5-cycle and a transposition, you conclude the
    group is S5.


    Two questions about the logical sequence of deductions here:

    1. For 5th degree equations, if by Eisenstein I know the equation is
    irreducible, don't I already know the Galois group is S5 immediately? If
    the group were anything smaller, wouldn't the polynomial be reducible?

    2. In the analysis above, we found the roots in order to find the Galois
    group. If my main goal is to find the roots, does Galois Theory give me
    any helpful computational tools? (Certainly it offers an explanation why
    the search for a general root finding algorithm fails, but is there
    any constructive way to use Galois Theory?)
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  10. #10
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    in repsonse to 1):

    not necessarily. if f(x) has only one real root, it is possible that its galois group is a subgroup of A5 (or A5 itself).

    for example, f(x) = x^5 + 20x + 16 is one such polynomial. note that (complex) conjugation acts on this polynomial

    by swapping roots in pairs (just because the galois group contains a 5-cycle doesn't mean the galois group is S5).

    2. only in certain cases. finding the roots of polynomials is hard, and the higher the degree, the harder it is. what galois theory gives us,

    is information about how the field in which the roots lie behaves. consider how large a group S5 is: it has 120 elements,

    and many subgroups (i think it has 156 different subgroups). and that is how many possible automorphisms we may have

    of some extension field of Q (for some polynomial we may wish to solve), which means that the extension field

    (the splitting field we're looking for) can be built from Q in a LOT of different ways.
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  11. #11
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    Thank you.

    Regards (1) - I've clearly confused the 2 concepts of irreducibility and unsolvability. It's back to the books for this one.

    Regards (2) - Do you have a concrete example how deducing something about the Galois group helps in finding the roots?
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