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Math Help - Galois Theory

  1. #1
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    Galois Theory

    1. The polynomail f(x) = x^5-5x+1 is irreducible in \mathbb Q [x]. Find the number if real roots of f(x).

    2. Let M be the splitting field for f(x) over \mathbb Q. Show that M/\mathbb Q is a Galois extension
    and determine its Galois group Gal(M/\mathbb Q). Is f(x) solvable by root?

    In the following it can be used without a proof that the Discrimentant of f(x)
    is -3\cdot5^6\cdot 17.

    3. Show that \mathbb Q(\sqrt{-51}) is contained in M and determine that f(x)
    is irreducible in \mathbb Q(\sqrt{-51})[x].

    4. Determine the Galois group Gal(M/ \mathbb Q(\sqrt{-51})).
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    1. The polynomail f(x) = x^5-5x+1 is irreducible in \mathbb Q [x]. Find the number if real roots of f(x).

    2. Let M be the splitting field for f(x) over \mathbb Q. Show that M/\mathbb Q is a Galois extension
    and determine its Galois group Gal(M/\mathbb Q). Is f(x) solvable by root?

    In the following it can be used without a proof that the Discrimentant of f(x)
    is -3\cdot5^6\cdot 17.

    3. Show that \mathbb Q(\sqrt{-51}) is contained in M and determine that f(x)
    is irreducible in \mathbb Q(\sqrt{-51})[x].

    4. Determine the Galois group Gal(M/ \mathbb Q(\sqrt{-51})).
    1. use rational root test
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  3. #3
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    rational root test to find real root? and then whT bout others?
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  4. #4
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    Quote Originally Posted by peteryellow View Post
    1. The polynomail f(x) = x^5-5x+1 is irreducible in \mathbb Q [x]. Find the number if real roots of f(x).
    Using analytic methods (or just graphing it) we see that f has three real roots, the other two roots are complex conjugates.

    2. Let M be the splitting field for f(x) over \mathbb Q. Show that M/\mathbb Q is a Galois extension
    and determine its Galois group Gal(M/\mathbb Q). Is f(x) solvable by root?
    Obviously, M/\mathbb{Q} is a Galois extension since it is a splitting field over a perfect field. We want to compute \text{Gal}(M/\mathbb{Q}). First of all we can identify the Galois group as a subgroup of S_5 because each automorphism of M is also a permutation of the roots of f. To show that \text{Gal}(M/\mathbb{Q})\simeq S_5 all we need to do is show that \text{Gal}(M/\mathbb{Q}) contains a permutation of order 5 and a transpotion.* Let \alpha be a root of f in M then \mathbb{Q}\subseteq \mathbb{Q}(\alpha)\subseteq M. Notice that [\mathbb{Q}(\alpha):\mathbb{Q}] = 5, therefore 5 divides |\text{Gal}(M/\mathbb{Q}) by the fundamental theorem of Galois theory. Thus, by Cauchy's theorem there is an element in \text{Gal}(M/\mathbb{Q}) of order 5 i.e. a 5-cycle if viewed as an element in S_5. And surly there is a transposition in \text{Gal}(M/\mathbb{Q}), just take \sigma: M\to M defined by \sigma (x) = \overline{x} i.e. complex conjugation. Then \sigma is a transposition if viewed as an element of S_5. Therefore, \text{Gal}(M/\mathbb{Q}) \simeq S_5. This symmetric group is not solvable therefore the polynomial is not solvable by radicals.


    *)A subgroup of S_5 generated by a 5-cycle and a 2-cycle must be S_5 itself.


    3. Show that \mathbb Q(\sqrt{-51}) is contained in M and determine that f(x)
    is irreducible in \mathbb Q(\sqrt{-51})[x].

    4. Determine the Galois group Gal(M/ \mathbb Q(\sqrt{-51})).
    Let \Delta = \sqrt{-3\cdot 5^6\cdot 17} then \mathbb{Q}(\Delta) corresponds to subgroup S_n\cap A_n = A_n. Notice that \mathbb{Q}(\Delta) = \mathbb{Q}(\sqrt{-51}), therefore \text{Gal}(M/\mathbb{Q}(\sqrt{-51})) \simeq A_n under this Galois correspondence.
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