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Math Help - Galois group

  1. #1
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    Question Galois group

    Who can give me an example of a field extension that has the Galois group isomorphic with Z_3, Z_p (p=prime number), S_p?
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  2. #2
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    Quote Originally Posted by Veve View Post
    Who can give me an example of a field extension that has the Galois group isomorphic with Z_3, Z_p (p=prime number), S_p?
    In general any finite group can be made isomorphic to a Galois group. However, you might be interested in finding the inverse Galois problem for \mathbb{Z}_p and S_p. Here is a start. If f(x) is an irreducible polynomial over \mathbb{Q} with prime degree such that it has exactly two non-real zeros then form the splitting field F. We argue that the group \mbox{Gal}(F/\mathbb{Q})\simeq S_p. First, note that \mbox{Gal}(F/\mathbb{Q}) is a subgroup of the permutation group of the zeros. Second, let \theta be complex-conjugation, then \theta \in \mbox{Gal}(F/K) and it leaves all real zeros intact, therefore, \theta can be viewed as a transposition. Third, let \phi be an automorphism which premutes the zeros in a cyclic fashion then \phi can be regarded as a cycle of length p. Thus, we have that \text{Gal}(F/\mathbb{Q})\leq S_p has a cycle of length p and a transpotion, thus in fact, \text{Gal}(F/\mathbb{Q}) = S_p.

    (The link provides how to get \mathbb{Z}_n as an inverse Galois problem).

    This is Mine 97th Post!!!
    Last edited by ThePerfectHacker; May 17th 2008 at 08:12 PM.
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  3. #3
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    Quote Originally Posted by Veve View Post
    Who can give me an example of a field extension that has the Galois group isomorphic with Z_3, Z_p (p=prime number), S_p?
    you should know that every finite extension of a finite field is Galois. so, for example, if q=2^n, then \text{Gal}(\mathbb{F}_q/\mathbb{F}_2) \simeq \mathbb{Z}_n.

    to answer your last question, i just answered the question for case p = 2, because S_2 \simeq \mathbb{Z}_2, or another example for this

    case is \mathbb{Q}(\sqrt{2})/\mathbb{Q}. so i will assume, from now on, that p > 2.

    recall that if f \in \mathbb{Q}[x] is irreducible and \deg f = p, where p is prime, and if f has exactly two non-real roots, then the

    Galois group of f is S_p. so we just need to find an example of such polynomials. here's one example:

    let f_{\alpha}(x)=x^p + \alpha p^2(x -1)(x-2) \ ... \ (x - p + 2) - p, with the integer \alpha > 0 to be determined! obviously, by Eisenstein

    lemma, f_{\alpha} is irreducible. now let x_j, \ 1 \leq j \leq p, be the roots of f_{\alpha}. then \sum x_j^2 = -2 \alpha p^2 < 0, and thus f_{\alpha} has at least

    two non-real roots. so the only thing we need to show now is that there exists \alpha > 0 such that f_{\alpha} has at least p-2

    real roots. to prove this, first see that f_{\alpha}(1) < 0 and f_{\alpha}(2) > 0. so f_{\alpha} has a root in the interval (1,2), which proves the

    claim for p = 3. so i will assume that p > 3. now let 1 \leq j \leq \frac{p-3}{2}, and fix 2j < t_j < 2j+1. choose \alpha_j > 0 large enough

    so that f_{\alpha}(t_j) < 0, \ \forall \alpha \geq \alpha_j. (why is it possible?) thus if \alpha \geq \alpha_j, then f_{\alpha} has at least 2 roots in the interval (2j, 2j+1),

    since f_{\alpha}(2j) > 0, \ f_{\alpha}(2j+1) > 0, and f_{\alpha}(t_j) < 0. now let \alpha \geq \max _j \alpha_j. then f_{\alpha} will have at least p-2 real roots. Q.E.D.
    Last edited by NonCommAlg; May 17th 2008 at 07:53 PM.
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    Here is a way to show any (finite) group can be thought of as a Galois group. Let F be a field and E=F(x_1,...,x_n) be the rational function field in n variables. Let K = F(f_1,...,f_n) where f_1,...,f_n are the elementary symettric polynomials. Now \mbox{Gal}(E/K)\simeq S_n because any permutation of x_1,...,x_n will leave K fixed. By Cayley's theorem every group can be regarded as a subgroup of the symettric group. Using the fundamental theorem we simply pick the appropriate subgroup H of S_n isomorphic to G and take the corresponding fixed field, i.e. take E^H and form the group N = \text{Gal}(F/E^H) then N\simeq G.
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  5. #5
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    Here is another way to get \mathbb{Z}_n. Let F=\mathbb{C}(t) the rational function field. Let x^n - t\in F[x] then x^n - t is irreducible by Eisenstein (since \mathbb{C}(t) is a quotient field of \mathbb{C}[t]). Let \alpha solve the equation x^n - t = 0 then \alpha , \zeta \alpha , ... , \zeta^{n-1}\alpha all solve this equation where \zeta = e^{2\pi i/n}. Thus, K = F(\alpha) and furthermore \mbox{Gal}(K/F) \simeq \mathbb{Z}_n since we can map \alpha \mapsto \zeta^k \alpha and so \alpha generates this group with order n.
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