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Math Help - Galois theory

  1. #1
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    Galois theory

    Can someone please please help me with the following question(s). I have been trying for hours and have got nowhere

    Thanks
    Edgar



    Let p1, p2, p3, …be distinct primes.

    1) prove by induction on m that if x є Q and √x є Q(√p1…..√pm ) then
    x = p1^α1…..pm^αm y^2 for some integers a1....am and some y є Q.

    2) Deduce that for any n √p n+1 (n+1 subscript) is not an element of
    Q(√p1…..√pn )

    3) hence prove by induction on n that [Q(√p1…..√pn ):Q] = 2^n
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  2. #2
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    Quote Originally Posted by edgar davids View Post
    3) hence prove by induction on n that [Q(√p1…..√pn ):Q] = 2^n
    If we can show p_n\not \in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n-1}} ) it means x^2 - p_n is minimal polynomial for \sqrt{p_n}. This means [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n-1}})][\mathbb{Q}(\sqrt{p_n}):\mathbb{Q}] = 2^{n-1}\cdot 2 = 2^n. So it remains to prove the first step.

    Start with the basic case that \sqrt{x} \in \mathbb{Q}(\sqrt{p_1}) it means \sqrt{x} = a + b\sqrt{p_1} for some rational numbers a,b. Thus, x = a^2+2ab\sqrt{p_1}+b^2p_1, for this expression to hold we need that a=0 or b=0 because RHS is irrational otherwise so x=a^2 or x=b^2p_1, which means x=p_1^{a_1} y^2.

    Now suppose that \sqrt{x} \in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k}) means x=p_1^{a_1}...p_k^{a_k}y^2. Say that \sqrt{x} \in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k})(\sqrt{p_{k+1  }}). It means \sqrt{x} = p+q\sqrt{p_{k+1}} where p,q\in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k}). Thus, x = p^2 + 2pq\sqrt{p_{k+1}} + q^2p_{k+1}. So that means by induction p or q is zero. Which means x=p_1^{a_1}...p_{k+1}^{a_{k+1}}y^2.
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  3. #3
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    need to find the glaios group of the extension Q(√p1.....√pn):Q and show that G is isomorphic with C2^n

    once this has been done then need to use this Galois correspondence to show that
    Q(√p1.....√pn) = Q(√p1 + √p2 + √p3 +....+√pn)
    It is easier to explain what is going on in a specific case, say, \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3})/\mathbb{Q} rather then in general n, but the pattern is the same in general.

    The Galois group is isomorphic to \mathbb{Z}_2^3 because if \sigma_i are the automorphisms \sigma_i(\sqrt{p_i}) = - \sqrt{p_i} for i=1,2,3 and any composition between them is too, which gives us 8 altogether, which is the size of the Galois group, and furthermore each automorphism which generates the Galois group has order two, so it is isomorphic to the group \mathbb{Z}_2^3. The picture below shows all those fields E such that \mathbb{Q}\subseteq E\subseteq \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3}), note there have to have 12 fields E just like the subgroups of \mathbb{Z}_2^3 (it happens to be impossible to draw a nice subfield diagram in 2 dimension, that has something to do with non-planar graphs).

    Now, of course, \mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\sqrt{p_3}) is a subfield. But it cannot be among any of those proper intermediate subfields. Thus, it can therefore only be the improper subfield, i.e. equal to \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3}) itself.

    Note, the same pattern happens when you are using n primes. It is just easier to explain the situation in a specific case. To actually prove that the Galois group for n primes is isomorphic to \mathbb{Z}_2^n is to note that the size of the Galois group has to be the degree over the field over \mathbb{Q} which is 2^n. Now we need to find its elements. If \sigma_i (\sqrt{p_i}) = -\sqrt{p_i} are automorphism for i=1,2,...,n then \left< \sigma_1,...,\sigma_n \right> generate a group with 2^n elements. This is of course isomorphic to \mathbb{Z}_2^n.
    Attached Thumbnails Attached Thumbnails Galois theory-picture.jpg  
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  4. #4
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    Quote Originally Posted by edgar davis
    Let p1, p2, p3..... be distinct primes

    1) Deduce that for any n, √ p (n+1) is not an element of Q (√ p1......√ pn)

    (fyi (n+1) is subscript)

    2) Show that the extension Q (√ p1......√ pn): Q is normal.
    For (1) note that if \sqrt{p_{n+1}}\in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}) then it would mean \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n},\sqrt{p_{n+1}  }) = \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}) and so [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n},\sqrt{p_{n+1}  }:\mathbb{Q}] = 2^n. But that is not true because we know this degree extension has to be 2^{n+1}.

    For (2) you have to use a theorem (or maybe some would call it a definition). Given fields F/K. We say F is normal (and finite) over K iff for every irreducible polynomial in K[x] that has a zero in F has all its zeros in F. It turns out that F/K is a normal (and finite) seperable extension if and only if F is the splitting field over K of some seperable polynomial. Now let F=\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}) and K=\mathbb{Q}. Note that F is the splitting field over (x^2-p_1)...(x^2-p_n). Furthermore, every (non-constant) polynomial over K is seperable because K is perfect (because \text{char}(K) = 0). Thus, F/K is a splitting field extension of a seperable polynomial so it is in fact a normal extension.
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