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Math Help - Subfield

  1. #1
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    Subfield

    Hi everybody. I have this exercise where I am to describe all the subfields K_1 and K_2 of L_1 and L_2 with [K_1:\mathbb{Q}]=2 and [K_2:\mathbb{Q}]=2, where L_1 and L_2 are the splitting fields for the polynomials g_1(x)=(x^3-2) and g_2=(x^4+1) over \mathbb{Q}.
    (Remark: L_2 is the cyclotomic field \mathbb{Q}_8)

    Now, I would suppose that you have to use the transitivity theorem to say that; [K_1:\mathbb{Q}]=[K_1:\mathbb{Q}(\epsilon_3)][\mathbb{Q}(\epsilon_3):\mathbb{Q}]

    [K_2:\mathbb{Q}]=[K_2:\mathbb{Q}(\epsilon_8)][\mathbb{Q}(\epsilon_8):\mathbb{Q}]

    and then argue that [K_1:\mathbb{Q}(\epsilon_3)]=[K_2:\mathbb{Q}(\epsilon_8)]=1 or did I get that wrong? I'm just not sure, then, how to argue that I then have found ALL the subfields with this dimension?
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  2. #2
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    Quote Originally Posted by Carl View Post
    Hi everybody. I have this exercise where I am to describe all the subfields K_1 and K_2 of L_1 and L_2 with [K_1:\mathbb{Q}]=2 and [K_2:\mathbb{Q}]=2, where L_1 and L_2 are the splitting fields for the polynomials g_1(x)=(x^3-2) and g_2=(x^4+1) over \mathbb{Q}.
    (Remark: L_2 is the cyclotomic field \mathbb{Q}_8)

    Now, I would suppose that you have to use the transitivity theorem to say that; [K_1:\mathbb{Q}]=[K_1:\mathbb{Q}(\epsilon_3)][\mathbb{Q}(\epsilon_3):\mathbb{Q}]

    [K_2:\mathbb{Q}]=[K_2:\mathbb{Q}(\epsilon_8)][\mathbb{Q}(\epsilon_8):\mathbb{Q}]

    and then argue that [K_1:\mathbb{Q}(\epsilon_3)]=[K_2:\mathbb{Q}(\epsilon_8)]=1 or did I get that wrong? I'm just not sure, then, how to argue that I then have found ALL the subfields with this dimension?
    I start you out with K_1 and you think about K_2. Since L_1 is the splitting of x^3 - 2 over \mathbb{Q} it means L_1 = \mathbb{Q}(\zeta,\sqrt[3]{2}). We will prove that \text{Gal}(L_1/\mathbb{Q}) \simeq S_3. If \sigma is an automorphism of L_1 then it is completely determined by \sigma (\zeta) and by \sigma (\sqrt[3]{2}). Now \zeta is a zero of the irreducible polynomial x^2+x+1, this polynomial has zeros \zeta,\zeta^2, thus \sigma (\zeta) = \zeta \text{ or }\zeta^2. By similar reasoning \sigma (\sqrt[3]{2}) = \sqrt[3]{2},\zeta\sqrt[3]{2},\text{ or }\zeta^2\sqrt[3]{2} since \sqrt[3]{2} is a zero of the irreducible polynomial x^3-2. By isomorphism extension theorem there is an automorphism \sigma so that \sigma (\zeta) = \zeta^2 and \sqrt(\sqrt[3]{2}) = \sqrt[3]{2}. Also, there is an automorphism \tau so that \tau (\zeta) = \zeta and \tau (\sqrt[3]{2}) = \zeta\sqrt[3]{2}. Notice that \sigma (\zeta) = \zeta^2 = \overline{\zeta} so \sigma is in fact complex-conjugation. We see that the order of \sigma in the Galois group is 2 and order of \tau in the Galois group is 3, furthermore \text{Gal}(L_1/\mathbb{Q}) = \left< \sigma,\tau\right>\simeq S_3. You want, as the problem says, to find all K_1 so that [K_1:\mathbb{Q}] = 2. Remember by the fundamental theorem of Galois theory [K_1:\mathbb{Q}] = [S_3:H] where H is the corresponding subgroup i.e. H = \text{Gal}(L_1/K_1). Thus, K_1 corresponds to the subgroup of S_3 which has index 2 i.e. that is H = A_3 and there is only one such subgroup. Therefore, there is only one subfield of L_1 which has [K_1:\mathbb{Q}]=2. It is clear to see that [\mathbb{Q}(\zeta):\mathbb{Q}]=2 and so K_1 = \mathbb{Q}(\zeta).
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  3. #3
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    Ahh, ok, thanks a lot. I think I got the idea.

    I will try to figure out K_2 then.
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  4. #4
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    Quote Originally Posted by Carl View Post
    Ahh, ok, thanks a lot. I think I got the idea.

    I will try to figure out K_2 then.
    The hint that I can give you is that \text{Gal}(L_2/\mathbb{Q}) = \mathbb{Z}_2\times \mathbb{Z}_2 and so there are 3 subgroups of index two, these will correspond to [K_2:\mathbb{Q}]=2. Thus, you will have three such subfields. For this problem this post may help you.
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  5. #5
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    Ok, I think I got it now:
    So, I say that Gal(L_2/ \mathbb{Q})=V
    And since V has 3 subgroups of order 2, which are \langle \sigma\rangle, \langle \tau \rangle and \langle \sigma\tau\rangle, then there are 3 \alpha such that [\mathbb{Q}(\alpha):\mathbb{Q}]=2 and those are \sqrt{2}i, i and \sqrt{2} since the splitting field for x^4+1 is \mathbb{Q}(\sqrt{2},i).
    And then I just have to check that those 3 are independent from each other and thereby I have found the 3 K_2's
    Did I get that right?
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  6. #6
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    Quote Originally Posted by Carl View Post
    Ok, I think I got it now:
    So, I say that Gal(L_2/ \mathbb{Q})=V
    And since V has 3 subgroups of order 2, which are \langle \sigma\rangle, \langle \tau \rangle and \langle \sigma\tau\rangle, then there are 3 \alpha such that [\mathbb{Q}(\alpha):\mathbb{Q}]=2 and those are \sqrt{2}i, i and \sqrt{2} since the splitting field for x^4+1 is \mathbb{Q}(\sqrt{2},i).
    And then I just have to check that those 3 are independent from each other and thereby I have found the 3 K_2's
    Did I get that right?
    Yes, , and it is easy to show that \mathbb{Q}(\sqrt{2}),\mathbb{Q}(i),\mathbb{Q}(i\sq  rt{2}) are distinct. It is easy to see they are distinct because \mathbb{Q}(i) = \{ a+bi|a,b\in \mathbb{Q}\}, \mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}|a,b\in \mathbb{Q}\}, and \mathbb{Q}(i\sqrt{2}) = \{ a + bi\sqrt{2}|a,b\in \mathbb{Q}\}. Now it is clear that these are distinct because \sqrt{2} is irrational.
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  7. #7
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    Allright, thanks a lot, you have been most helpful
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