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Thread: Subfield

  1. #1
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    Subfield

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

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

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

    and then argue that $\displaystyle [K_1:\mathbb{Q}(\epsilon_3)]=[K_2:\mathbb{Q}(\epsilon_8)]=1$ 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 $\displaystyle K_1$ and $\displaystyle K_2$ of $\displaystyle L_1$ and $\displaystyle L_2$ with $\displaystyle [K_1:\mathbb{Q}]=2$ and $\displaystyle [K_2:\mathbb{Q}]=2$, where $\displaystyle L_1$ and $\displaystyle L_2$ are the splitting fields for the polynomials $\displaystyle g_1(x)=(x^3-2)$ and $\displaystyle g_2=(x^4+1)$ over $\displaystyle \mathbb{Q}$.
    (Remark: $\displaystyle L_2$ is the cyclotomic field $\displaystyle \mathbb{Q}_8$)

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

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

    and then argue that $\displaystyle [K_1:\mathbb{Q}(\epsilon_3)]=[K_2:\mathbb{Q}(\epsilon_8)]=1$ 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 $\displaystyle K_1$ and you think about $\displaystyle K_2$. Since $\displaystyle L_1$ is the splitting of $\displaystyle x^3 - 2$ over $\displaystyle \mathbb{Q}$ it means $\displaystyle L_1 = \mathbb{Q}(\zeta,\sqrt[3]{2})$. We will prove that $\displaystyle \text{Gal}(L_1/\mathbb{Q}) \simeq S_3$. If $\displaystyle \sigma$ is an automorphism of $\displaystyle L_1$ then it is completely determined by $\displaystyle \sigma (\zeta)$ and by $\displaystyle \sigma (\sqrt[3]{2})$. Now $\displaystyle \zeta$ is a zero of the irreducible polynomial $\displaystyle x^2+x+1$, this polynomial has zeros $\displaystyle \zeta,\zeta^2$, thus $\displaystyle \sigma (\zeta) = \zeta \text{ or }\zeta^2$. By similar reasoning $\displaystyle \sigma (\sqrt[3]{2}) = \sqrt[3]{2},\zeta\sqrt[3]{2},\text{ or }\zeta^2\sqrt[3]{2}$ since $\displaystyle \sqrt[3]{2}$ is a zero of the irreducible polynomial $\displaystyle x^3-2$. By isomorphism extension theorem there is an automorphism $\displaystyle \sigma$ so that $\displaystyle \sigma (\zeta) = \zeta^2$ and $\displaystyle \sqrt(\sqrt[3]{2}) = \sqrt[3]{2}$. Also, there is an automorphism $\displaystyle \tau$ so that $\displaystyle \tau (\zeta) = \zeta$ and $\displaystyle \tau (\sqrt[3]{2}) = \zeta\sqrt[3]{2}$. Notice that $\displaystyle \sigma (\zeta) = \zeta^2 = \overline{\zeta}$ so $\displaystyle \sigma$ is in fact complex-conjugation. We see that the order of $\displaystyle \sigma$ in the Galois group is $\displaystyle 2$ and order of $\displaystyle \tau$ in the Galois group is $\displaystyle 3$, furthermore $\displaystyle \text{Gal}(L_1/\mathbb{Q}) = \left< \sigma,\tau\right>\simeq S_3$. You want, as the problem says, to find all $\displaystyle K_1$ so that $\displaystyle [K_1:\mathbb{Q}] = 2$. Remember by the fundamental theorem of Galois theory $\displaystyle [K_1:\mathbb{Q}] = [S_3:H]$ where $\displaystyle H$ is the corresponding subgroup i.e. $\displaystyle H = \text{Gal}(L_1/K_1)$. Thus, $\displaystyle K_1$ corresponds to the subgroup of $\displaystyle S_3$ which has index $\displaystyle 2$ i.e. that is $\displaystyle H = A_3$ and there is only one such subgroup. Therefore, there is only one subfield of $\displaystyle L_1$ which has $\displaystyle [K_1:\mathbb{Q}]=2$. It is clear to see that $\displaystyle [\mathbb{Q}(\zeta):\mathbb{Q}]=2$ and so $\displaystyle K_1 = \mathbb{Q}(\zeta)$.
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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 $\displaystyle K_2$ 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 $\displaystyle K_2$ then.
    The hint that I can give you is that $\displaystyle \text{Gal}(L_2/\mathbb{Q}) = \mathbb{Z}_2\times \mathbb{Z}_2$ and so there are $\displaystyle 3$ subgroups of index two, these will correspond to $\displaystyle [K_2:\mathbb{Q}]=2$. 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 $\displaystyle Gal(L_2/ \mathbb{Q})=V$
    And since V has 3 subgroups of order 2, which are $\displaystyle \langle \sigma\rangle, \langle \tau \rangle$ and $\displaystyle \langle \sigma\tau\rangle$, then there are 3 $\displaystyle \alpha$ such that $\displaystyle [\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and those are $\displaystyle \sqrt{2}i, i $ and $\displaystyle \sqrt{2}$ since the splitting field for $\displaystyle x^4+1$ is $\displaystyle \mathbb{Q}(\sqrt{2},i)$.
    And then I just have to check that those 3 are independent from each other and thereby I have found the 3 $\displaystyle K_2$'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 $\displaystyle Gal(L_2/ \mathbb{Q})=V$
    And since V has 3 subgroups of order 2, which are $\displaystyle \langle \sigma\rangle, \langle \tau \rangle$ and $\displaystyle \langle \sigma\tau\rangle$, then there are 3 $\displaystyle \alpha$ such that $\displaystyle [\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and those are $\displaystyle \sqrt{2}i, i $ and $\displaystyle \sqrt{2}$ since the splitting field for $\displaystyle x^4+1$ is $\displaystyle \mathbb{Q}(\sqrt{2},i)$.
    And then I just have to check that those 3 are independent from each other and thereby I have found the 3 $\displaystyle K_2$'s
    Did I get that right?
    Yes, , and it is easy to show that $\displaystyle \mathbb{Q}(\sqrt{2}),\mathbb{Q}(i),\mathbb{Q}(i\sq rt{2})$ are distinct. It is easy to see they are distinct because $\displaystyle \mathbb{Q}(i) = \{ a+bi|a,b\in \mathbb{Q}\}$, $\displaystyle \mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}|a,b\in \mathbb{Q}\}$, and $\displaystyle \mathbb{Q}(i\sqrt{2}) = \{ a + bi\sqrt{2}|a,b\in \mathbb{Q}\}$. Now it is clear that these are distinct because $\displaystyle \sqrt{2}$ 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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