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Thread: Kummer Theory

  1. #1
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    Kummer Theory

    Let $\displaystyle K/F$ be a Kummer $\displaystyle n$-extension. Write $\displaystyle K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$. If we define $\displaystyle \text{KUM}(K/F) = \{a\in K^{\times} | a^n \in F^{\times}\}$ and $\displaystyle \text{kum}(K/F) = \text{KUM}(K/F)/F^{\times}$ then it can be shown that $\displaystyle \text{Gal}(K/F) \simeq \text{kum}(K/F)$. Therefore, the degree of the extension, $\displaystyle [K:F]$, is $\displaystyle |\text{kum}(K/F)|$. Define the embedding $\displaystyle \phi: \text{kum}(K/F) \mapsto F^{\times}/(F^{\times})^n$ by $\displaystyle \phi(\alpha F^{\times}) = \alpha^n (F^{\times})^n$. Then, $\displaystyle |\text{im}(\phi)| = |\text{kum}(K/F)| = [K:F]$. It turns out that $\displaystyle \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$ (this notation means 'group generated by'). By the embedding it means $\displaystyle [K:F] = |\left< a_1(F^{\times})^n,...,a_r(F^{\times})^n \right>|$. And this gives us a way to find the degree of a Kummer extension. What I do not see is why $\displaystyle \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. I tried out some examples with $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3})$ and convinced myself that the only way to get an element when squared is when the element is a product of the $\displaystyle n$-th roots. But I cannot find a nice proof to this.
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    Quote Originally Posted by ThePerfectHacker View Post
    What I do not see is why $\displaystyle \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$.
    that's a fairly quick result of this theorem (before i explain why, you need to tell me if you basically know this theorem or not):

    Theorem: the map $\displaystyle H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n$ onto the class of all Kummer

    n-extensions of $\displaystyle F.$
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    Quote Originally Posted by NonCommAlg View Post
    that's a fairly quick result of this theorem (before i explain why, you need to tell me if you basically know this theorem or not):

    Theorem: the map $\displaystyle H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n$ onto the class of all Kummer

    n-extensions of $\displaystyle F.$
    No, I am not familar with this theorem. I do not even understand the notation you are using.

    However, I did think of an elementary way of proving $\displaystyle \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. We note that every element in $\displaystyle K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ has to be expressed in terms of powers and linear combinations of $\displaystyle \sqrt[n]{a_i}$'s. For example, consider $\displaystyle K=F(\sqrt{2},\sqrt{3})$ (here $\displaystyle F=\mathbb{Q}$) then $\displaystyle \{1,\sqrt{2}\}$ is basis for $\displaystyle F(\sqrt{2})/F$ and $\displaystyle \{1,\sqrt{3}\}$ is basis for $\displaystyle F(\sqrt{2},\sqrt{3})/F(\sqrt{2})$. Thus, basis for $\displaystyle K/F$ is $\displaystyle \{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}$. Therefore, any element can be written as $\displaystyle a+b\sqrt{2}+c\sqrt{3}+d\sqrt{2}\cdot \sqrt{3}$. And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the $\displaystyle n$-th roots. Let $\displaystyle \{x_1,...,x_m\}$ be a basis for $\displaystyle K= F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ where $\displaystyle x_i$'s are products of the $\displaystyle \sqrt[n]{a_j}$'s by above. If $\displaystyle \alpha \in K$ then we can write $\displaystyle \alpha = f(x_1,...,x_m)$ where $\displaystyle f$ is a polynomial over $\displaystyle F$ with $\displaystyle m$ variables. Assume that $\displaystyle \alpha$ has the property that $\displaystyle \alpha^n \in F$. We wish to argue that all terms of $\displaystyle f(x_1,...,x_m)$ must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the $\displaystyle n$-th roots. Suppose this was not the case and there were two coefficients for $\displaystyle x_i,x_j$ which were non-zero. Then in the expansion of $\displaystyle \alpha^n = f^n (x_1,...,x_m)$ we would have a non-zero "cross term" $\displaystyle x_ix_j$. But then this means $\displaystyle \alpha^n \not \in F$ by the uniqueness of the representation of an element in terms of a basis.
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    Quote Originally Posted by ThePerfectHacker View Post
    No, I am not familar with this theorem. I do not even understand the notation you are using.

    However, I did think of an elementary way of proving $\displaystyle \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. We note that every element in $\displaystyle K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ has to be expressed in terms of powers and linear combinations of $\displaystyle \sqrt[n]{a_i}$'s. For example, consider $\displaystyle K=F(\sqrt{2},\sqrt{3})$ (here $\displaystyle F=\mathbb{Q}$) then $\displaystyle \{1,\sqrt{2}\}$ is basis for $\displaystyle F(\sqrt{2})/F$ and $\displaystyle \{1,\sqrt{3}\}$ is basis for $\displaystyle F(\sqrt{2},\sqrt{3})/F(\sqrt{2})$. Thus, basis for $\displaystyle K/F$ is $\displaystyle \{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}$. Therefore, any element can be written as $\displaystyle a+b\sqrt{2}+c\sqrt{3}+d\sqrt{2}\cdot \sqrt{3}$. And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the $\displaystyle n$-th roots. Let $\displaystyle \{x_1,...,x_m\}$ be a basis for $\displaystyle K= F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ where $\displaystyle x_i$'s are products of the $\displaystyle \sqrt[n]{a_j}$'s by above. If $\displaystyle \alpha \in K$ then we can write $\displaystyle \alpha = f(x_1,...,x_m)$ where $\displaystyle f$ is a polynomial over $\displaystyle F$ with $\displaystyle m$ variables. Assume that $\displaystyle \alpha$ has the property that $\displaystyle \alpha^n \in F$. We wish to argue that all terms of $\displaystyle f(x_1,...,x_m)$ must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the $\displaystyle n$-th roots. Suppose this was not the case and there were two coefficients for $\displaystyle x_i,x_j$ which were non-zero. Then in the expansion of $\displaystyle \alpha^n = f^n (x_1,...,x_m)$ we would have a non-zero "cross term" $\displaystyle x_ix_j$. But then this means $\displaystyle \alpha^n \not \in F$ by the uniqueness of the representation of an element in terms of a basis.
    this proof is not complete! you can't just extend things from a simple extension such as $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3})$ to the general case! ok, let's see:

    i think you chose your basis to be all possible products of powers of $\displaystyle \sqrt[n]{a_j}.$ then you need to show that this set is linearly independent

    over F. then when you choose an element of $\displaystyle \alpha$ (which by the way would be just a simple linear combination of $\displaystyle x_j$ and you don't need

    to consider it as a polynomial $\displaystyle f(x_1, \ ... \ , x_m)$), then in computing $\displaystyle \alpha^n$ the charactersitic of F will also get involved!
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    Quote Originally Posted by NonCommAlg View Post
    this proof is not complete! you can't just extend things from a simple extension such as $\displaystyle \mathbb{Q}(\sqrt{2},\sqrt{3})$ to the general case!
    Consider this. Let $\displaystyle K=F(\sqrt[4]{a_1},\sqrt[4]{a_2},\sqrt[4]{a_3})$. Then $\displaystyle \{1,\sqrt[4]{a_1},\sqrt[4]{a_1}^2,\sqrt[4]{a_1}^3\}$ will spam $\displaystyle F(\sqrt[4]{a_1})/F$. It is not necessarily a basis so $\displaystyle \{1,\sqrt[4]{a_1}^2\}$ is a basis - say for simplicity sake. Now $\displaystyle \{1,\sqrt[4]{a_2},\sqrt[4]{a_2}^2,\sqrt[4]{a_3}^3\}$ will spam $\displaystyle F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F(\sqrt[4]{a_1})$. But it is not necessarily a basis. Say $\displaystyle \{1,\sqrt[4]{a_2}\}$ is this basis. Thus, $\displaystyle \{1,\sqrt[4]{a_1}^2,\sqrt[4]{a_2},\sqrt[4]{a_1}^2\sqrt[4]{a_2}\}$ is basis for $\displaystyle F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F$. And so on. The same idea for $\displaystyle K/F$. Thus, any element is a linear combination of the extraction of roots in general.

    to consider it as a polynomial $\displaystyle f(x_1, \ ... \ , x_m)$), then in computing $\displaystyle \alpha^n$ the charactersitic of F will also get involved!
    But the charachteristic of $\displaystyle F$ should not come into play, should it? Because since $\displaystyle F$ has all $\displaystyle n$-roots of unity it means the characheristic of $\displaystyle F$ does not divide $\displaystyle n$. Thus, we do not get multiple of $\displaystyle n$ when we expand $\displaystyle f(x_1,...,x_m)^n$. Do you buy that?
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    Quote Originally Posted by ThePerfectHacker View Post
    Consider this. Let $\displaystyle K=F(\sqrt[4]{a_1},\sqrt[4]{a_2},\sqrt[4]{a_3})$. Then $\displaystyle \{1,\sqrt[4]{a_1},\sqrt[4]{a_1}^2,\sqrt[4]{a_1}^3\}$ will spam $\displaystyle F(\sqrt[4]{a_1})/F$. It is not necessarily a basis so $\displaystyle \{1,\sqrt[4]{a_1}^2\}$ is a basis - say for simplicity sake. Now $\displaystyle \{1,\sqrt[4]{a_2},\sqrt[4]{a_2}^2,\sqrt[4]{a_3}^3\}$ will spam $\displaystyle F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F(\sqrt[4]{a_1})$. But it is not necessarily a basis. Say $\displaystyle \{1,\sqrt[4]{a_2}\}$ is this basis. Thus, $\displaystyle \{1,\sqrt[4]{a_1}^2,\sqrt[4]{a_2},\sqrt[4]{a_1}^2\sqrt[4]{a_2}\}$ is basis for $\displaystyle F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F$. And so on. The same idea for $\displaystyle K/F$. Thus, any element is a linear combination of the extraction of roots in general.
    then you need to prove that in general K has a basis chosen from the set of products of powers of $\displaystyle \sqrt[n]{a_j}$ which is

    closed under multiplication ... and proving the linear independence in general is not an easy task!

    But the charachteristic of $\displaystyle F$ should not come into play, should it? Because since $\displaystyle F$ has all $\displaystyle n$-roots of unity it means the characheristic of $\displaystyle F$ does not divide $\displaystyle n$. Thus, we do not get multiple of $\displaystyle n$ when we expand $\displaystyle f(x_1,...,x_m)^n$. Do you buy that?
    i know that the char(F) doesn't divide n, but char(F) can still divide some of the coefficients in the expansion of $\displaystyle \alpha^n.$
    Last edited by NonCommAlg; Jul 9th 2008 at 11:12 PM.
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    Quote Originally Posted by NonCommAlg View Post
    then you need to prove that in general K has a basis chosen from the set of products of powers of $\displaystyle \sqrt[n]{a_j}$ which is

    closed under multiplication ... and proving the linear independence in general is not an easy task!
    Let $\displaystyle E_2/E_1$ be an algebraic extension and let $\displaystyle \alpha \in E_2$. It is a known result that if $\displaystyle \deg(\alpha,E_2) = l$ then $\displaystyle \{1,\alpha,...,\alpha^{l-1}\}$ is a basis for $\displaystyle E_1(\alpha)/E_1$. Furthermore, if $\displaystyle E_1\subseteq E_2\subseteq E_3$ are fields with $\displaystyle E_3/E_2$ having basis $\displaystyle \{\gamma_i : i \in I\}$ and $\displaystyle E_2/E_1$ having basis $\displaystyle \{\delta_j : j\in J\}$ then $\displaystyle \{ \gamma_i\delta_j\}$ is basis for $\displaystyle E_3/E_1$. Now let us apply these facts to $\displaystyle K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$. Let $\displaystyle \alpha_k = \sqrt[n]{a_k}$. Now define $\displaystyle L_0=F,L_1=F(\alpha_1),L_2=F(\alpha_1,\alpha_2),... ,L_r = F(\alpha_1,...,\alpha_r) = K$. We have the tower of fields $\displaystyle L_0\subseteq L_1\subseteq L_2\subseteq ... \subseteq L_r$. Let $\displaystyle d_i = [L_i : L_{i-1}]$ for $\displaystyle 1\leq i\leq n$. Because $\displaystyle x^n - a_i$ has $\displaystyle \alpha_i$ as a root it means $\displaystyle d_i\leq n$. This means $\displaystyle \{1,\alpha_i,...,\alpha_i^{d_i}\}$ is a basis for $\displaystyle L_i/L_{i-1}$. Thus, $\displaystyle B=\{ \alpha_1^{k_1}\alpha_2^{k_2}\cdot ... \cdot \alpha_r^{k_r}| 0\leq k_i\leq d_i \}$ is a basis for $\displaystyle K=L_r/L_0=F$. Let $\displaystyle x_1,...,x_m$ be all the elements of $\displaystyle B$. This means if $\displaystyle \beta \in K$ then we can write $\displaystyle \beta = \Sigma_{j=1}^m c_jx_j$.

    If $\displaystyle \beta^n \in L_0$ then all $\displaystyle c_j$'s are zero expect for possibly one of them. We will assume $\displaystyle \text{char}(L_0) =0$. If this was not the case then there are $\displaystyle x_1,x_2$ (by relabeling if necessary) in the linear expansion of $\displaystyle \beta$ such that the coefficients of $\displaystyle x_1,x_2$ are non-zero. Then in the expansion of $\displaystyle (c_1x_1+...+c_mx_m)^n$ we will have a 'cross term' $\displaystyle x_1x_2$ which a non-zero coefficient. Let $\displaystyle x_1 = \alpha_1^{k_1}\cdot ... \cdot \alpha_r^{k_r}$ and $\displaystyle x_2 = \alpha_1^{s_1}\cdot ... \cdot \alpha_r^{s_r}$ where $\displaystyle 0\leq k_j,s_j\leq d_j$. Then the cross term $\displaystyle x_1x_2 = \alpha_1^{k_1+s_1}\cdot ... \cdot \alpha_r^{k_r+s_r}$. Now realize that $\displaystyle 0\leq k_j+s_j\leq 2d_j$. For this to be an element of $\displaystyle L_0$ we need that $\displaystyle k_j+s_j$ are multiples of $\displaystyle d_j$. Note that if $\displaystyle k_j+s_j>n$ then we can write $\displaystyle \alpha_j^{k_j+s_j} = \alpha_j^n \alpha_j^{k_j+s_j-n}$ with $\displaystyle \alpha_j^n$. Because of the constrainst it is impossible for all $\displaystyle k_j+s_j$ to be multiples of $\displaystyle d_j$ unless $\displaystyle x_1=x_2$ which is not the case. It seems that the problem is that that we do not know if there is a triple cross term, for example, $\displaystyle x_1x_2x_3$ whose basic reduced form (of the exponents) is not the same as $\displaystyle x_1x_2$. And therefore it is possible we get cancellation from the cross terms.
    Last edited by ThePerfectHacker; Jul 11th 2008 at 10:34 AM.
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    Quote Originally Posted by NonCommAlg View Post
    Theorem: the map $\displaystyle H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n$ onto the class of all Kummer
    How do you prove the generating fact using this theorem?
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    Quote Originally Posted by ThePerfectHacker View Post
    How do you prove the generating fact using this theorem?
    let $\displaystyle H_1=(K^{\times})^n \cap F^{\times},$ and $\displaystyle H_2=<(F^{\times})^n,a_1,a_2, \ ... \ , a_r>.$ then both $\displaystyle H_1$ and $\displaystyle H_2$ are subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n.$

    Claim: $\displaystyle F(H_1^{\frac{1}{n}})=F(H_2^{\frac{1}{n}})=K.$

    Proof: since $\displaystyle \sqrt[n]{a_j} \in H_1^{\frac{1}{n}},$ and $\displaystyle \sqrt[n]{a_j} \in H_2^{\frac{1}{n}},$ we must have $\displaystyle K \subseteq F(H_1^{\frac{1}{n}}),$ and $\displaystyle K \subseteq F(H_2^{\frac{1}{n}}).$ now if $\displaystyle x \in H_1^{\frac{1}{n}},$ then $\displaystyle x^n =\alpha^n,$ $\displaystyle \alpha \in K^{\times}.$

    thus $\displaystyle x=\alpha \xi,$ for some n-th root of unity $\displaystyle \xi.$ hence $\displaystyle x \in K.$ so $\displaystyle F(H_1^{\frac{1}{n}}) \subseteq K.$ finally if $\displaystyle x \in H_2^{\frac{1}{n}},$ then $\displaystyle x^n=u^n a_1^{t_1}a_2^{t_2} \ ... \ a_r^{t_r},$ for some

    $\displaystyle u \in F^{\times},$ which gives us $\displaystyle x=u\sqrt[n]{a_1^{t_1}} \sqrt[n]{a_2^{t_2}} \ ... \ \sqrt[n]{a_r^{t_r}} \xi,$ for some n-th root of unity $\displaystyle \xi.$ so $\displaystyle x \in K,$ and thus $\displaystyle F(H_2^{\frac{1}{n}}) \subseteq K. \ \ \ \ \square$

    so by the theorem $\displaystyle H_1=H_2.$ now it's obvious that $\displaystyle \text{kum}(K/F)=<\sqrt[n]{a_1}F^{\times}, \sqrt[n]{a_2}F^{\times}, \ ... \ , \sqrt[n]{a_r}F^{\times}>.$
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    Are you defining $\displaystyle H_1^{1/n} = \{ x\in H_1 | x^n \in H_1 \}$?
    I just never seen that notation before.

    EDIT: Yes it is. Sorry, I was going to delete this but I see you are replying.
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    Quote Originally Posted by ThePerfectHacker View Post
    Are you defining $\displaystyle H_1^{1/n} = \{ x\in H_1 | x^n \in H_1 \}$?
    I just never seen that notation before.
    $\displaystyle x \in H_1??$ no! $\displaystyle x \in H_1^{\frac{1}{n}},$ if, in some extension of $\displaystyle H_1$, it's a root of $\displaystyle t^n - \alpha = 0,$ for some $\displaystyle \alpha \in H_1.$
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    Quote Originally Posted by NonCommAlg View Post
    Theorem: the map $\displaystyle H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n$ onto the class of all Kummer n-extensions of $\displaystyle F.$
    I would restate the theorem as: $\displaystyle H\mapsto F(H^{1/n})$ is a bijection from the set of all subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n$ onto the set of all Kummer n-extension of $\displaystyle F$ which are contained in $\displaystyle \bar F$ (the algebraic closure of $\displaystyle F$). Is this okay (this does not damage the theorem in any way) ?

    Quote Originally Posted by NonCommAlg View Post
    $\displaystyle x \in H_1??$ no! $\displaystyle x \in H_1^{\frac{1}{n}},$ if, in some extension of $\displaystyle H_1$, it's a root of $\displaystyle t^n - \alpha = 0,$ for some $\displaystyle \alpha \in H_1.$
    That was stupid of me. When I looked at your proof I used $\displaystyle \{x\in \bar F | x^n \in H_1\}$.


    Quote Originally Posted by NonCommAlg View Post
    let $\displaystyle H_1=(K^{\times})^n \cap F^{\times},$ and $\displaystyle H_2=<(F^{\times})^n,a_1,a_2, \ ... \ , a_r>.$ then both $\displaystyle H_1$ and $\displaystyle H_2$ are subgroups of $\displaystyle F^{\times}$ which contain $\displaystyle (F^{\times})^n.$

    Claim: $\displaystyle F(H_1^{\frac{1}{n}})=F(H_2^{\frac{1}{n}})=K.$

    Proof: since $\displaystyle \sqrt[n]{a_j} \in H_1^{\frac{1}{n}},$ and $\displaystyle \sqrt[n]{a_j} \in H_2^{\frac{1}{n}},$ we must have $\displaystyle K \subseteq F(H_1^{\frac{1}{n}}),$ and $\displaystyle K \subseteq F(H_2^{\frac{1}{n}}).$ now if $\displaystyle x \in H_1^{\frac{1}{n}},$ then $\displaystyle x^n =\alpha^n,$ $\displaystyle \alpha \in K^{\times}.$

    thus $\displaystyle x=\alpha \xi,$ for some n-th root of unity $\displaystyle \xi.$ hence $\displaystyle x \in K.$ so $\displaystyle F(H_1^{\frac{1}{n}}) \subseteq K.$ finally if $\displaystyle x \in H_2^{\frac{1}{n}},$ then $\displaystyle x^n=u^n a_1^{t_1}a_2^{t_2} \ ... \ a_r^{t_r},$ for some

    $\displaystyle u \in F^{\times},$ which gives us $\displaystyle x=u\sqrt[n]{a_1^{t_1}} \sqrt[n]{a_2^{t_2}} \ ... \ \sqrt[n]{a_r^{t_r}} \xi,$ for some n-th root of unity $\displaystyle \xi.$ so $\displaystyle x \in K,$ and thus $\displaystyle F(H_2^{\frac{1}{n}}) \subseteq K. \ \ \ \ \square$

    so by the theorem $\displaystyle H_1=H_2.$ now it's obvious that $\displaystyle \text{kum}(K/F)=<\sqrt[n]{a_1}F^{\times}, \sqrt[n]{a_2}F^{\times}, \ ... \ , \sqrt[n]{a_r}F^{\times}>.$
    I think I understand your proof if I do it in the following manner: Given $\displaystyle K$ and $\displaystyle F$. Construct $\displaystyle \bar F$ so that $\displaystyle K\subset \bar F$. Define $\displaystyle H_1^{1/n} = \{x\in \bar F | x^n \in H_1\}$ and $\displaystyle H_2^{1/n} = \{ x\in \bar F | x^n \in H_2\}$. So this is basically what you are doing except I picked a big extension field to work under - this makes me feel safe. And now I just carry through the exact same argument. I think that is okay to do.

    My last question is how do you prove this bijection theorem between subgroups and Kummer n-extensions. In fact, I do not even see why $\displaystyle F(H^{1/n})$ is a Kummer n-extension over $\displaystyle F$. The definition I am using is that $\displaystyle K/F$ is a Kummer n-extension when $\displaystyle K/F$ is Galois and $\displaystyle F$ has primitive $\displaystyle n$-th root of unity and $\displaystyle \text{Gal}(K/F)$ is a finite abelian group whose exponent divides $\displaystyle n$. I can see that $\displaystyle F(H^{1/n})$ is a splitting field over $\displaystyle \{ x^n - h | h\in H\}$ and each polynomial is seperable over $\displaystyle F$. However, I do not see why $\displaystyle \text{Gal}(F(H^{1/n})/F)$ is a finite abelian group whose exponent divides $\displaystyle n$.
    Note, I am familar with the theorem that $\displaystyle B: \text{Gal}(K/F)\times \text{kum}(K/F)\to \{1,\zeta,...,\zeta^{n-1}\}$ defined by $\displaystyle B(\sigma, \alpha F^{\times}) = \sigma(\alpha)/\alpha$ is a non-degenerate pairing. And so $\displaystyle \text{kum}(K/F) \simeq \text{hom}(\text{Gal}(K/F), \{1,...,\zeta^{n-1}\})$. Maybe your bijection theorem is somehow related to it, but I do not see the connection.

    Congratulations on your 1th post!
    Last edited by ThePerfectHacker; Jul 17th 2008 at 09:05 PM.
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    Here is another approach that my professor showed me without using that theorem.

    We know that $\displaystyle ((F^{\times})^{1/n} \cap K^{\times})/F^{\times}$ (the Kummer group) is isomorphic to $\displaystyle \text{Gal}(K/F)$. Let $\displaystyle H = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times} \right> $ and $\displaystyle U$ be the $\displaystyle n$-th roots of unity. Define $\displaystyle B: H\times \text{Gal}(K/F)\to U$ by $\displaystyle B(\alpha F^{\times},\sigma) = \sigma(\alpha)/\alpha$. This is a bilinear pairing, since it is non-degenerate it means $\displaystyle H\simeq \text{hom}(\text{Gal}(K/F),U)$. But $\displaystyle \text{hom}(\text{Gal}(K/F),U)\simeq \text{hom}(\text{Gal}(K/F),\mathbb{C})\simeq \text{Gal}(K/F)$ since $\displaystyle \text{Gal}(K/F)$ is a finite abelian group it is isomorphic to its charachter group. And this means that $\displaystyle H$ is isomorphic to the Kummer group, since $\displaystyle H$ is a subgroup of the Kummer group it means that $\displaystyle H$ in fact the Kummer group.
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