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Math Help - Kummer Theory

  1. #1
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    Kummer Theory

    Let K/F be a Kummer n-extension. Write K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r}). If we define \text{KUM}(K/F) = \{a\in K^{\times} | a^n \in F^{\times}\} and \text{kum}(K/F) = \text{KUM}(K/F)/F^{\times} then it can be shown that \text{Gal}(K/F) \simeq \text{kum}(K/F). Therefore, the degree of the extension, [K:F], is |\text{kum}(K/F)|. Define the embedding \phi: \text{kum}(K/F) \mapsto F^{\times}/(F^{\times})^n by \phi(\alpha F^{\times}) = \alpha^n (F^{\times})^n. Then, |\text{im}(\phi)| = |\text{kum}(K/F)| = [K:F]. It turns out that \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 [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 \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>. I tried out some examples with \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 n-th roots. But I cannot find a nice proof to this.
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  2. #2
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    Quote Originally Posted by ThePerfectHacker View Post
    What I do not see is why \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 H \mapsto F(H^{\frac{1}{n}}) is a bijection from the class of all subgroups of F^{\times} which contain (F^{\times})^n onto the class of all Kummer

    n-extensions of 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 H \mapsto F(H^{\frac{1}{n}}) is a bijection from the class of all subgroups of F^{\times} which contain (F^{\times})^n onto the class of all Kummer

    n-extensions of 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 \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>. We note that every element in K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r}) has to be expressed in terms of powers and linear combinations of \sqrt[n]{a_i}'s. For example, consider K=F(\sqrt{2},\sqrt{3}) (here F=\mathbb{Q}) then \{1,\sqrt{2}\} is basis for F(\sqrt{2})/F and \{1,\sqrt{3}\} is basis for F(\sqrt{2},\sqrt{3})/F(\sqrt{2}). Thus, basis for K/F is \{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}. Therefore, any element can be written as 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 n-th roots. Let \{x_1,...,x_m\} be a basis for K=  F(\sqrt[n]{a_1},...,\sqrt[n]{a_r}) where x_i's are products of the \sqrt[n]{a_j}'s by above. If \alpha \in K then we can write \alpha = f(x_1,...,x_m) where f is a polynomial over F with m variables. Assume that \alpha has the property that \alpha^n \in F. We wish to argue that all terms of 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 n-th roots. Suppose this was not the case and there were two coefficients for x_i,x_j which were non-zero. Then in the expansion of \alpha^n = f^n (x_1,...,x_m) we would have a non-zero "cross term" x_ix_j. But then this means \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 \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>. We note that every element in K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r}) has to be expressed in terms of powers and linear combinations of \sqrt[n]{a_i}'s. For example, consider K=F(\sqrt{2},\sqrt{3}) (here F=\mathbb{Q}) then \{1,\sqrt{2}\} is basis for F(\sqrt{2})/F and \{1,\sqrt{3}\} is basis for F(\sqrt{2},\sqrt{3})/F(\sqrt{2}). Thus, basis for K/F is \{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}. Therefore, any element can be written as 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 n-th roots. Let \{x_1,...,x_m\} be a basis for K= F(\sqrt[n]{a_1},...,\sqrt[n]{a_r}) where x_i's are products of the \sqrt[n]{a_j}'s by above. If \alpha \in K then we can write \alpha = f(x_1,...,x_m) where f is a polynomial over F with m variables. Assume that \alpha has the property that \alpha^n \in F. We wish to argue that all terms of 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 n-th roots. Suppose this was not the case and there were two coefficients for x_i,x_j which were non-zero. Then in the expansion of \alpha^n = f^n (x_1,...,x_m) we would have a non-zero "cross term" x_ix_j. But then this means \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 \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 \sqrt[n]{a_j}. then you need to show that this set is linearly independent

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

    to consider it as a polynomial f(x_1, \ ... \ , x_m)), then in computing \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 \mathbb{Q}(\sqrt{2},\sqrt{3}) to the general case!
    Consider this. Let K=F(\sqrt[4]{a_1},\sqrt[4]{a_2},\sqrt[4]{a_3}). Then \{1,\sqrt[4]{a_1},\sqrt[4]{a_1}^2,\sqrt[4]{a_1}^3\} will spam F(\sqrt[4]{a_1})/F. It is not necessarily a basis so \{1,\sqrt[4]{a_1}^2\} is a basis - say for simplicity sake. Now \{1,\sqrt[4]{a_2},\sqrt[4]{a_2}^2,\sqrt[4]{a_3}^3\} will spam F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F(\sqrt[4]{a_1}). But it is not necessarily a basis. Say \{1,\sqrt[4]{a_2}\} is this basis. Thus, \{1,\sqrt[4]{a_1}^2,\sqrt[4]{a_2},\sqrt[4]{a_1}^2\sqrt[4]{a_2}\} is basis for F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F. And so on. The same idea for K/F. Thus, any element is a linear combination of the extraction of roots in general.

    to consider it as a polynomial f(x_1, \ ... \ , x_m)), then in computing \alpha^n the charactersitic of F will also get involved!
    But the charachteristic of F should not come into play, should it? Because since F has all n-roots of unity it means the characheristic of F does not divide n. Thus, we do not get multiple of n when we expand f(x_1,...,x_m)^n. Do you buy that?
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    Quote Originally Posted by ThePerfectHacker View Post
    Consider this. Let K=F(\sqrt[4]{a_1},\sqrt[4]{a_2},\sqrt[4]{a_3}). Then \{1,\sqrt[4]{a_1},\sqrt[4]{a_1}^2,\sqrt[4]{a_1}^3\} will spam F(\sqrt[4]{a_1})/F. It is not necessarily a basis so \{1,\sqrt[4]{a_1}^2\} is a basis - say for simplicity sake. Now \{1,\sqrt[4]{a_2},\sqrt[4]{a_2}^2,\sqrt[4]{a_3}^3\} will spam F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F(\sqrt[4]{a_1}). But it is not necessarily a basis. Say \{1,\sqrt[4]{a_2}\} is this basis. Thus, \{1,\sqrt[4]{a_1}^2,\sqrt[4]{a_2},\sqrt[4]{a_1}^2\sqrt[4]{a_2}\} is basis for F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F. And so on. The same idea for 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 \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 F should not come into play, should it? Because since F has all n-roots of unity it means the characheristic of F does not divide n. Thus, we do not get multiple of n when we expand 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 \alpha^n.
    Last edited by NonCommAlg; July 9th 2008 at 11:12 PM.
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  7. #7
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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 \sqrt[n]{a_j} which is

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

    If \beta^n \in L_0 then all c_j's are zero expect for possibly one of them. We will assume \text{char}(L_0) =0. If this was not the case then there are x_1,x_2 (by relabeling if necessary) in the linear expansion of \beta such that the coefficients of x_1,x_2 are non-zero. Then in the expansion of (c_1x_1+...+c_mx_m)^n we will have a 'cross term' x_1x_2 which a non-zero coefficient. Let x_1 = \alpha_1^{k_1}\cdot ... \cdot \alpha_r^{k_r} and x_2 = \alpha_1^{s_1}\cdot ... \cdot \alpha_r^{s_r} where 0\leq k_j,s_j\leq d_j. Then the cross term x_1x_2 = \alpha_1^{k_1+s_1}\cdot ... \cdot \alpha_r^{k_r+s_r}. Now realize that 0\leq k_j+s_j\leq 2d_j. For this to be an element of L_0 we need that k_j+s_j are multiples of d_j. Note that if k_j+s_j>n then we can write \alpha_j^{k_j+s_j} = \alpha_j^n \alpha_j^{k_j+s_j-n} with \alpha_j^n. Because of the constrainst it is impossible for all k_j+s_j to be multiples of d_j unless 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, x_1x_2x_3 whose basic reduced form (of the exponents) is not the same as x_1x_2. And therefore it is possible we get cancellation from the cross terms.
    Last edited by ThePerfectHacker; July 11th 2008 at 10:34 AM.
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  8. #8
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    Quote Originally Posted by NonCommAlg View Post
    Theorem: the map H \mapsto F(H^{\frac{1}{n}}) is a bijection from the class of all subgroups of F^{\times} which contain (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 H_1=(K^{\times})^n \cap F^{\times}, and H_2=<(F^{\times})^n,a_1,a_2, \ ... \ , a_r>. then both H_1 and H_2 are subgroups of F^{\times} which contain (F^{\times})^n.

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

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

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

    u \in F^{\times}, which gives us 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 \xi. so x \in K, and thus F(H_2^{\frac{1}{n}}) \subseteq K. \ \ \ \ \square

    so by the theorem H_1=H_2. now it's obvious that \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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  10. #10
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    Are you defining 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 H_1^{1/n} = \{ x\in H_1 | x^n \in H_1 \}?
    I just never seen that notation before.
    x \in H_1?? no! x \in H_1^{\frac{1}{n}}, if, in some extension of H_1, it's a root of t^n - \alpha = 0, for some \alpha \in H_1.
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    Quote Originally Posted by NonCommAlg View Post
    Theorem: the map H \mapsto F(H^{\frac{1}{n}}) is a bijection from the class of all subgroups of F^{\times} which contain (F^{\times})^n onto the class of all Kummer n-extensions of F.
    I would restate the theorem as: H\mapsto F(H^{1/n}) is a bijection from the set of all subgroups of F^{\times} which contain (F^{\times})^n onto the set of all Kummer n-extension of F which are contained in \bar F (the algebraic closure of F). Is this okay (this does not damage the theorem in any way) ?

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


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

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

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

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

    u \in F^{\times}, which gives us 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 \xi. so x \in K, and thus F(H_2^{\frac{1}{n}}) \subseteq K. \ \ \ \ \square

    so by the theorem H_1=H_2. now it's obvious that \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 K and F. Construct \bar F so that K\subset \bar F. Define H_1^{1/n} = \{x\in \bar F | x^n \in H_1\} and 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 F(H^{1/n}) is a Kummer n-extension over F. The definition I am using is that K/F is a Kummer n-extension when K/F is Galois and F has primitive n-th root of unity and \text{Gal}(K/F) is a finite abelian group whose exponent divides n. I can see that F(H^{1/n}) is a splitting field over \{ x^n - h | h\in H\} and each polynomial is seperable over F. However, I do not see why \text{Gal}(F(H^{1/n})/F) is a finite abelian group whose exponent divides n.
    Note, I am familar with the theorem that B: \text{Gal}(K/F)\times \text{kum}(K/F)\to \{1,\zeta,...,\zeta^{n-1}\} defined by B(\sigma, \alpha F^{\times}) = \sigma(\alpha)/\alpha is a non-degenerate pairing. And so \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; July 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 ((F^{\times})^{1/n} \cap K^{\times})/F^{\times} (the Kummer group) is isomorphic to \text{Gal}(K/F). Let H = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times} \right> and U be the n-th roots of unity. Define B: H\times \text{Gal}(K/F)\to U by B(\alpha F^{\times},\sigma) = \sigma(\alpha)/\alpha. This is a bilinear pairing, since it is non-degenerate it means H\simeq \text{hom}(\text{Gal}(K/F),U). But \text{hom}(\text{Gal}(K/F),U)\simeq \text{hom}(\text{Gal}(K/F),\mathbb{C})\simeq \text{Gal}(K/F) since \text{Gal}(K/F) is a finite abelian group it is isomorphic to its charachter group. And this means that H is isomorphic to the Kummer group, since H is a subgroup of the Kummer group it means that H in fact the Kummer group.
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