1. 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.

2. Originally Posted by ThePerfectHacker
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 $F.$

3. Originally Posted by NonCommAlg
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.

4. Originally Posted by ThePerfectHacker
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!

5. Originally Posted by NonCommAlg
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?

6. Originally Posted by ThePerfectHacker
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 $\alpha^n.$

7. Originally Posted by NonCommAlg
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.

8. Originally Posted by NonCommAlg
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?

9. Originally Posted by ThePerfectHacker
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 $\text{kum}(K/F)=<\sqrt[n]{a_1}F^{\times}, \sqrt[n]{a_2}F^{\times}, \ ... \ , \sqrt[n]{a_r}F^{\times}>.$

10. 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.

11. Originally Posted by ThePerfectHacker
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 $\alpha \in H_1.$

12. Originally Posted by NonCommAlg
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) ?

Originally Posted by NonCommAlg
$x \in H_1??$ 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 $\{x\in \bar F | x^n \in H_1\}$.

Originally Posted by NonCommAlg
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.

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.