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I am trying to prove the following:

Let $K=\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$. Show that $K/\mathbb{Q}$ is Galois with Galois group $(\mathbb{Z}/2\mathbb{Z})^n$.

I have attached my proof below not as a means of verifying it, only as a means of making my question as clear as possible.

My question is:

Is it correct to say that that $\tau_i$'s generate $\operatorname{Gal}(K/\mathbb{Q})$, and therefore that we can decompose elements of $\operatorname{Gal}(K/\mathbb{Q})$ into compositions of these functions.

My proof is as follows:

To show $K/\mathbb{Q}$ is Galois we need to show it is the splitting field of a separable polynomial. The field $\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$ is obtained by an $n$ step process where step $i$ consists of adjoining $\sqrt{p_i}$ to the field $Q_{i-1}=\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_{i-1}})$ and $Q_0$ is defined to be $\mathbb{Q}$. This gives the following tower of extensions

$$\mathbb{Q}\subset \mathbb{Q}(\sqrt{p_1})\subset \mathbb{Q}(\sqrt{p_1},\sqrt{p_2})\subset \cdots\subset \mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}).$$

Note that $Q_i\neq Q_{i+1}$, because $\sqrt{p_{i+1}} \not \in Q_{i}$ for all $i$; the fact the $\sqrt{p_1},\ldots,\sqrt{p_n}$ are distinct primes insures this. Since none of the $p_j$ are squares (because they are prime) we know $\sqrt{p_j} \not \in \mathbb{Q}$. Furthermore, the distinctness of the $p_j$ insure that $\sqrt{p_i} \neq \sqrt{p_j}$ for all $i \neq j$.

The minimal polynomial of the extension $Q_{i+1}/Q_i$ is $f_{i+1}(X)=X^2-p_{i+1}$, which is separable with roots $\pm\sqrt{p_{i+1}}$. From this we see that the polynomial we are concerned with is

$$f(X)=f_1(X)\cdots f_n(X)=(X^2-p_1)\cdots(X^2-p_n),$$

also a separable polynomial. $K$ is a splitting field for this separable polynomial over $\mathbb{Q}$, and thus $K/\mathbb{Q}$ is Galois. From the tower of fields above we see they $[K:\mathbb{Q}]=2^n$, since the degree of each $Q_i/Q_{i-1}$ is 2. From this we can deduce that $\operatorname{Gal}(K/\mathbb{Q})=2^n$. As with before, $\operatorname{Gal}(K/\mathbb{Q})$ is determined by its action on the roots of $f(X)$, which are

$$\{\pm \sqrt{p_1},\ldots,\pm \sqrt{p_n}\},$$

or, more correctly, on the roots of the minimal polynomial of each of these roots. Consider the map \begin{equation*} \tau_{(a_1,\ldots,a_n)}= \begin{cases} \sqrt{p_1}& \mapsto \pm \sqrt{p_1},\\ \sqrt{p_2}& \mapsto \pm \sqrt{p_2},\\ & \vdots\\ \sqrt{p_n}& \mapsto \pm \sqrt{p_n}, \end{cases} \end{equation*} where $a_i\in \{0,1\}$ and $a_i=0$ means $\sqrt{p_i} \mapsto \sqrt{p_i}$ and $a_i=1$ means $\sqrt{p_i} \mapsto \sqrt{p_i}$. This gives $2n$ possible maps which take roots of the $f_i(X)$ to roots of $f_i(X)$ (note that these are all such maps as well). Since $|\operatorname{Gal}(K/\mathbb{Q})|=2^n$ all of these maps must be automorphisms of $K$.

What remains to be seen is that $\operatorname{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^n$. We can decompose $\tau_{(a_1,\ldots,a_n)}$ into a composition of the maps $\tau_i:K\rightarrow K$ defined by \begin{equation*} \tau_i = \begin{cases} \tau_i\left(\sqrt{p_j}\right) = -\sqrt{p_i} & \text{for } i=j,\\ \tau_i\left(\sqrt{p_j}\right) = \sqrt{p_j} & \text{for } i\neq j, \end{cases} \end{equation*} where $\tau_{(a_1,\ldots,a_n)}=\tau_1^{a_1}\circ \cdots \circ \tau_n^{a_n}$. We see that every element of $\operatorname{Gal}(K/\mathbb{Q})$ can be obtained through the composition of $\tau_i$'s and therefore $\{\tau_1,\ldots,\tau_n\}$ generates $\operatorname{Gal}(K/\mathbb{Q})$.

Let $\tau \in \operatorname{Gal}(K/\mathbb{Q})$ such that $\tau_1^{a_1}\circ \cdots \circ \tau_n^{a_n}$. Define the map $\chi: \operatorname{Gal}(K/\mathbb{Q}) \rightarrow \{\pm 1\}^n$ by first defining the action of $\chi$ on each $\tau_i$ by

$$\chi(\tau_i) \mapsto \tau_i(\sqrt{p_i})/\sqrt{p_i},$$

and then defining \begin{align*} \chi(\tau)&=\chi(\tau_1^{a_1}\circ \cdots \circ \tau_n^{a_n})\\ &=(\chi(\tau_1^{a_1}),\ldots,\chi(\tau_n^{a_n})). \end{align*} Once we show this map is injective we are done, as any injective map between finite sets of the same cardinality is surjective. Assume $(a_1,\ldots,a_n)=(b_1,\ldots,b_n)$. Then clearly $\tau_1^{a_1}\circ \cdots\circ \tau_n^{a_n}=\tau_1^{b_1}\circ \cdots \circ \tau_n^{b_n}$, since $a_i=b_i$ for all $i$.

Now, $\{\pm 1\}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, as the map $\varphi: \{\pm 1\} \rightarrow \mathbb{Z}/2\mathbb{Z}$ by $\varphi(-1)=1$ and $\varphi(1)=0$ is an injective homormorphism. Therefore

$$\operatorname{Gal}(K/\mathbb{Q})\cong (\mathbb{Z}/2\mathbb{Z})^n.$$

Attempted proof of Martin Brandenburg's Claim:

Claim: For all $i<n$ and $i<k_1<\ldots<k_s\leq n$, $\sqrt{p_{k_1}\ldots p_{k_s}} \not \in Q_i$.

Proceed by induction on $n$. If $i=0$, then we must show that $\sqrt{p_{k_1}\ldots p_{k_s}} \not \in \mathbb{Q}$, where $0<k_1<\ldots<k_s\leq n$. This is equivalent to showing that $p_{k_1}\ldots p_{k_s}$ is not a square, which is clear since $p_{k_1}, \ldots ,p_{k_s}$ are prime numbers. Assume the result is true for all $i<n-1$ and consider $i=n-1$. We need to show $\sqrt{p_n}\not \in Q_{n-1}$. Assume the contrary, that $\sqrt{p_n} \in Q_{n-1}$. Then

$$\sqrt{p_n}=\sum_{(\alpha_1,...,\alpha_{n-1})} a_{(\alpha_1,...,\alpha_{n-1})} \sqrt{{p_1}^{\alpha_1} \ldots p_{n-1}^{\alpha_{n-1}}}, \quad \dagger$$

where $\alpha_i \in \{0,1\}$ and $a_{(\alpha_1,...,\alpha_{n-1})}\in \mathbb{Q}$. Multiplying both sides of $(\dagger)$ by $\sqrt{p_n}$ gives

$$p_n=\sum_{(\alpha_1,...,\alpha_{n-1})} a_{(\alpha_1,...,\alpha_{n-1})} \sqrt{{p_1}^{\alpha_1} \ldots p_{n-1}^{\alpha_{n-1}}p_n},$$

contradicting the fact $\sqrt{{p_1}^{\alpha_1} \ldots p_{n-1}^{\alpha_{n-1}}p_n} \not \in \mathbb{Q}$. Therefore $\sqrt{p_n} \not \in Q_{n-1}$.

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4  
It is not as easy as you seem to think it is to show that, say, $\sqrt{11}$ is not in ${\bf Q}(\sqrt2,\sqrt3,\sqrt5,\sqrt7)$. –  Gerry Myerson May 28 '12 at 12:32
    
@gerrymyerson thanks for pointing that out. I will have to work on that part. –  Holdsworth88 May 28 '12 at 12:35
1  
It is yes for your question. Once you have showed that $\tau_i$ is really (well defined) a ring map, it is equivalent to show the fact as Gerry Myerson mentioned. And there is a typo, every $2n$ should be $2^n$, right? –  wxu May 28 '12 at 13:41
    
@wxu you are right, thank you for pointing that out. –  Holdsworth88 May 28 '12 at 13:46
2  
\begin{cases} a \\ b \end{cases}, with a comma after "\end{cases}", gives you this: $\begin{cases} a \\ b \end{cases},$ ${}\qquad{}$. But if you type \begin{cases} a, \\ b, \end{cases} with the commas inside, you get this: $\begin{cases} a, \\ b, \end{cases}$ ${}\qquad{}$. I changed your "array" to "cases" (simpler and often gives better results) and rearranged the commas. And did some other copy-editing. If you use $\LaTeX$ normally (as opposed to its use on web pages, where results vary), you will find that "..." does not look the same as \ldots. And \cdots looks good in some places. –  Michael Hardy May 28 '12 at 16:41

1 Answer 1

up vote 4 down vote accepted

The crucial point is: It is not trivial at all to see that $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree exactly $2^n$ (only $\leq 2^n$ is clear) over $\mathbb{Q}$, or equivalently, that $p_i \notin \mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_{i-1}})$ for all $i \leq n$.

Here is what you can do: Let $m_1,\dotsc,m_r$ be coprime and square free numbers $>1$ (for example distinct primes). Then I claim that $K_r := \mathbb{Q}(\sqrt{m_1},\dotsc,\sqrt{m_r})$ has degree $2^r$ over $\mathbb{Q}$.

It is enough to prove $\sqrt{m_{i+1}} \notin K_i$ for all $i < r$. But it turns out one should show something stronger: For $i<r$ and $i < j_1 < \dotsc < j_s \leq r$ we have $\sqrt{m_{j_1} \cdot \dotsc \cdot m_{j_s}} \notin K_i$. Now this can be shown by induction on $i$; I leave it for you as an exercise.

Now the rest is easy: $K_r$ is the splitting field of $(x^2-m_1) \cdot \cdots \cdot (x^2-m_r)$, hence normal, and everything in characteristic $0$ is separable. Thus $K_r$ is Galois over $\mathbb{Q}$; let $G$ be the Galois group. Since the conjugates of $\sqrt{m_i}$ are $\pm \sqrt{m_i}$, there is a homomorphism $\alpha : G \to \{\pm 1\}^r$ given by $\alpha(g) \sqrt{m_i} = g(\sqrt{m_i})$. Clearly it is injective. Since $G$ has order $2^r$, it is also surjective. Thus we have an isomorphism $G \cong \{\pm 1\}^r$. Explicitly, the Galois automorphisms are exactly as you've described them, they are given by $\sqrt{m_i} \mapsto \varepsilon_i \sqrt{m_i}$, where $\varepsilon_i = \pm 1$.

Remark: There are shorter proofs, using Kummer Theory.

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I added a proof of your claim, although I think it is incorrect. I feel I am missing something, perhaps messing the induction up (which feels quite stupid after the thousand induction proofs I have done). If you have a second could you look at it? –  Holdsworth88 May 30 '12 at 8:27

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