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I'm trying to answer the following question:

"Suppose that $F=K(x_1,\dots,x_n)$ where $0\neq x_i^2=a_i\in K$. Show that $F/K$ is a Galois extension, with Galois group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^m$ for some $m\leq n$."

My first thought was show that $[F:K]=2^m$ and then find $2^m$ K-homs (and hence automorphisms) from $F$ to $F$ and thus the Galois group.

I'm wondering if this is the "right" approach, since it seems we need to define maps sending $x_i$ to $\pm x_i$, and then prove that these are homs by considering $F$ as a $K$-vector space, which, at first sight, appears messy and unenlightening.

Is there a conceptually clearer way of doing things here?

Thanks.

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2  
The clearest way to do this is to induct on $n$. –  Qiaochu Yuan Apr 26 '11 at 22:41
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Hum. I take back my previous comment; there's actually a much more direct solution. (Show that every element of the Galois group squares to zero.) –  Qiaochu Yuan Apr 26 '11 at 23:18
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Note that in general, if $F_1/K$ and $F_2/K$ are Galois with Galois groups $G_1$ and $G_2$ and if $F_1\cap F_2=K$, then $F_1F_2/K$ is Galois with Galois group isomorphic to $G_1\times G_2$. The proof is very easy, you will have no difficulties writing it down. After that, following Qiaochu's hint becomes a triviality. –  Alex B. Apr 27 '11 at 3:19

2 Answers 2

up vote 8 down vote accepted

First we need to assume that $K$ has characteristic $\neq 2$. Under this hypothesis, $F/K$ is separable, and to prove that it is Galois we need to show that it is normal. As a matter of fact, we put all the roots of the polynomials $X^2-a_i$ (the $\pm x_i$) in $F$.

Finally, each element of the Galois group maps $x_i$ to $\pm x_i$, so we get a morphism $\mathrm{Gal}(F/K) \rightarrow \left( \{ \pm 1 \} \right)^n \simeq \left(\mathbb{Z}/2\mathbb{Z} \right)^n$, $\sigma \mapsto (\sigma(x_i)/x_i)_{1 \leq i \leq n}$, and it is easily shown to be injective. We are left with the following statement to prove: a subgroup of $\left(\mathbb{Z}/2\mathbb{Z} \right)^n$ is isomorphic to $\left(\mathbb{Z}/2\mathbb{Z} \right)^m$ with $m \leq n$, and this follows from the fact that a subgroup of $\left(\mathbb{Z}/2\mathbb{Z} \right)^n$ is a sub-$\mathbb{Z}/2\mathbb{Z}$-vector space.

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Great! That's even more direct than what I was thinking. –  Qiaochu Yuan Apr 26 '11 at 23:29
    
Very slick. A tiny cavil: $\mathbb{Z}/2\mathbb{Z}$ looks like an additive group to me: I think you want $\{ \pm 1\}$ instead. –  Pete L. Clark May 11 '11 at 20:12
    
Thank you, corrected. –  Plop May 11 '11 at 20:17

Let $K_i = K(\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_i}) = K_{i-1}(\sqrt{a_i})$.

Each time you go from $K_i$ to $K_{i+1}$ you're either doing a quadratic extension or not doing anything (in the case where $\sqrt{a_{i+1}} \in K_i$).

The theorem that fits this together is proved here. So the chain of groups is $(\mathbb Z/2 \mathbb Z)^{m}$ for $m$ being how many quadratic fields added were independent.

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5  
The "theorem" you cite is wrong. For example, an $S_3$ extension can always be written as a quadratic extension followed by a cubic extension, but $S_3$ is not isomorphic to $C_2 \times C_3$. The most you can say in general is that $\text{Gal}(C/A)$ contains $H$ as a subgroup, and the quotient is isomorphic to $G$. –  Qiaochu Yuan Apr 26 '11 at 23:08
    
@Qiaochu, thanks for that important point and counterexample - I put some notes about what I meant. –  quanta Apr 26 '11 at 23:28
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Your notes are still pretty confused. $G$ is a priori a quotient of $\text{Gal}(B(\sqrt{a})/A)$ and not a subgroup, so the statement $\sigma\eta = \eta\sigma$ for $\eta\in G$ doesn't make any sense as written. –  Alex B. Apr 27 '11 at 3:17
    
@Alex, thanks again, I've corrected it now and just noticed that I am carrying out exactly the suggestion in the comment you made above! –  quanta Apr 27 '11 at 15:14

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