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Let ${\mathbb K} \subseteq {\mathbb C}$ be a finite extension of $\mathbb Q$, and let $n=[{\mathbb K} : {\mathbb Q}]$. Let $X_{\mathbb K}$ denote the set of all “components” (i.e., real and imaginary parts) of elements in ${\mathbb K}$. Let ${\mathbb L}=\mathbb Q(X_{\mathbb K})$ ; this is a subfield of $\mathbb R$.

It is easy enough to see that $[{\mathbb L}:{\mathbb Q}]$ is finite, and in fact, it is $\leq 2n^2$ (see proof below). Denote by $f(n)$ the best possible bound ; I have computed that $f(2)=2$ and $f(3)=6$, and I ask : what is $f(n)$ in general ?

Proof of the upper bound

Let ${\mathbb M}_1$ denote the image of $\mathbb K$ by complex conjugation, ${\mathbb M}_2$ the smallest subfield of $\mathbb C$ containing both $\mathbb K$ and ${\mathbb M}_1$ (the so-called “compositum”), and finally ${\mathbb M}_3={\mathbb M}_2[i]$.

By construction, ${\mathbb M}_3$ contains $i$ and is invariant by complex conjugation, so it is also invariant with respect to taking real and imaginary parts. Since it contains $\mathbb K$, we see that ${\mathbb L} \subseteq {\mathbb M}_3$.

Now $[{\mathbb M}_2:{\mathbb Q}]$ is smaller than the product $[{\mathbb K}:{\mathbb Q}][{\mathbb M}_1:{\mathbb Q}]=n^2$, and in turn $[{\mathbb M}_3:{\mathbb Q}]$ is smaller than $2[{\mathbb M}_2:{\mathbb Q}] $, qed.

Update at 17:39 (in answer to Qiaochu’s comment) : it is not true that $f(n)=2n$ for every $n \geq 3$. Indeed, I can show that $f(4) \geq 12$. To check this, let $P$ be any rational polynomial of degree $4$ with no real roots, no purely imaginary root, and Galois group $S_4$ (for example, $P=X^4 - 6X^3 + 15X^2 - 19X + 13$ will do). Then take ${\mathbb K}={\mathbb Q}(\lambda)$ where $\lambda$ is any root of $P$. Using Galois theory, it is easy to see that in this case $[{\mathbb L}:{\mathbb Q}]$ is 12.

Another way to put it : under those hypotheses, there is an automorphism $\sigma$ of $\mathbb C$ fixing $i$ and $\lambda$ but not $\bar{\lambda}$. This $\sigma$ acts as the identity on $\mathbb L$, but acts non-trivially on the real part of ${\mathbb K}[i]$.

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I'm not sure if [number-fields] a good tag. Perhaps [algebraic-number-theory] or so is enough. –  Asaf Karagila Nov 18 '12 at 10:12
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I might be missing something, but isn't $L$ just the real subfield of $\mathbb{K}[i]$? If that's true we should just have $f(n) = 2n$ for $n \ge 3$, right? –  Qiaochu Yuan Nov 18 '12 at 10:25
    
(oops, I guess you are right) –  Hagen von Eitzen Nov 18 '12 at 10:35
    
@Asaf : I corrected the tag. –  Ewan Delanoy Nov 18 '12 at 11:36
    
@QiaochuYuan : I think not, and I believe I have a counterexample. I’ll update the OP about this when I have the time. –  Ewan Delanoy Nov 18 '12 at 11:37

1 Answer 1

up vote 3 down vote accepted

Pick a primitive element $x \in \mathbb K$. Then $\mathbb M_2 = \mathbb K[\overline{x}]$. But $\overline{x}$ has the same minimal polynomial as $x$ over $\mathbb Q$, which factors as $(X-x)Q$ with $Q$ of degree $n-1$ in $\mathbb K [X]$. Thus the extension $\mathbb K \subset \mathbb M_2$ is at most of degree $n-1$. This gives you a better upper bound, $n(n-1)$, for $[\mathbb M_2 : \mathbb Q]$ (which is attained if and only if the action of the Galois group on the conjugates of $x$ is $2$-transitive).

Next, you have $\mathbb L[i] = \mathbb M_2[i]$, and $[\mathbb L[i] : \mathbb L] = 2$ because $\mathbb L$ is real. This implies that $[\mathbb L : \mathbb Q] = [\mathbb M_2 : \mathbb Q]$ or $[\mathbb M_2 : \mathbb Q]/2$ (depending if $i \in \mathbb M_2$ or not). So $[\mathbb L : \mathbb Q] \le n(n-1)$.

The last thing to do is to show that the bound can be obtained. Suppose $n \ge 4$, that the Galois group of the Galois closure of $\mathbb K$ over $\mathbb Q$ is $S_n$, and that $i \in \mathbb Q[x_1,x_2]$ for a pair of distinct conjugates of $x$. Since $\mathbb Q \subset \mathbb Q[i]$ is a Galois extension and $S_n$ is $2$-transitive, we must have $i \in \mathbb Q[x_i,x_j]$ for any pair of distinct conjugates of $x$. Since the permutations with two or more fixpoints generate $S_n$, we must have $i \in \mathbb Q$, which is false. Thus $i$ can't be in $\mathbb Q[x_1,x_2]$. Finally, if $x$ has a nonreal conjugate, then we have that $i \notin \mathbb Q[x,\overline{x}]$, and so $[\mathbb L : \mathbb Q] = n(n-1)$

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For even $n$, your bound is clearly optimal, it suffices to generalize the example at the end of the OP : take a polynomial $P$ with no real roots, no purely imaginary roots, and Galois group $S_n$. For odd $n$, this is not so obvious. –  Ewan Delanoy Nov 18 '12 at 18:02
    
@EwanDelanoy : I think it works whenever the Galois group is $S_n$ (see my edit).I wonder if it is enough for the Galois group to be $2$-transitive –  mercio Nov 18 '12 at 19:58
    
it seems indeed that the only property you need for the Galois group is that of being $2$-transitive. –  Ewan Delanoy Nov 19 '12 at 7:39
    
My proof doesn't work for $S_3$ because even though it is $2$-transitive, $S_3$ isn't generated by the elements with two fixpoints. –  mercio Nov 19 '12 at 7:42

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