Suppose that $L \subseteq \mathbb{C}$ is the splitting field of a polynomial $f \in \mathbb{Q}[x]$, and suppose that $f$ has non-real roots. Then complex conjugation (let's call it $\sigma$) is a non-trivial element of $\mathrm{Gal}(L/\mathbb{Q})$.

The fixed subfield of $\langle\sigma\rangle$ is $L \cap \mathbb{R}$. But is there anyway we can express this in terms of the roots of $f$? I ask this because I noticed, for example, that for the choices of $f$ I have considered, this fixed subfield is simply the field obtained by adjoining the real part of each root of $f$ to $\mathbb{Q}$. I can't prove nor disprove this. In the event that this is false, is there any other way we can express this fixed subfield in terms of the roots of $f$ (e.g. further adjoin products of complex conjugate roots of $f$, etc)?


This isn't always true, but we can show that it's not far off. Consider $f(X) = (X^2+1)(X^2+2)$. Then the splitting field is $\mathbb{Q}(i,i\sqrt{2})$, but adjoining the real parts of each root (i.e. $0$) doesn't give the fixed field of $\sigma$. In fact, what we have is the following: suppose $L$ is the splitting field for $f$, with roots $\alpha_1\pm i\beta_1,\dots,\alpha_n\pm i\beta_n$. Then the fixed field of complex conjugation is:

$$\mathbb{Q}(\alpha_1,\dots,\alpha_n,\{\beta_j\beta_k\}_{1\leq j,k\leq n}).$$

You can describe the fixed field of $\sigma$ abstractly as the unique real subfield $K$ such that $[L:K] = 2$, and this will always contain $K'=\mathbb{Q}(\alpha_1,\dots,\alpha_n)\subset \mathbb{R}$. But since $i\beta_j$ lies in $L$ for each $j$, $K$ also contains $\beta_j\beta_k$ for any $j,k$ i.e. $K$ contains $K''=K'(\{\beta_j\beta_k\}_{1\leq j,k\leq n})\subset\mathbb{R}$. We now show that in fact $K'' = K$.

Since we have: $$\frac{-\beta_j\beta_k}{i\beta_j}= i\beta_k, $$ for any $j$ with $\beta_j \not = 0$, $L = K''(i\beta_j)$. But since for any non-zero $j, \beta_j^2 \in K''$, and so $X^2+\beta_j^2\in K''[X]$. Thus $i\beta_j$ is degree $2$ over $K''$, so $[L:K''] = 2$, so $K=K''$.


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