Let $K \subseteq \mathbb{C}$ be a number field (I'm fixing an embedding), and assume $K/\mathbb{Q}$ is Galois with Galois group $G$. Let $\tau \in G$ denote complex conjugation. This question concerns the condition, let's call it (*), that $\tau$ is in the center of $G$. In other words, the condition says that $\sigma(\overline{\alpha}) = \overline{\sigma(\alpha)}$ whenever $\sigma \in G$ and $\alpha \in K$.

Some observations:

-(*) is satisfied if $K \subseteq \mathbb{R}$, since $\tau$ is trivial then, and

-the condition (*) is very restrictive on the structure of $G$ if $\tau$ is not trivial.

I feel like I have seen this condition in literature before, but I can't remember where, hence this question:

Can you provide any references in the literature to this condition (such as theorems where it is a hypothesis or conclusion)?

See also this slightly related post: Centraliser of the complex conjugation in the absolute Galois group

One may also ask whether (*) holds for every embedding $K \hookrightarrow \mathbb{C}$, so references to this condition would also be appreciated.


Your condition is that $K$ be a kroneckerian field, namely either a totally real (as you mention) field or a totally imaginary quadratic extension of a totally real field: in this second case $K$ is said to be a CM field.

Since you have already treated in your question the case when $K$ is totally real, let me assume $K$ is CM. In this case, one has to notice that since a compositum of totally real fields is again totally real, we can speak of the maximal totally real subextension, call it $K^+$ of $K$. By assumption, $K^+\subsetneq K$ and since there is a totally real subfield $F\subseteq K$ such that $[K:F]=2$ we deduce $F=K^+$. Since you assumed $K$ to be Galois over $\mathbb{Q}$, I claim that $K^+/\mathbb{Q}$ is also Galois: this boils down to the statement that all embeddings $\iota:K^+\hookrightarrow\mathbb{R}$ have the same set-theoretic image, because this would imply that all roots of a minimal polynomial ok $K^+/\mathbb{Q}$ already lie in $K^+$. To check this, call $\mathbb{K}$ the subfield of $\mathbb{C}$ which is the image of one, and hence all by normality, embeddings of $K$ into $\mathbb{C}$: let $\mathbb{K}^+=\mathbb{K}\cap\mathbb{R}$. Then, for every $\iota:K^+\hookrightarrow\mathbb{C}$ we have $\iota(K^+)=\mathbb{K}^+$ since the image has to be contained in $\mathbb{R}\cap\mathbb{K}$ and needs be maximal. We have thus proven that $\langle\tau\rangle$ is a normal subgroup in $G$. This means that for each $g\in G$ we have $g\tau g^{-1}\in\langle\tau\rangle$ and since $\tau$ has order $2$ the only possibility is $g\tau g^{-1}=\tau$, namely $g\tau=\tau g$, proving that $\tau$ is central.

The other direction is obvious: if $\tau$ is central, then $\langle\tau\rangle$ is normal and if we set $K^+:=K^{\tau}$ we find a normal extension of $\mathbb{Q}$ (hence, either totally real or totally imaginary) that is fixed by at least one complex conjugation, hence is not totally imaginary and therefore is totally real. As $[K:K^+]=\#\langle\tau\rangle=2$, $K$ is a CM field.

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  • $\begingroup$ Thanks for showing this argument. About your observation: "complex conjugation in $G$" is meaningful regardless of the situation, only it may depend on the choice of embedding, as I implied. Clearly (although I wasn't thinking this when I posted originally), (*) is equivalent to "complex conjugation is well-defined indep. of a choice of embedding," and the second way of saying it more clearly shows that $K$ must be Kroneckerian. $\endgroup$ – Bobby Grizzard Apr 15 '15 at 20:03
  • $\begingroup$ You're perfectly right, I had overlooked your assumption while answering. I erased the portion of text in question. $\endgroup$ – Filippo Alberto Edoardo Apr 16 '15 at 6:43
  • $\begingroup$ Would it be possible to argue that $K^+$ is Galois because it is $K \cap \mathbb Q^{\mathrm{tr}}$, where $\mathbb Q^{\mathrm{tr}}$ is the set of algebraic numbers whose minimal polynomial is split over $\mathbb R$, and that both $K/\mathbb Q$ and $\mathbb Q^{\mathrm{tr}}/\mathbb Q$ are Galois? $\endgroup$ – PseudoNeo Jan 13 '18 at 18:04
  • $\begingroup$ Yes, I think that this makes perfect sense and it is a quicker way of proving the statement. Thanks. $\endgroup$ – Filippo Alberto Edoardo Jan 14 '18 at 11:15

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