ramification of archimedean places

Throughout this entry, if $\alpha$ is a complex number, we denote the complex conjugate of $\alpha$ by $\overline{\alpha }$.

Definition 1.

Let $K$ be a number field.

1. An archimedean place of $K$ is either a real embedding $\varphi :K\to ℝ$ or a pair of complex-conjugate embeddings $(\psi ,\overline{\psi })$, with $\overline{\psi }\ne \psi$ and $\psi :K\to ℂ$. The archimedean places are sometimes called the infinite places (cf. place of field).
   

2. The non-archimedean places of $K$ are the prime ideals in ${𝒪}_K$, the ring of integers of $K$ (see non-archimedean valuation (http://planetmath.org/Valuation)). The non-archimedean places are sometimes called the finite places.

Notice that any archimedean place $\varphi :K\to ℂ$ can be extended to an embedding $\widehat{\varphi }:\overline{\mathbb{Q}}\to ℂ$, where $\overline{\mathbb{Q}}$ is a fixed algebraic closure of $\mathbb{Q}$ (in order to prove this, one uses the fact that $ℂ$ is algebraically closed and also Zorn’s Lemma). See also this entry (http://planetmath.org/PlaceAsExtensionOfHomomorphism). In particular, if $F$ is a finite extension of $K$ then $\varphi$ can be extended to an archimidean place $\widehat{\varphi }:F\to ℂ$ of $F$.

Next, we define the decomposition and inertia group associated to archimedean places. For the case of non-archimedean places (i.e. prime ideals) see the entries decomposition group and ramification.

Let $F/K$ be a finite Galois extension of number fields and let $\varphi$ be a (real or a pair of complex) archimedean place of $K$. Let ${\varphi }_1$ and ${\varphi }_2$ be two archimedean places of $F$ which extend $\varphi$. Notice that, since $F/K$ is Galois, the image of ${\varphi }_1$ and ${\varphi }_2$ are equal, in other words:

$${\varphi }_1(F)={\varphi }_2(F)\subset ℂ.$$

Hence, the composition ${\varphi }_1^{-1}\circ {\varphi }_2$ is an automorphism of $F$ (here ${\varphi }_1^{-1}$ denotes the inverse map of ${\varphi }_1$, restricted to ${\varphi }_1(F)$). Thus, ${\varphi }_1^{-1}\circ {\varphi }_2=\sigma \in \mathrm{Gal}(F/K)$ and

$${\varphi }_2={\varphi }_1\circ \sigma$$

so ${\varphi }_1$ and ${\varphi }_2$ differ by an element of the Galois group. Similarly, if $({\psi }_1,\overline{{\psi }_1})$ and $({\psi }_2,\overline{{\psi }_2})$ are complex embeddings which extend $\varphi$, then there is $\sigma \in \mathrm{Gal}(F/K)$ such that

$$({\psi }_2,\overline{{\psi }_2})=({\psi }_1,\overline{{\psi }_1})\circ \sigma$$

meaning that either ${\psi }_2={\psi }_1\circ \sigma$ (and thus $\overline{{\psi }_2}=\overline{{\psi }_1}\circ \sigma$) or $\overline{{\psi }_2}={\psi }_1\circ \sigma$ (and thus ${\psi }_2=\overline{{\psi }_1}\circ \sigma$). We are ready now to make the definitions.

Definition 2.

Let $F\mathrm{/}K$ be a Galois extension of number fields and let $w$ be an archimedean place of $F$ lying above a place $v$ of $K$. The decomposition and inertia subgroups for the pair $w\mathrm{|}v$ are equal and are defined by:

$$D(w|v)=T(w|v)=\{\sigma \in \mathrm{Gal}(F/K):w\circ \sigma =w\}.$$

Let $e\mathrm{=}e\mathrm{(}w\mathrm{|}v\mathrm{)}\mathrm{=}\mathrm{|}T\mathrm{(}w\mathrm{|}v\mathrm{)}\mathrm{|}$ be the size of the inertia subgroup. If $e\mathrm{>}\mathrm{1}$ then we say that the archimedean place $v$ is ramified in the extension $F\mathrm{/}K$.

The ramification in the archimedean case is much simpler than the non-archimedean analogue. One readily proves the following proposition:

Proposition 1.

The inertia subgroup $T\mathrm{(}w\mathrm{|}v\mathrm{)}$ is nontrivial only when $v$ is real, $w\mathrm{=}\mathrm{(}\psi \mathrm{,}\overline{\psi }\mathrm{)}$ is a complex archimedean place of $F$ and $\sigma$ is the “complex conjugation” map which has order $\mathrm{2}$. Therefore $e\mathrm{(}w\mathrm{|}v\mathrm{)}\mathrm{=}\mathrm{1}$ or $\mathrm{2}$ and ramification of archimedean places occurs if and only if there is a complex place of $F$ lying above a real place of $K$.

Proof.

Suppose first that $w=\varphi :F\to ℝ$ is a real embedding. Then $\varphi$ is injective and $\varphi \circ \sigma =\varphi$ implies that $\sigma$ is the identity automorphism and $T(w|v)$ would be trivial. So let us assume that $w=(\psi ,\overline{\psi })$ is a complex archimedean place and let $\sigma \in \mathrm{Gal}(F/K)$ such that

$$(\psi ,\overline{\psi })=(\psi ,\overline{\psi })\circ \sigma .$$

Therefore, either $\psi =\psi \circ \sigma$ (which implies that $\sigma$ is the identity by the injectivity of $\psi$) or $\psi =\overline{\psi }\circ \sigma$. The latter implies that $\sigma =\overline{{\psi }^{-1}}\circ \psi$, which is simply complex conjugation:

$$\overline{{\psi }^{-1}}\circ \psi (k)=\overline{{\psi }^{-1}(\psi (k))}=\overline{k}.$$

Finally, since $w$ is an extension of $v$, the equation $w\circ \sigma =w$ restricts to $\overline{v}=v$, thus $v$ must be real.∎

Corollary 1.

Suppose $L\mathrm{/}K$ is an extension of number fields and assume that $K$ is a totally imaginary (http://planetmath.org/TotallyRealAndImaginaryFields) number field. Then the extension $L\mathrm{/}K$ is unramified at all archimedean places.

Proof.

Since $K$ is totally imaginary none of the embeddings of $K$ are real. By the proposition, only real places can ramify.∎