Let $L/K$ be an extension of number fields. Is there any sharp inequality between $[L:K], \#M_{K}^{\infty}$ and $\#M_{L}^{\infty}$? Here $\#M_{K}^{\infty}$ is a number of infinite (archimedean) places of $K$. If $r_{K}, s_{K}, r_{L}, s_{L}$ are number of real and complex embeddings of $K$ and $L$, then we have $$ [K:\mathbb{Q}] = r_{K} +2s_{K}, \quad \#M_{K}^{\infty} = r_{K} + s_{K} $$ and same for $L$.

  • $\begingroup$ @Lubin Actually I'm confusing about the relation between $r_{K}, s_{K}, r_{L}$ and $s_{L}$. How to prove your inequality? It seems to be easy but I'm so confused now... $\endgroup$ – Seewoo Lee Feb 12 at 7:04
  • $\begingroup$ I just realized that my inequalities were wrong. Should be $\#M_K^\infty\le\#M_L^\infty\le[L:K]\#M_K^\infty$. The situation is really more subtle than I realized last night, so let me think for a while and actually write up a proper Answer. (Maybe I’ll get lucky and a real number-theorist will beat me to it.) $\endgroup$ – Lubin Feb 12 at 14:55

It's just a question of book keeping, but to avoid confusion, let us repeat the conventions about the places of a number field $K$. Two absolute values ${\mid .\mid}_1$ and ${\mid .\mid}_2$ on $K$ are called equivalent iff the topological spaces which they define on $K$ are homeomorphic, iff there exits a strictly positive real constant $\lambda$ s.t. ${\mid .\mid}_2={{\mid .\mid}_1}^{\lambda}$. A place is then an equivalence class of absolute values.

1) Let us now stick to the archimedean places. The number field $K$ admits $r_K$ real embeddings ( = $\mathbf Q$- isomorphisms of $K$ into $\mathbf R$) and $s_K$ pairs of conjugate complex embeddings, so that $[K:\mathbf Q]=r_K +2s_K$ and the number of places of $K$ is $r_K +s_K$. Recall that for a real embedding $\sigma$, the absolute value ${\mid .\mid}_{\sigma}$ is defined by ${\mid x\mid}_{\sigma}={\mid \sigma(x)\mid}$ for all $x\in K$, where ${\mid .\mid}$ is the usual absolute value on $\mathbf R$. Whereas for a complex embedding $\sigma$, one defines ${\mid x\mid}_{\sigma}={\mid \sigma(x)\mid}^2={\mid x\mid}_{c\sigma}$, where $c$ denotes the complex conjugation and ${\mid .\mid}$ the usual modulus on $\mathbf C$. Here the reason for taking squares is to ensure the so called product formula $\prod_v {\mid x\mid}_v=1$, the product bearing over all places of $K$.

2) Given an extension $L/K$ of degree $n$, it is obvious that a complex embedding of $K$ remains a complex embedding of $L$. Whereas a real embedding of $K$ a priori gives rise to $\rho_1$ real embeddings and $2\rho_2$ conjugate complex embeddings of $L$, so that $r_L=\rho_1+\rho_2, s_L=\rho_2$ and $[L:K]=\rho_1+2\rho_2$. Note that the formulas in 1) and 2) are coherent because $r_{\mathbf Q}=1$ and $s_{\mathbf Q}=0$.

NB: Because of the squares in the definition of complex absolute values, on usually says that a real place which becomes complex in $L$ is ramified (as in the case of $p$-adic places). But when looking at the number of embeddings, it is not less natural to say that the real place of $K$ splits in $L$. Perhaps the term complexification of a real place would be the most appropriate.

  • $\begingroup$ Thanks for taking most of the burden off my shoulders. I’ll just add a bit to what you said. $\endgroup$ – Lubin Feb 12 at 22:32

In addition to the answer of Nguyen Quang Do, let me say a few words.

Look at any particular archimedean place $\mathfrak p$ of the subfield $K$. The case that $\mathfrak p$ is complex is easiest to handle. Then there will be exactly $[L:K]$ places $\mathfrak P_1,\mathfrak P_2,\cdots,\mathfrak P_n$ “above” $\mathfrak p$, all complex, of course. So in case $K$ is “totally complex” (i.e. has no real embeddings), we then get $\#M_L=[L:k]\#M_K$ .

In case your given archimedean place $\mathfrak p$ of the subfield $K$ is real, however, the situation becomes more complicated, somewhat less so if the extension is Galois. For in the Galois case, either all extensions of $\mathfrak p$ are real, or all are complex. In the former case, the extensions are $[L:K]$ in number; in the latter, there are $[L:K]/2$ of them.

In the remaining case that $\mathfrak p$ is real but $L$ is not Galois over $K$, all we can say is that there will be $m_1$ real extensions and $m_2$ complex extensions of the archimedean absolute value of $\mathfrak p$, each of these last coming from a pair of complex embeddings of $L$, so that $m_1+2m_2=[L:K]$.

I think you can figure out how to combine all these facts to get some idea of how the number of archimedean places of the big field can compare with the the number for the little field.


I asked the same question to my friend, and here's the answer that I got from him.

Assume that $L \neq K$. We have $$ \sharp M_{K}^{\infty} = r_{K} + s_{K} \leq [K:\mathbb{Q}] \leq \frac{1}{2}[L:\mathbb{Q}] = \frac{1}{2}(r_{L} + 2s_{L}) \leq \sharp M_{L}^{\infty}. $$ For the second inequality, note that we have $r_{L} \leq [L:K] r_{K}$ since there exist at most $[L:K]$-many extension of $\sigma : K\hookrightarrow \mathbb{R}$ to $\tilde{\sigma}:L\hookrightarrow \mathbb{R}$. Then $$ \sharp M_{L}^{\infty} = \frac{1}{2}([L:\mathbb{Q}] + r_{L}) \leq \frac{1}{2}[L:K]([K:\mathbb{Q}] + r_{K}) = [L:K]\sharp M_{K}^{\infty}. $$


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