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I've just read the following exercise:

"Determine for some (or all) $n\leq 10$ the prime decomposition of $2, 3, 5$ and $\infty$ in $\mathbb{Q}(\zeta_{12})$, where $\zeta_{12}$ is a primitive $12$-th root of unity. In particular, determine the different places above $2, 3, 5$ and $\infty$, their ramication indices and their inertia degrees."

What is "prime decomposition of $\infty$" supposed to mean here?

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    $\begingroup$ It would be helpful to know the book or website where you "read the following exercise". Readers may be better able to help knowing the author and date of publication of your exercise. $\endgroup$ – hardmath Feb 21 '17 at 1:57
  • $\begingroup$ it was from an exercise sheet the professor gave us. I've just edited to show the full question, hope it helps. Anyway, thanks in advance! $\endgroup$ – rmdmc89 Feb 21 '17 at 2:03
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Mysterious with no context, isn’t it. Here’s the context, though:

In Arithmetic, one often tries to handle the archimedean absolute value of $\Bbb Q$ in a manner as nearly parallel as possible to the treatment of the nonarchimedean absolute values. These latter are in one-to-one correspondence with the prime integers, while for consistency’s sake, one refers to the archimedean absolute value as “infinity”, $\infty$, and one may even call it “the infinite prime”.

Now, in a finite extension $K$ of $\Bbb Q$, each finite prime $p$ splits into a certain number $n_p$ of primes; this $n_p$ may be interpreted as the number of different $p$-adic absolute values on $K$ that extend the $p$-adic absolute value on $\Bbb Q$. In the same way, one asks for the number of archimedean absolute values that there are on $K$. The simple way of looking at it is that if $f(X)$ is the minimal polynomial for a generating element of $K$ over $\Bbb Q$, you factor $f$ into real-irreducible factors. There will be $r_1$ linears, and $r_2$ quadratic ones. (In case the extension is normal, one of these two numbers will be zero.) At any rate, $[K:\Bbb Q]=r_1+2r_2$.

Finally, one says that an archimedean absolute value is ramified in an extension $K\supset F$ if it’s complex in $K$ but restricts to a real absolute value in $F$. That is, the respective completions are $\Bbb C$ and $\Bbb R$. The ramification degree is $2$ in this case, $1$ in all other cases.

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An infinite prime is defined as an embedding of your field into $\mathbb{R}$ or $\mathbb{C}$. We count conjugate embeddings into $\mathbb{C}$ to be the same prime. Then an infinite prime splits in an extension if the upper field can be extended to an embedding in the same field as the base, $\mathbb{R}$ or $\mathbb{C}$, and it ramifies if the extension field can be embedded in $\mathbb{C}$ not fully in $\mathbb{R}$, extending an embedding of the base field in $\mathbb{R}$.

For example: $\mathbb{Q}$ has a single infinite real prime, because it has a single embedding in the reals. But if we extend it to $\mathbb{Q}(\sqrt[3]{2})$, this field can be embedded in $\mathbb{C}$ in three different ways, two of which are conjugate complex embeddings and one of which is a real embedding. So the real prime of $\mathbb{Q}$ partially ramifies and partially splits. The ramification index of the complex prime is 2 and the ramification index of the real prime is 1, and there is no inertia in infinite primes.

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    $\begingroup$ Thanks for this, giving a different outlook on the situation than mine, and especially for pointing out that there’s no inertia in the archimedean case. $\endgroup$ – Lubin Feb 21 '17 at 13:51
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The "prime decomposition of $\infty$" in a number field $K$ is more or less a question of convention in order to parallel the same phenomenon for prime ideals. But the question remains what convention to choose. It will take long to make things more precise, but let me first recall the definition(s) of a place of $K$ :

(1) Starting from absolute values of $K$ as in the answer of @Lubin, a place of $K$ is an equivalence class of non trivial absolute values (archimedean or not) of $K$, two absolute values being equivalent iff the topological spaces that they define on $K$ are homeomorphic. The set of places $Pl_K$ is determined from $Pl_{\mathbf Q}$ by explicit formulas :

  • if $P$ is a prime ideal of $K$, then $|x|_P$ := $N(P)^{-v_P (x)}$ , where $N(P)$ is the absolute norm of $P$ and $v_P (x)$ is the power to which $P$ appears in the ideal factorization of $(x)$
  • the archimedean absolute values are of two types : the real ones, indexed by the $r_1$ embedding $\sigma : K \to \mathbf R$ , defined by $|x|_{\sigma} = |\sigma x|$ ; the complex ones, indexed by the $r_2$ pairs of conjugate embeddings $\tau : K \to \mathbf C$, defined by $|x|_{\tau} = |\tau x|^2$ . Note that the square accounts for the fact that for each pair of conjugate $\tau$ 's, one picks only one of the two conjugate $\tau (x)$' s

(2) To define archimedean places starting from the $\mathbf Q$-embeddings of $K$ into an algebraic closure of $\mathbf Q$ as in the answer of @Bob Jones, we must first (for a fixed prime $p$) look at the $\mathbf Q_p$-embeddings of $K$ into the completion $\mathbf C_p$ of an algebraic closure of $\mathbf Q_p$. Given any $p$-adic valuation $v$ and any embedding $i_v : K \to K_v$ (the $v$-completion of $K$), it is easy to see that $K_v = i_v (K)\mathbf Q_p$. Two $\mathbf Q_p$-embeddings $\sigma , \tau : K \to \mathbf C_p$ will be called equivalent if $\mathbf Q_p \sigma(K)$ and $\mathbf Q_p \tau(K)$ are $\mathbf Q_p$-conjugate, and a new definition will be that a $p$-place = an equivalence class of such a $\mathbf Q_p$-embedding $\sigma$ = {$\sigma \tau.i_v$} for a chosen $i_v$ and for $\tau$ running through the $\mathbf Q_p$-isomorphisms of $K_v$ into $\mathbf C_p$. The coincidence with the first definition comes from the formula $|x|_v = |N_{K_v /\mathbf Q_p }(x)|_p$. The analogous definition of an archimedean complex place will obviously come from the formula $N_{\mathbf C / \mathbf R} (i_\infty(x)) = i_\infty(x)$. conjugate $i_\infty(x)$, just as previously in (1).

The presence of the square in the formula defining a complex achimedean place is perhaps at the origin of the widely accepted convention (since Hasse) that in a relative extension $K/F$, a real place of $F$ which become complex in $K$ is called ramified, with ramification index 2. But this is not undisputable. Instead of "ramification", some authors (e.g. G. Gras in his book CFT: from theory to practice, Springer 2003) advocate the more neutral word "complexification" (of a real place). In analogy with the vocabulary for a local extension $L_w / K_v$ obtained by completing a global $L/K$, there is no doubt that the case ${\mathbf R / \mathbf R}$ for the place $\infty$ should be called "totally decomposed". But the conventional systematic terminology "ramified" for the case ${\mathbf C / \mathbf R}$ leads to undue complications, e.g. in the formulation of the main theorems of CFT. Consider for instance $K = \mathbf Q (\zeta_m)$ , of which the "natural" conductor should be $m$, whereas in the classical formulation of CFT in terms of ray class fields, this conductor is actually the modulus $(m). \infty$. More seriously, in the part of CFT which deals with abelian extensions with restricted ramification, more precisely unramified outside $S$ and totally decomposed inside $T$, $S$ and $T$ being two finite disjoint sets of places of the base field (op. cit. chapter III), the so called Spiegelungsatz (reflection theorem) exchanges among other things the real infinite places and the real infinite places outside $S$, which renders the statements a bit messy when sticking to the usual convention.

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