# what does it mean for a prime at infinity to ramify?

I understand what it means for a prime number to ramify in a ring of integers of a number field. However, an infinite prime is an archimedean valuation, what does it mean for an archimedean valuation to ramify in a number field?

• Look at the archimedean valuation that are induced by the different embeddings of the number field into $\mathbb{C}$; each embedding yields an archimedean valuation that "extends" (or "lies over") an archimedean valuation of the base field; the archimedean valuation splits if all these extensions are different, and ramifies if some of them are the same (just as when you look at the primes $\mathfrak{q}_i$ lying over $\mathfrak{p}$). Consider how to extend the absolute value of $\mathbb{Q}$ to $\mathbb{Q}(i)$, and to $\mathbb{Q}(\sqrt{2})$; the first is an example of ramification. – Arturo Magidin Feb 18 '11 at 4:46
• Another way of saying it is that a real place (i.e. an embedding of your number field into $\mathbb{C}$ with image in $\mathbb{R}$) ramifies if it extends to a complex place (an embedding into $\mathbb{C}$ with non-real image). A complex place can never ramify in an extension. For an Archimedean place, splitting and being unramified are the same. – Keenan Kidwell Feb 18 '11 at 5:50
• @Keenan I think you should leave that as an answer, simply in the interest of not having unanswered questions. – Alex B. Feb 18 '11 at 8:56

On Alex's request I'm leaving my comment as an answer. Let $L/K$ be a finite extension of number fields. A real place $v$ of $K$ (which can be thought of as an embedding of $K$ into $\mathbb{C}$ with image contained in $\mathbb{R}$) is said to ramify in $L$ if it extends to an embedding of $L$ into $\mathbb{C}$ with non-real image. If all extensions of $v$ to places of $L$ are real (the associated embeddings have real image), then $v$ is unramified (also said to be split) in $L$. A complex place of $K$ (an embedding into $\mathbb{C}$ with non-real image) is always unramified. So, the extension $L/K$ is unramified at $\infty$ if all the real places stay real.

For example, in the extension $\mathbb{Q}(\zeta_p)/\mathbb{Q}(\zeta_p+\zeta_p^{-1})$, where $\zeta_p$ is a primitive $p$-th root of unity for some odd prime $p$, the base field is totally real (all its Archimedean places are real) while the top field is totally imaginary, so all real places ramify in the extension.

• The above definition of ramification for real places is the usual one, justified e.g. by the ramification index 2 which appears in a complex valuation over a real one (see Joequinn's answer). However the same phenomenon could also be interpreted as the splitting of the real place under the complex one. In many arithmetic problems, especially in CFT, 2 does not behave like the odd primes, and the classical definition requires special statements at the prime 2. This inconvenience is removed when using the second definition, see e.g. G. Gras' book "CFT. From theory to practice", Springer, 2003. – nguyen quang do Apr 29 '16 at 5:58

I think it's worth adding that while Keenan's answer is a good one, that this is not usually what's given as the definition. I think this issue can be confusing in the literature because in the case of number fields a lot of the relevant constructions become trivial, but it's often not mentioned what the punch line is. So, although this question is old maybe this will still help someone.

The Archimedean valuations on a number field $K$ come from the possible embeddings $\sigma:K\rightarrow\mathbb{C}$ (which you can also identify with $Gal(K,\mathbb{Q})$). For each such $\sigma$, if its image lies in $\mathbb{R}$, the valuation is $v_{\sigma}:x\mapsto |\sigma(x)|$, but if its image does not lie entirely in $\mathbb{R}$, then $\sigma$ and $\overline{\sigma}$ both yield the same valuation $v_{\sigma}:x\mapsto|\sigma(x)|^2$.

Now fix an infinite place $v$ on $K$, let $L$ be a finite field extension of $K$, and let $w$ be an extension of $v$ to $L$. The extension is said to ramify at $w$ iff $\#\{\tau\in Gal(L,K)\mid w\circ\tau=w\}>1$. But in reality this all simplifies to what Keenan said. The only possibilities for $\tau$ satisfying $w\circ\tau=w$ are the identity map and complex conjugation.

Moreover, if $v$ is a real embedding and $w$ is not, you will always have the option of $\tau$ being complex conjugation, so the extension will always be ramified there. And if the situation is any of the other possibilities ($v$ real & $w$ real; or $v$ non-real & $w$ non-real), then $\tau$ can only be the identity, hence not ramified.

• I realize I'm resurrecting an ancient post here, but do you happen to know a reference for the definition of ramification you give here? If I'm understanding correctly you're saying that the extension ramifies at $w$ if the size of the decomposition group is greater than $1$. I find this strange because for finite places which are unramified the size of the decomposition group can certainly be larger than $1$. It's rather the size of the inertia subgroup that tells you whether the place ramifies. – wgrenard Jun 4 at 18:13
• @wgrenard I'm not sure of the exact reference right now, since it's been a while, but recall liking Keith Conrad's way of writing about this. I recommend searching some of his content; he's pretty generous about sharing articles online. – j0equ1nn Jun 5 at 18:46