what does it mean for a prime at infinity to ramify?
Solution 1:
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.
Solution 2:
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.