This question has been edited in light of some helpful comments from @AdamHughes below.

Let $F$ be a totally real number field, i.e. $F=\mathbb{Q}(t)/m(t)$ where $m(t)\in\mathbb{Z}[t]$ and all roots of $m(t)$ lie in $\mathbb{R}$.

Let $K$ be a quadratic extension of $F$ so that $K=F(t)/(t^2+d)$ where $d\in F^+$. Then $[K:F]=2$ and $[K:\mathbb{Q}]=2[F:\mathbb{Q}]$, which also equals the number of embeddings of $K$ into $\mathbb{C}$. The places of $K$ over the basefield $\mathbb{Q}$ are identified with these embeddings, except that we count complex conjugates as the same place.

It seems to me that it is not possible for $K$ to embed into $\mathbb{R}$ because under any embedding, $d\in F$ is still required to have a negative square root. This leads me to believe that all the places contributed by $[F:\mathbb{Q}]$ become non-real embeddings. This is around where I feel like I am missing something, so let me stop here.

The question is, under what conditions does $K$ have a unique complex place? It seems to me that this would only be when $F=\mathbb{Q}$. But that begs the question of why some of the books I'm reading talk about "an imaginary quadratic extension of a totally real number field, having a unique complex place," when they could just say "a quadratic field."

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    $\begingroup$ Never, $\sqrt{-d}^2=-d$ is its defining property, and real field cannot have elements which square to negatives. $\endgroup$ – Adam Hughes Mar 27 '16 at 23:32
  • $\begingroup$ @AdamHughes Okay good, that makes sense. How about the other question? Can there be a unique complex place with $F\neq\mathbb{Q}$? $\endgroup$ – j0equ1nn Mar 27 '16 at 23:34
  • $\begingroup$ Places are (almost) just elements of the Galois group, so yes, there is a unique complex place which fixes the base field since complex embeddings come in conjugate pairs, and places are equivalence classes of absolute values. $\endgroup$ – Adam Hughes Mar 27 '16 at 23:37
  • $\begingroup$ @AdamHughes What is confusing me is when we think of $\mathbb{Q}$ as the base field, and $F\neq\mathbb{Q}$. Let's say $[K:\mathbb{Q}]=4$. Then since $[K:F]=2$ we have $[F:\mathbb{Q}]=2$. The first $2$ corresponds to identity/conjugation. The second $2$ corresponds to permutations of $F$ fixing $\mathbb{Q}$. But then doesn't that induce 2 new complex embeddings of $K$? $\endgroup$ – j0equ1nn Mar 27 '16 at 23:41
  • $\begingroup$ Not exactly. It seems your confusion is beyond just comments, so I'll post an answer instead. $\endgroup$ – Adam Hughes Mar 27 '16 at 23:44

This portion of the post is to address the revised question. For the original, scroll down.

The degree $[F(\sqrt{-d}):F]=2$ and since $F$ is totally real, $F\cap\Bbb Q(\sqrt{-d})=\Bbb Q$ shows that

$$\operatorname{Gal}(F(\sqrt{-d})/\Bbb Q)\cong\operatorname{Gal}(F/\Bbb Q)\times\operatorname{Gal}(\Bbb Q(\sqrt{-d})/\Bbb Q)$$

Now the field $\Bbb F(\sqrt{-d})$ has no real embeddings, since they symbol $\sqrt{-d}$ will always square to $-d$ no matter what, and so has $s=[F(\sqrt{-d}):\Bbb Q]=[F:\Bbb Q]$ complex embedding pairs, all of which are induced by the $s$ real-embeddings of $F$, in fact you can list out all the places of $F(\sqrt{-d})$ if you know that $\operatorname{Gal}(F/\Bbb Q)=\{\sigma_i\}_{i=1}^s$. They are


where $|\cdot |$ is the absolute value given by any fixed choice of embedding of $F$ into $\Bbb C$. In particular if you look at the original answer, that makes sense since you chose the embedding of $F$ you only had the one, but when you let $F$ be more abstract, you get this nice listing instead. So the short answer is never unless $F=\Bbb Q$.

This is the original response

To the first question: no: the symbol $\sqrt{-d}\in F(\sqrt{-d})$ is defined by the property that $\sqrt{-d}^2=-d<0$ which cannot happen in a real field.

To the second, you want to note that since you have already fixed an embedded $F\subseteq \Bbb C$, you don't get any action there, your base field elements are fixed. So if say $F=\Bbb Q(\sqrt 2)$ where the symbol $\sqrt 2$is the unique positive real number squaring to $2$, then your embedding has $F$ fixed that way, the abstract field $\Bbb Q(\sqrt 2)\cong\Bbb Q[x]/(x^2-2)$ has multiple choices for the symbol, but you've said you've chosen the embedding already.

Now, what happens when I extend $F$? Well, $F(\sqrt{-d})$ has degree $1$or $2$ over $F$, and obviously we're assuming for simplicity that the degree is $2$, since $1$ is boring. Then what you're asking is really about $F[x]/(x^2+d)$, but the embeddings, and therefore the places are just $x\mapsto\pm\sqrt{-d}$ where here we overload the $\sqrt{-d}$ notation and mean it to be the unique upper-half plane element of $\Bbb C$ which squares to $-d$. So what you're asking about is $\operatorname{Gal}(F(\sqrt{-d})/F)$ which only has degree $2$.

So elements of this extension are of the form $a+b\sqrt{-d}$ where $a,b\in F$. Then the two possible absolute values are just

$$|a+b\sqrt{-d}|, |a-b\sqrt{-d}|$$

where $|\cdot |$ is the absolute value already present on $\Bbb C$ with $F$ as an embedded subfield. If $F\subseteq \Bbb R$ is real, then it's clear the two absolute values are complex conjugates of one another, hence induce equivalent places. In short: they always induce a unique complex place.

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    $\begingroup$ I thank you for taking the time to write this but I understand how the extension of $F$ works, as a fixed real field. In the context I'm coming from I want to know about how this goes over $\mathcal{Q}$, but I see from your explanation that my question makes that nonsensical. Allow me to edit the question then to be more clear about what I'm puzzled with here.. $\endgroup$ – j0equ1nn Mar 28 '16 at 0:05
  • $\begingroup$ @j0equ1nn I've updated my answer. Cheers. $\endgroup$ – Adam Hughes Mar 28 '16 at 0:31
  • $\begingroup$ This is very clear now and much appreciated. You pretty much just saved me from saying something very stupid to a number theory audience! (I am more of a topologist.) $\endgroup$ – j0equ1nn Mar 28 '16 at 0:51

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