2
$\begingroup$

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."

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

$$|\sigma_i(a)+\sigma_i(b)\sqrt{-d}|$$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.