Do finite places of a number field also correspond to embeddings?

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Something that seems to be pretty standard in every introductory treatment is that the infinite places correspond to embeddings into $\mathbb{C}$. Do the finite places correspond to embeddings as well? I can envision two possibilities. My first guess is that the primes sitting above $p \in \mathbb{Q}$ correspond to embeddings into $\overline{\mathbb{Q}_p}$, and thus also to embeddings into $\mathbb{C}$ by some messy non-canonical field isomorphism. My second guess, which I think would imply the first, is that the places of $\mathbb{Q}[\alpha]$ above $p \in \mathbb{Q}$ correspond to embeddings into $\mathbb{Q}_p[\alpha]$. I've never been able to find a precise statement about this in any of the texts I've been studying (mostly Milne's notes and Frohlich & Taylor) and would appreciate if anyone could let me know where to learn more about this -- or if I'm just plain wrong.

One other thing is that the embeddings into $\mathbb{C}$ play a central role in analyzing the basic structure of a number field by way of Minkowski theory. Is there some analog for the finite places, or does that even make any sense?

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One way to make your second question precise is Leopoldt's conjecture (whose status is currently up in the air, at least last I heard; maybe someone here has an update); see e.g. its Wikipedia page. – D. Savitt Jan 28 2010 at 11:13
Thanks everybody for all the great responses. Every one of them is helpful and I wish I could accept more than one. However, I am going to wait a bit before accepting any, so as not to discourage any other good answers. That is, if there's even any more to be said at this point anyway.. – Jon Yard Jan 29 2010 at 6:50

6 Answers

The Archimedean places of a number field K do not quite correspond to the embeddings of K into $\mathbb{C}$: there are exactly $d = [K:\mathbb{Q}]$ of the latter, whereas there are $r_1 + r_2$ Archimedean places, where:

if $K = \mathbb{Q}[t]/(P(t))$, then $r_1$ is the number of real roots of $P$ and $r_2$ is the number of complex-conjugate pairs of complex roots of $P$. In other words, $r_1$ is the number of degree $1$ irreducible factors and $r_2$ is the number of degree $2$ irreducible factors of $P(t) \in \mathbb{R}[t]$.

There is a perfect analogue of this description for the non-Archimedean places. Namely, the places of $K$ lying over the $p$-adic place on $\mathbb{Q}$ correspond to the irreducible factors of `$\mathbb{Q}_p[t]/(P(t))$`; or equivalently, to the prime ideals in the finite-dimensional `$\mathbb{Q}_p$`-algebra $K \otimes_{\mathbb{Q}} \mathbb{Q}_p$.

More generally: if $L = K[t]/P(t)/K$ is a finite degree field extension and $v$ is a place of $K$ (possibly Archimedean), then the places of $L$ extending $v$ correspond to the prime ideals in $L \otimes_K K_v$ or, if you like, to the distinct irreducible factors of $P(t)$ in $K_v$, where $K_v$ is the completion of $K$ with respect to $v$.

See e.g. Section 9.9 of Jacobson's Basic Algebra II.

By coincidence, this is exactly the result I'm currently working towards in a course I'm teaching at UGA. I'll post my lecture notes when they are finished. (But I predict they will bear a strong resemblance to the treatment in Jacobson's book.)

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Thanks Pete. I didn't know the formulation using prime ideals in the tensor product algebra.  Looking forward to seeing your lecture notes when they're done. – Jon Yard Feb 1 2010 at 5:08

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I'm not sure why Pete stopped short of coming back to the original poster's description in terms of embeddings. The point is that the infinite places of a number field $K$ correspond to embeddings of $K$ into $\mathbb{C}$ up to equivalence, where two embeddings are considered equivalent if one can be obtained from the other by post-composing with an element of `$\textrm{Aut}(\mathbb{C}/\mathbb{R})$`. In exactly the same way the places of $K$ lying over $p$ correspond to embeddings of $K$ into $\overline{\mathbb{Q}}_p$ counted up to equivalence, where two embeddings are considered equivalent if one can be obtained from the other by post-composing with an element of $\textrm{Aut}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$.

This is not hard to derive this from what Pete wrote. (Consider `$K = \mathbb{Q}[t]/P(t)$` and the description of the places above $p$ as irreducible factors of $P(t)$ over $\mathbb{Q}_p$. An embedding of $K$ into $\overline{\mathbb{Q}}_p$ amounts to choosing a root of $P(t)$ in $\overline{\mathbb{Q}}_p$, and two embeddings will be equivalent if and only if the corresponding roots are roots of the same irreducible factor of $P(t)$.)

And of course by the same argument there's a similar description of the places of L lying over a place of K.

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"I'm not sure why..." Because I was laboring under the delusion that if I stopped typing up answers on Math Overflow, I might get some sleep? Also, because I saw something in the question that reminded me so strongly of what's already on my mind, I sort of ignored the rest. And finally, I'm not completely sure I knew explicitly what you said [although if you told me that I did, I would believe you]. I think you have given me an exercise for my course: +1. – Pete L. Clark Jan 28 2010 at 11:26
OK, yes, upon further reflection, I did know this. But thanks for reminding me! – Pete L. Clark Jan 28 2010 at 16:48
Thanks David. I really like the notion of embeddings up to equivalence. It makes complete sense. – Jon Yard Feb 1 2010 at 4:27

Another way to think of a place of a number field is in terms of equivalence classes of absolute values. You should work out how this definition is the same as the one given by David Savitt.

Let $k$ be any field. An absolute value on $k$ is a homomorphism $|\ |:k^\times\to\mathbb{R}^{\times\circ}$ from the multiplicative group of $k$ into the ordered group of stritctly positive reals which is not trivial and such that the triangular inequality $$|x+y|\le |x|+|y|$$ is satisfied for all $x,y\in k$, with the convention that $|0|=0$.

An absolute value $|\ |$ gives rise to a distance $d(x,y)=|x-y|$ on $k$, making it into a metric space. It can thus be completed to a field $k_{|\ |}$ in which $k$ is a dense subfield. Two absolute values on $k$ are equivalent if they induce the same topology on $k$, and thus give rise to the same completion.

Ostrowski (1918) determined all absolute values on a number field, and Artin (1932) gave a beautifully simple proof of his theorem. You can read all about it in many places, including my online notes arXiv:0903.2615

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Thanks for the link, Chandan.  Your notes will be of use to me for the class I'm teaching. – Pete L. Clark Jan 28 2010 at 16:47
Regarding DLS: he hath revealed unto us his true name.   – Pete L. Clark Jan 29 2010 at 8:38
Ah !  He turns out to be a friend !  Welcome, David. – Chandan Singh Dalawat Jan 29 2010 at 9:40
Hi!  Likewise, good to see you here. – D. Savitt Jan 29 2010 at 11:51
Thanks to the pointer to your notes Chandan. – Jon Yard Feb 1 2010 at 4:29

A more precise way of expressing your second guess is as follows.

Given a number field $K$ (a finite extension of $\mathbb{Q}$), and a place $v$ of $\mathbb{Q}$ (so that $v$ is either a prime number of the sybol $\infty$, corresponding to the completion $\mathbb{R}$ of $\mathbb{Q}$), we may ask what $K\otimes\mathbb{Q}_v$ is, where $\mathbb{Q}_v$ is the completion of $\mathbb{Q}$ at $v$.

Pete beat me to the rest.

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Just to tack something on about your final question: There exists a rather concrete "yes" answer to your last question in the form of Arakelov theory. One defines an Arakelov divisor as a finite formal linear combination of places, with integral coefficients on the finite places and real coefficients on the infinite places. You map $K^\times$ into the divisor group via the natural definition of principal Arakelov divisors, and if you compose this map with the projection on to the infinite components of the divisor group, you recover the logarithm map from Minkowski theory. So at the very least, ignoring the extra stuff you just chucked in doesn't prevent you from recovering Minkoswki theory. But there's much more to be gained, namely, by considering the quotient of the Arakelov divors by the principal ones and dubbing this the Arakelov class group.

The power of this construction, at least as I see it, is to simultaneously deal with two of the most powerful ideas in basic algebraic number theory -- the class group and the Minkowski embedding -- within the context of a single object, the Arakelov class group. This line of thought is pursued in depth beautifully in Chapter 3 of Neukirch's Algebraic Number Theory. Roughly, the Arakelov class group looks a lot more like the divisor class groups arising in function fields, and motivated by this analogy, you can define line bundles, Euler characteristics, genus, canonical bundles, Chern classes, etc., for number fields, culminating with a pretty compelling analogue for Riemann-Roch for number fields. (Again, see Neukirch).

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Thanks to pointing me to Neukirch for the Arakelov theory.  I knew (loosely) about the analog between number and function fields, but didn't know where to look to find out more.  This was actually part of the motivation for my question; since places are the points of a topological space, it seems they should be treated on the same footing as much as possible. – Jon Yard Feb 1 2010 at 4:33

I would say that the finite primes of $K$ play a very strong role in analyzing the structure of a number field! So-called local methods are one of the most powerful tools in developing basic algebraic number theory! It may be that they are not 'essential' in developing the theory, but I think they bring a great deal of clarity to the picture. There are a bunch of good answers already, but I want to throw a couple things on the pile.

To begin with, one can think about the primes of $K$ as being given by equivalence classes of metrics on $K$. Equivalently, you can think of them as being given by all the equivalence classes of embeddings of $K$ into complete fields. We know that studying the embeddings of $K$ into $\mathbb{C}$ leads to Minkowkski theory and, among other things, a proof of Dirichlet's Unit Theorem, finiteness of the classgroup, etc. What can we get out of studying the embeddings of $K$ into $K_v$, the completion of $K$ at a finite prime $v$?

One nice thing is that when you complete $K$, you also complete $\mathcal{O}_K$, the ring of integers in $K$. If $\mathcal{O}_v$ is this completion than it turns out that $\mathcal{O}_v$ is always a complete discrete valuation ring and as such, questions about ideals in this ring are simpler than they are in $\mathcal{O}_K$. The theory that one develops allows you to easily pass back and forth between questions about ideals in $\mathcal{O}_K$ and ideals in $\mathcal{O}_v$, at the cot of having to consider all of the finite primes. Off the top of my head, this technique makes it very easy to prove functoriality properties of the different associated to an extension of number fields.

Also, when working with complete objects, the techniques of analysis become available! An important step in at least one proof of class field theory involves using $p$-adic analogues of the logarithm and exponential functions to relate the structure of the multiplicative group of units in $\mathcal{O}_v$ with the structure of the additive group of $\mathcal{O}_v$.

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"using $p$-adic analogues of the logarithm and exponential functions to relate the structure of the multiplicative group of units in $\mathfrak{o}_v$ with the structure of the additive group of $\mathfrak{o}_v$". You can do without. See Part III of arXiv:0711.3878. – Chandan Singh Dalawat Jan 29 2010 at 10:30
Thanks. I hadn't realized that completing the field completes its ring of integers as well, but I guess this makes complete sense in hindsight. – Jon Yard Feb 1 2010 at 4:35
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