Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

By the fundamental theorem of algebra, the algebraic closure $\mathbb{K}$ of $\mathbb{Q}$ decomposes as $\mathbb{K} = F \oplus i F$ where $F = \mathbb{R} \cap \mathbb{K}$ (the intersection is in $\mathbb{C}$). I want to know if there is a purely algebraic way to characterize $F$, i.e. without invoking any analysis, topology, or transcendental number theory. I am asking this because I noticed that it is often convenient when working with examples in characteristic 0 algebraic number theory to give preference to the real roots of a polynomial, and I am wondering if there is a canonical algebraic way to formulate this preference. It doesn't seem like an object built out of lots and lots of transcendental extensions should be so fundamental to purely algebraic examples.

Here are some specific questions that I have been playing with.

Is there a purely algebraic way to distinguish between the splitting fields of $x^2 + 2$ and $x^2 - 2$?

Is there a purely algebraic way to distinguish the real root among the three roots of $x^3 - 2$ in a splitting field?

Of course, the relevant algebraic structures can't be invariant under $\mathbb{Q}$ automorphisms. But I don't see why one can't just be a little bit imaginative.

(Examples edited, changing 1 to 2)

share|improve this question
    
I guess a better question is whether one can easily distinguish which automorphism of the splitting field is complex conjugation. –  François G. Dorais May 24 '10 at 1:13
2  
François: Only for CM fields (totally imaginary quadratic extensions of a totally real number field) is there truly a canonical complex conjugation. That is, if you embed a Galois extension of Q into C, apply complex conjugation, and then pull it back to the original field you get an order 2 aut. of the field (we assume it's not a totally real number field). When is this aut. independent of the embedding? Only for CM fields. –  KConrad May 24 '10 at 1:15
    
Thanks KConrad! I now have a better idea what CM fields are all about. –  François G. Dorais May 24 '10 at 1:26
    
KConrad: Am I right in thinking that it is always possible to extend to a CM field? But I guess there is no canonical way to do that. –  François G. Dorais May 24 '10 at 1:33
    
Arg... to obtain the examples that I was actually thinking about, replace "1" with "2" everywhere. I don't even know why I typed 1. I am editing the question. –  Paul Siegel May 24 '10 at 1:33

5 Answers 5

up vote 14 down vote accepted

Paul: you ask if there is a way to algebraically characterize the field of real algebraic numbers. As a specific field in $\mathbf C$, no there's not a good algebraic characterization, but as an abstract field yes there is a characterization. This field is one particular example (and the only concrete one at that) of a real closure of $\mathbf Q$. Any two real closures of $\mathbf Q$ are isomorphic to each other.

If you pick a number field $K$ other than the rationals, you can contemplate its real closures: maximal algebraic extensions of $K$ which admit an ordering. Assuming this is possible at least once (e.g., ${\mathbf Q}(i)$ has no real closure), then you can ask if the real closures of $K$ are all isomorphic to each other respecting the embedding of $K$ into them. Nope.

Consider $K = {\mathbf Q}(a)$ where $a^2 = 2$. If we stuff $K$ into the real algebraic numbers by sending $a$ to $\sqrt{2}$ then $a$ is a square in the real algebraic numbers (it's the square of $\sqrt[4]{2}$, which is a real algebraic number: we're talking about concrete real numbers that are algebraic). But if we stuff $K$ into the real algebraic numbers by sending $a$ to $-\sqrt{2}$ then $a$ is not a square in the real algebraic numbers (all squares in the real numbers are positive). Therefore these two embeddings of $K$ into the real alg. numbers are not compatible with each other as extensions of $K$. That is, there is no automorphism of the real algebraic numbers which commutes with these two embeddings of $K$ into it. In other words, real closures of $K$ are not all isomorphic as extensions of $K$.

Theorem: Let $K$ be a number field. Every real closure of $K$, up to isomorphism as an extension of $K$, looks like the real algebraic numbers using some real embedding of $K$, and different real embeddings lead to non-isomorphic real closures as extensions of $K$.

I think I got that right. If I screwed up I'm sure BCnrd will let me know. :)

share|improve this answer
    
Perfect. This answers the question that I wanted to ask, and you gave some great insight into the related question that I actually did ask in your comments. Thanks a ton! –  Paul Siegel May 24 '10 at 4:14
    
Just a side comment: the fact that any two real-closures of $\mathbb{Q}$ are isomorphic (and hence $\mathbb{Q}$-isomorphic!) is "lucky" -- i.e., it is absolutely not a general property of real-closures: consider the case of $\mathbb{Q}(t)$. I have been recently thinking that the name "real-closure" it itself somewhat misleading for these reasons... –  Pete L. Clark May 24 '10 at 11:45
    
Pete: I already wrote that for number fields K other than Q with a real embedding, its real closures are not isomorphic as extensions of K. What you are saying about real closures of Q(t) which is different? –  KConrad May 24 '10 at 19:52
    
@Pete: I had always assumed that "real closure" is something you do to an ordered field, rather than just a field, in which case it is unique (I think?). –  Hurkyl Feb 17 at 18:05
    
@Hurkyl: it can be used either way, and in the case of ordered fields it is unique. The double usage and the proof of uniqueness of real closure in the ordered field case occurs (e.g.) in my field theory notes:math.uga.edu/~pete/FieldTheory.pdf. –  Pete L. Clark Feb 19 at 0:32

The answer to your title question is "no", assuming "purely algebraic criterion" means something like a first order statement using only ring operations. This is because there are ring-theoretic automorphisms of the algebraic numbers that take the real algebraic numbers to some other real closed subfield (e.g., choose one that permutes the cube roots of 2 transitively, i.e., any element of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ whose image in $Gal(\mathbb{Q}[\sqrt[3]{2}, \frac{1+\sqrt{-3}}{2}]/\mathbb{Q})$ has order 3).

share|improve this answer
    
Using the fact that the automorphism group of $\bar{\mathbb{Q}}$ has cardinality $2^{\aleph_{0}}$ and that the field of real algebraic numbers is rigid, it follows that there are $2^{\aleph_{0}}$ copies of the real algebraic numbers in $\bar{\mathbb{Q}}$. –  Simon Thomas May 24 '10 at 2:28
    
I am certainly willing to entertain more general algebraic criteria, if it helps. For example, one idea I had was to characterize the field of real algebraic numbers as the maximal ordered subfield of the algebraic closure of $\mathbb{Q}$, but I have no reason to believe that this gives the correct answer. –  Paul Siegel May 24 '10 at 3:02
    
@Paul: The criterion of admitting a maximal linear ordering in an algebraic closure is (one of many criteria that are) equivalent to being real closed, and as Simon Thomas mentioned, there are a lot of real closed subfields of an algebraic closure of the rationals. –  S. Carnahan May 24 '10 at 3:14
    

The real root of $x^3-1$ satisfies $x=1$, which the non-real roots do not, but I cannot imagine any purely algebraic way to distinguish the real root of $x^3-2$ from the others. This may say more about my imagination than it does about the cube roots of two, but it seems to me that any purely algebraic equation satisfied by any one of the roots is satisfied by all three, making it impossible to distinguish them in any purely algebraic way.

share|improve this answer
4  
Paul, along the lines of what Gerry writes you should think about the cube roots of 2 over the 5-adic numbers. The polynomial x^3 - 2 over Q_5 has one linear factor and one quadratic factor, so it has a unique root in Q_5. One might say "this root is different from the other two", but there is no way you can compare the unique root in Q_5 with the unique real root. In fact, the 3 roots of x^3 - 2 in C are not comparable in a reasonable way with the 3 roots of x^3 - 2 in the alg. closure of Q_5. It's like trying to compare a choice of generators in two different cyclic groups of order 9. –  KConrad May 24 '10 at 1:06
    
Hmm, the example of $x^2 - 2$ over $\mathbb{Q}_5$ is pretty persuasive. Does the unique root really have nothing to do with the real cube root of 2 (other than the obvious)? Is there any good way to describe the 5-adic cube root of 2? –  Paul Siegel May 24 '10 at 1:44
    
Paul: You meant x^3 - 2. –  KConrad May 24 '10 at 2:05
    
That there are three roots of x^3 - 2 in C and in the alg. closure of Q_5 really forces you to understand better the theorem "a poly. of degree n over a field has at most n roots". Apparently I have given 6 roots? Well, the theorem is really that there are at most n roots in any specific field extension. Let's consider x^2 - 11, which has two real roots and two 5-adic roots. Can we identify the real and 11-adic roots with one another any natural way? No. The fields R and Q_5 are two fields containing Q, but there is no natural embedding of either field in the other. –  KConrad May 24 '10 at 2:09
    
The correct way to deal with this situation is to think about the two real embeddings of Q(sqrt(11)) and the two 11-adic embeddings of Q(sqrt(11)) as separate gadgets. –  KConrad May 24 '10 at 2:10

One of your questions does have a positive answer.

  • Is there a purely algebraic way to distinguish between the splitting fields of $x^2+2$ and $x^2−2$?

The field $Q(\sqrt 2)$ can be ordered while the field $Q(i\sqrt 2)$ can not since -1 is a sum of squares.

share|improve this answer
1  
The purely algebraic way to distinguish between these fields is that they are not isomorphic as fields, since $-1$ is not a square in $\mathbb{Q}$! The question seems rather to be about the existence of formally real but not totally real algebraic numbers... –  Pete L. Clark May 24 '10 at 8:58

As other correspondents have pointed out there is no algebraic way to distinguish the elements of $\mathbb{Q}^{alg}\cap\mathbb{R}$ but there is an algebraic way of distinguishing totally real algebraic numbers (those whose conjugates are all real). An algebraic number $\alpha$ is totally real iff $n-r=1$ where $n$ is the degree of $K=\mathbb{Q}(\alpha)$ and $r$ is the rank of the unit group of the integral closure of $\mathbb{Z}$ in $K$. This follows from Dirichlet's units theorem.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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