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I've been reading recently about algebraically closed fields and complete metric spaces, and it seems to me that they are very similar ideas. Is there some more general mathematical concept which these two ideas are both instantiations of?

Also, I am sure that $\mathbb R$ is a complete metric space, but is not algebraically closed, while $\mathbb C$ is both. Is it possible to construct an algebraically closed metric space which is not complete?

Thanks in advance!

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When you say "algebraically closed metric space," you're implicitly talking about a set which is equipped both with the structure of a field and with the structure of a metric space, and you need to specify what kind of compatibility you want these structures to have (e.g. in the case of $\mathbb{R}$ the field operations are all continuous w.r.t. the metric). –  Qiaochu Yuan Dec 23 '11 at 19:43
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3 Answers 3

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The processes of algebraic closure and metric completion are actually less similar than they first appear. Metric completion is canonical in the sense that doing it requires making no arbitrary choices. You simply "adjoin limits of all Cauchy sequences." (More precisely - and this is not going to make sense to you, but I include it for the sake of completeness - there is an inclusion functor from the category of complete metric spaces to the category of metric spaces, and metric completion is its left adjoint.)

Algebraic closure, however, requires making certain arbitrary choices. It may not seem like it does, since you just "adjoin roots of all polynomials," but Cauchy sequences have unique limits and polynomials do not have unique roots; you (naively) have to pick some arbitrary order in which to adjoin the roots of a particular polynomial even though there's no canonical way to choose such an order. Moreover, you (naively) have to pick an order on the set of polynomials; you can't just adjoin all of their roots at once like you can with Cauchy sequences, since adjoining some roots will make adjoining the roots of other polynomials redundant. (More precisely - again, for the sake of completeness - there is an inclusion functor from the category of algebraically closed fields to the category of fields, and it does not have a left adjoint. Moreover, algebraic completion can't be extended to a functor at all.)

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A very illuminating answer! –  Lubin Dec 23 '11 at 20:01
    
Between this answer and the answer below in regards to phrasing the operations in terms of closure, I think I understand things a bit better. Thanks! :-) –  Dan M. Katz Dec 23 '11 at 21:56
    
@Dan: I assume that you mean my answer. I'm glad it helps. –  Asaf Karagila Dec 23 '11 at 22:09
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The algebraic closure of the rational numbers is an algebraically closed field, however it is only countable, so it is not $\mathbb C$.

However, there is still a Cauchy sequence converging to $\pi$ (since the rationals are dense in $\mathbb R$) but alas $\pi$ is not algebraic over $\mathbb Q$ so the field is not a complete metric space.

You may be interested in reading about the notions of real closed field and formally closed field which are somewhat related to your question.

On a general note, it seems that you see the similarity in "closure under property X", either Cauchy limits or polynomials. This is not a strange concept, and it is very common to take a certain property and ask yourself what happens when you close under it.

If you take the natural numbers, what happens when you close it under subtraction? You get $\mathbb Z$; when you close that under division you get $\mathbb Q$; and you can keep going and find richer structures (exponentiation, continuity, etc.)

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The algebraic closure of $\mathbb{Q}$ is algebraically closed. Considering it as a subset of $\mathbb{C}$ it has an induced metric. It is not complete in this metric, for example because it contains $Q[i]$ which is dense in $\mathbb{C}$, but it is different from $\mathbb{C}$ (for example, because it is countable).

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