# Relationship between algebraically closed fields and complete metric spaces?

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?

• 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). Commented Dec 23, 2011 at 19:43

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.)

• A very illuminating answer! Commented Dec 23, 2011 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! :-) Commented Dec 23, 2011 at 21:56
• @Dan: I assume that you mean my answer. I'm glad it helps. Commented Dec 23, 2011 at 22:09

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.)

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).

tl;dr Both metric completion and taking algebraic closure are (generalized) closure operators.

Harry Barber has mentioned that

Metric completion and algebraic closure kinda feel the same until you dig in and realise they share no properties.

(Thanks Harry for redirecting me to this question)

This makes me wonder why they look the same in the first place. Existing answers focus on explaining their differences. Therefore, my answer will only explain why they look the same.

Wikipedia suggests there's something called a closure operator. However, the definition given is too restrictive. (Our domain and codomain are proper class instead of set. Also, subset and equality are too strong.) Therefore, we tweak the given definition to make it better suit our purposes:

A (generalized) closure operator on a concrete category $$\mathscr{C}$$ is defined to be a mapping $$\operatorname{cl}: \mathscr{C} \to \mathscr{C}$$ satisfying: $$X \hookrightarrow \operatorname{cl}(X) \tag{extensive}$$ $$X \hookrightarrow Y \implies \operatorname{cl}(X) \hookrightarrow \operatorname{cl}(Y) \tag{increasing}$$ $$\operatorname{cl}(\operatorname{cl}(X)) \cong \operatorname{cl}(X) \tag{idempotent}$$ for each $$X, Y \in \mathscr{C}$$ where $$\hookrightarrow$$ denotes $$\text{embeds into}"$$ and $$\cong$$ denotes $$\text{is isomorphic to}"$$.

Now we claim that metric completion is a closure operator on the category of isometries (with metric spaces as objects) and taking algebraic closure is a closure operator on the category of fields (with ring homomorphism as maps). Before proceeding, recall that in both categories, all maps are injective, so no checking is required.

Proof (taking algebraic closure)

Extensivity follows from algebraic closure being an algebraic extension of the original field.

Idempotency holds because an algebraically closed field has no proper algebraic extension. Therefore, an algebraic closure of an algebraically closed field is forced to be isomorphic to itself.

To see that taking algebraic closure is increasing, suppose $$K \hookrightarrow L$$. Then $$K \hookrightarrow \overline{L}$$ by composition of extension. Now $$\{\alpha \in \overline{L} \mid \alpha \text{ is algebraic over } K\}$$ is an algebraic closure over $$K$$. Therefore, $$\overline{K} \cong \{\alpha \in \overline{L} \mid \alpha \text{ is algebraic over } K\} \subseteq \overline{L}$$. $$\tag*{\square}$$

Proof (metric completion)

To show extensivity, consider the map $$r \mapsto [(r)]$$ which takes $$r \in M$$ to the equivalent class of the constant sequence $$(r) \in \overline{M}$$. This is an isometry because $$d([(r)], [(s)]) = \lim_n d(r, s) = d(r, s)$$

For idempotency, it suffices to show the map $$r \mapsto [(r)]$$ is surjective when $$M$$ is complete. Take any $$[(r_n)] \in \overline{M}$$. By completeness of $$M$$, $$(r_n) \to r$$ for some $$r \in M$$. This shows $$r \mapsto [(r)] = [(r_n)]$$.

To see metric completion is increasing, note that the map $$\phi: M \to N$$ induces another map $$\overline{\phi}: \overline{M} \to \overline{N}$$ via $$[(r_n)] \mapsto [(\phi(r_n))]$$. Suppose $$(r_n), (s_n) \in [(r_n)]$$. Then $$d([(r_n)], [(s_n)]) = 0 \implies \lim_n d(r_n, s_n) = 0$$ and \begin{align} d([(\phi(r_n))], [(\phi(s_n))]) &= \lim_n d(\phi(r_n), \phi(s_n)) \\ &= \lim_n d(r_n, s_n) \end{align} since $$\phi$$ is an isometry. Hence $$d([(\phi(r_n))], [(\phi(s_n))]) = 0 \implies [(\phi(r_n))] = [(\phi(s_n))]$$ This shows $$\overline{\phi}$$ is well-defined. To see $$\overline{\phi}$$ is an isometry. Note that \begin{align} d([(\phi(r_n))], [(\phi(s_n))]) &= \lim_n d(\phi(r_n), \phi(s_n)) \\ &= \lim_n d(r_n, s_n) \\ &= d([r_n], [s_n]) \end{align} using the fact that $$\phi$$ is an isometry. $$\tag*{\square}$$