toad and I were talking on IRC, and he brought up some questions I don’t know the answers to, so maybe someone reading this can help.

- What complete fields are there besides the real line and complex plane?
- Is every finite field complete in a metric other than the discrete metric?
- What about matrices over a finite field? Since a matrix is a linear operator we can give it in a norm, but does being in a finite field screw with things?

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2. The discrete metric is basically the only metric on a finite field. Start by making a topology, obviously we want individual points to be closed sets. This makes finite unions of points also closed sets. However ever subset of a finite set is a finite union of points, hence ever set is closed (and hence every set is open). This implies that we have the discrete topology and hence the discrete metric.

The answer to 2. implies that finite fields are examples of 1. since the only sequences converge to one of the points in the field. Also, the algebraic closure of finite fields is almost certainly complete, but I don’t know of a proof or even a nice topology off the top of my head.

Comment by L. Zoel — July 16, 2008 @ 4:17 pm |

I guess I have a few terminology things that I need to be clarified. Complete has more than one meaning. I’m assuming also by “field” you mean “ordered field” in which case I will assume complete means “having the least upper bound property.” (The problem is that you talk about metric in 2, in which case complete usually means that every Cauchy sequence converges).

The complex plane is not an ordered field, so I may be misinterpreting one of the definitions. So for 1, often times the real line is called “the” complete ordered field, because essentially every complete ordered field must behave like R. Any subfield of R is totally ordered, but as you have seen Q is not complete. Also, any finite field is not an ordered field, so complete in the sense of order doesn’t apply there.

I’ll hold off for further clarification before posting more.

Comment by hilbertthm90 — July 16, 2008 @ 4:20 pm |

By complete I mean that every Cauchy sequence converges to a point in the space. By field I mean the algebraic structure, with or without order; a set with two operators, multiplication and addition, such that the set under addition is an Abelian group, the set without zero under multiplication is an Abelian group, and the distributive properties hold.

The question arises from the post on all finite-dimensional normed spaces on R or C being Banach. Finite dimensional normed spaces over a general field may not be complete; take a vector space over Q for instance. The question is what other fields can I take a normed space over where the space will be complete whenever the space is finite dimensional.

Comment by cjohnson — July 16, 2008 @ 4:28 pm |

Let me try again. It is sort of odd to speak so generally, because it seems as if this partially comes from the norm and not the field. In other words, (I don’t have an example, so this may not be true) it seems as if you could have two different norms on the same finite dimensional vector space over some field, but one is complete and the other is not. I haven’t thought about this carefully, though, so it may not be true.

The other thought I had was that if the proof does come down to relying on the completeness of the underlying metric space, then a key fact is that every complete metric space (with no isolated points) is uncountable. This rules out Q (which was known) or finite fields.

Comment by hilbertthm90 — July 16, 2008 @ 8:54 pm |

I don’t believe you can have two different norms and be complete with one and not complete with the other, since all norms on the same finite-dimensional vector space are equivalent, any two norms or only “off” from one another by multiplication of a constant, which doesn’t change the limit of a sequence, or the sequence’s “Cauchy-ness.”

Hmm… I haven’t seen a proof that a complete space with no isolated points is uncountable, but I that sounds reasonable.

I’m wondering now though if this is what you were saying, and I was just too dense to get it: a “field” is an algebraic structure, but “completeness” is an analytic thing that depends on having some metric at hand, so is there a problem with saying “a field is complete” ?

Comment by cjohnson — July 16, 2008 @ 9:14 pm |

That is precisely what was confusing me at first! Sorry, I was unclear.

Ah, you’re right about that equivalence of norms on finite-dimensional vector spaces. I decided to completely ignore the “finite-dimensional” part of the question for some reason.

If you know the Baire Category Theorem, then the proof is quite simple.

Let X be a countable complete metric space. Then . In other words, X is a union of all the elements written as singletons. This is a countable union by assumption. Now as long as no x is isolated, each singleton is nowhere dense. This violates the Baire category theorem that says you cannot write a complete metric space as a countable union of nowhere dense sets.

Comment by hilbertthm90 — July 16, 2008 @ 10:15 pm |

1. I assume you are asking if there are other fields that are also metric spaces which are complete. The answer is, yes, definitely. The p-adics are a good example. That said, one can take *any* field that is a metric space, and its (metric space) completion will have a natural field structure, and will, obviously, be complete. Are you familiar with what I mean by a (metric space) completion? The procedure is exactly analogous to constructing R from Q (by taking Cauchy sequences and modding out by the equivalence relation ~ defined by (a_n)~(b_n) if (a_n-b_n)->0).

2. Any finite metric space is discrete, so the answer is trivially no.

3. I’m not sure what norm you’re giving these matrices. Do you mean the determinant? Remember that there are two different notions of norm here–a (topological) norm is a map into R with the required properties, whereas a (matrix) norm, e.g. the determinant, is a map into the underlying field.

Comment by Daniel — July 17, 2008 @ 11:28 am |

Or, re: 3 I suppose you may have meant the operator norm. But remember, finite metric spaces are discrete. As the space of n x n matrices is finite, the same thing happens.

Comment by Daniel — July 17, 2008 @ 11:35 am |

Yes, I’m familiar with the completion of a metric space (I was actually planning on writing the proof of that up here soon), but I’m still a bit confused by your response to #1. We can always turn a field (or any set) into a metric space, so if we have a field F, give it a metric, we can then find a complete space that, say G, that has a dense subset isometric to F. My question now though is how do you know that G is itself a field? (I’m absolutely not an algebra person, so that might be a trivial question).

For #2, what about taking a the set of all n-tuples over Z_2 (so n-tuples where each entry is a 0 or 1), and using the Hamming distance: the number of “spots” where the n-tuples disagree? I.e., d((0, 0, 1), (1, 0, 0)) = 2. This isn’t the discrete metric I was referring, but wouldn’t the space be complete with this metric? All distance are going to be a natural number, or zero, so for a sequence to be Cauchy all its entries must become constant at some point, right?

I did mean the operator norm, yes.

Comment by cjohnson — July 17, 2008 @ 12:00 pm |

There is the brute force way to show G is a field. I’ve worked enough out to convince myself, but there is probably a much easier way to see it.

We need to extend the operations to G. Let a be in G. By denseness we have a sequence that converges to a. Let b be in G. Likewise . The sum of two convergent sequences and the product of two convergent sequences is the sum and product of the limits respectively (Use completeness to show existence and uniqueness). So define + on G by and * on G by .

Now we have two operations on G (and they correspond properly with F). We just need to check whether they satisfy the field properties. These seem mostly obvious. Just pass to the limit definition and invoke limit properties. I haven’t checked them, though, as it would be more work than it is worth.

Easier methods would be greatly appreciated!

Comment by hilbertthm90 — July 17, 2008 @ 1:17 pm |