Algebraic completion (closure)

[Ben]

Since this is supposed to a Physics AND Math forum, I am going to try and represent the mathematicians here. So.

 

This question arose in another thread

Could one

abstractly define [imath]\mathbb{C}[/imath] as being THE uncountable field which is also algebraically complete?

 

There might be theorems out there by which one could say so...

 

Off the top of my head, I had thought the answer would be "yes, up to isomorphism".

 

But serendipitously revising some Galois Theory (a torture I suppose physicists don't need to endure), I am not so sure. In what follows, I am more-or-less thinking aloud, so bear with me.

 

First some definitions: A field [math]K[/math] is algebraically complete (closed) iff every polynomial with coefficients in [math]K[/math] has roots in [math]K[/math].

 

Let [math]K[/math] be a field. Then by definition it has an additive identity [math]0[/math] and a multiplicative identity [math]1[/math].

 

I will define the characteristic of the field [math] K[/math] to be [math]0[/math] iff the additive identity is the integer zero.

 

In the case, say, of some field [math]Z_n \equiv Z/nZ[/math] the additive identity really isn't an integer, rather it's the equivalence class [math][0] = \{\cdots,-3n,-2n, -n, 0, n, 2n,3n \cdots \}[/math].

 

In which case, one says that the this field has characteristic [math]=n[/math].

 

Discussion point: Why does this imply that any field whatever has characteristic zero or p, where p is prime?

 

Another definition: the field of real numbers [math]R[/math] is uncountable and not algebraically complete (closed). The algebraic of completion of [math] R[/math] is found by adjoining the imaginary unit [math]R(i) \equiv C[/math]. This is called an extension of [math]R[/math], where, by definition that [math] R\subsetneq C[/math]

 

Claim: Any two (or more) uncountable fields which are algebraically complete AND of the same characteristic are isomorphic.

 

Proof? (Lacking at present, bear with me)

 

Assuming I can get a proof, the clear implication here is that uncountable algebraic fields of different characteristic need not be isomorphic.

 

So another discussion point: does this mean we should accept that the complex numbers [math]C[/math] are not the only uncountable algebraically complete field.

[Ben]

I will define the characteristic of the field [math] K[/math] to be [math]0[/math] iff the additive identity is the integer zero.

 

In the case, say, of some field [math]Z_n \equiv Z/nZ[/math] the additive identity really isn't an integer, ................... In which case, one says that the this field has characteristic [math]=n[/math].

 

Discussion point: Why does this imply that any field whatever has characteristic zero or p, where p is prime?

Now you guys really aren't paying attention; you failed to spot I effed up here.

 

First I have just proved that the RING [math]Z_n[/math] will be a field iff [math]n[/math] is prime. I can bore you with the details if you want (you won't!)

 

Second, the right formulation for the characteristic of a field is this:

 

Define the prime subfield of the field [math]K[/math] to be the smallest field contained in [math]K[/math], i.e. the intersection of all its subfields.

 

Now the prime subfield of any field is isomorphic either to [math]Q[/math], the field of rational numbers, or to [math]Z_p[/math], the field of integers modulo some prime number [math]p[/math].

 

I also have a proof of this. Go on, dare me.....

 

So if [math]K[/math] has a prime subfield isomorphic to [math]Q[/math] one says that [math]K[/math] is of characteristic zero, essentially for the reason I gave. Likewise, if [math]K[/math] has a prime subfield isomorphic to [math]Z_p[/math] one says it has characteristic [math]p[/math]

 

Sadly I don't yet have a proof of my main contention, that algebraically complete uncountable fields are isomorphic iff they have the same characteristic.

[Nootropic]

You should be careful with your definitions. What we are discussing is the algebraic closure of a field and "completion" is an entirely different subject.

 

While I know the proofs to things you've posted (Z/nZ is a field iff n is prime and the prime subfield of a field is isomorphic to either Q or Z/pZ), it'd be a good idea if you posted them so that your ideas can be reviewed.

 

Now onto your main conjecture. This flat out false. And it is fairly easy to construct a counterexample. Let F be the field of rational functions over the real numbers in countably many variables, thus F (as a vector space) over R has infinite dimension and is clearly uncountable (in particular it contains R), thus it is an uncountable field. Now we know that algebraic closures exist (this is not at all trivial and requires some rather "tedious" bookkeeping using Zorn's Lemma, for a proof, see Dummit and Foote), so simply let K be the algebraic closure of F, then K contains F, so it is uncountable and is not isomorphic to C because it is not a finite dimensional over R as a vector space.

 

Here are some results that may interest you PlanetMath: cardinality of algebraic closure

[Ben]

You should be careful with your definitions. What we are discussing is the algebraic closure of a field and "completion" is an entirely different subject.

Sure, but the question I was originally asked used the term "algebraic completion". I don't like it, not because it need cause confusion with, say, Cauchy completion, rather that the concept of the algebraic closure of a field has a nice parallel with the concept of the closure of a topological space.

 

While I know the proofs to things you've posted ........, it'd be a good idea if you posted them so that your ideas can be reviewed.

And who, specifically, would be reviewing them? It would be be magic if you would be so kind, but who else? I seem to have a habit of boring the pants off forum members with excessive maths

Now onto your main conjecture. This flat out false......... so simply let K be the algebraic closure of F, then K contains F, so it is uncountable and is not isomorphic to C because it is not a finite dimensional over R as a vector space.

Then you mis-read my "main conjecture". It is true that, as I said at the head, that I had started by assuming that all algebraically closed fields are isomorphic to [math]C[/math], but then expressed my doubts.

 

So, I was in the process of trying to construct a proof that not every uncountable algebraically closed field is isomorphic to [math]C[/math]. Up to the point you joined this thread, that is where I was at.

 

Anyhoo, I thank you for your input. Lemme see what I can do with it.

[Nootropic]

Sure, but the question I was originally asked used the term "algebraic completion". I don't like it, not because it need cause confusion with, say, Cauchy completion, rather that the concept of the algebraic closure of a field has a nice parallel with the concept of the closure of a topological space.

 

Well, when I say be careful I'm saying it's wrong to use the term "algebraic completion", as this could possibly be confused with the completion of a ring or group and it is also not a mainstream term, as algebraic closure is.

 

And who, specifically, would be reviewing them? It would be be magic if you would be so kind, but who else? I seem to have a habit of boring the pants off forum members with excessive maths

 

It would hardly bore me. My favorite thing is discussing mathematics and I wouldn't mind seeing your proofs. Send them in an email if you prefer not to post them.

 

Then you mis-read my "main conjecture". It is true that, as I said at the head, that I had started by assuming that all algebraically closed fields are isomorphic to C, but then expressed my doubts.

 

So, I was in the process of trying to construct a proof that not every uncountable algebraically closed field is isomorphic to C. Up to the point you joined this thread, that is where I was at.

 

Anyhoo, I thank you for your input. Lemme see what I can do with it.

 

I'm not sure what else there is left to do? My example is one of many (see the link I left) and it is possibly to construct a field for the cardinalities mentioned in the link in a much more general fashion. There are certainly other modifications (ie. take F to be the field of rational functions over Z/pZ in uncountably many variables), but I believe my example settles the question at hand.

[Qfwfq]

Ben! I'm glad my query set you in motion, I feared my bun throwing had induced your flight from these forums.... :)

 

I can't linger here all that much, else I'd throw a few buns here too. :hyper: I'll just say there's a reason why I said 'complete' (not knowing, I confess, the mainstream term) and this reason causes me to cast a shade of doubt on the definition:

 

What I would tend to mean by algebraically "complete" is that, if [imath]n[/imath] is the degree of the polynomial, then it should have exactly [imath]n[/imath] roots counting multiplicity. Aside from the use of the two terms in topology, I see complete as being a fitting one, although I don't find it surprising if those darn mathematicians choose to say closed instead. :hihi:

 

See ya both later (I hope). :)

[Nootropic]

Another reason why not to use "algebraically complete" http://www.springerlink.com/content/u5866417715051q0/fulltext.pdf?page=1

[Qfwfq]

Errrrr... I don't like to insist myself with nitpicking but, in the paper you link to Nootropic, the very first definition uses the term complete in much the same manner as we've been meaning it.

 

In any case, as you seem to have busted the idea of these two properties being sufficient to single out [imath]\mathbb{C}[/imath] as a field, my musing now becomes to asks what further properties are requisite?

[maddog]

I have got admit my confusion also. I have always heard [imath]\mathbb{C}[/imath] as "uncountable" and as "algebraically complete". The Uncountability I can see trivially since [imath]\mathbb{C}[/imath] is a composition of

[imath]\mathbb{R}[/imath] which is already uncountable. I took algebraically complete to mean every complex polonomial has a solvable root.

 

Is this Not the case ?? :confused: :eek:

:xx::blink:

 

maddog

[Nootropic]

Perhaps this is a bit of "overseas miscommunication" if you will. Ben is from Great Britain (correct me if I'm wrong), so perhaps there (over in Europe) they refer to algebraic closure as algebraic completion, in much the same way that in French the term we use for a field is "body" (le corps). So I apologize if that's the mainstream terminology in Europe. I suppose maybe this is older terminology, but I can't say for sure, as most mathematicians I've met use "closure" and "closed".

 

Errrrr... I don't like to insist myself with nitpicking but, in the paper you link to Nootropic, the very first definition uses the term complete in much the same manner as we've been meaning it.

 

Similar, yes-- maybe it's my ignorance on the subject of the paper--but, not the same. Perhaps they are the same if I knew more about lattices, what exactly they meant by "parameters" (most likely coefficients, the obvious guess) and universal algebras. Maybe it's a generalization of what you have been using, but I can't say for sure.

 

In any case, as you seem to have busted the idea of these two properties being sufficient to single out C as a field, my musing now becomes to asks what further properties are requisite?

 

Not entirely sure what you mean by this? Are you asking under what conditions is an algebraically closed field isomorphic to C?

 

Is this Not the case ??

 

This is most definitely the case. The matter at hand was whether or not the complex numbers were the only (up to isomorphism) uncountable algebraically closed field. However, this is not true, as my example (in the first post) shows and in the link (also in the first post) constructs [uncountable] algebraically closed fields in a much more general fashion.

[Nootropic]

There seems to be numerous problems with the proof of "Theorem 3" in the first link I posted, but I still believe this statement to be true (I suppose that's what I get for using things like Planet Math), but a proof (by me) requires slightly more knowledge of transfinite cardinals than I have at the moment. One problem is that the "field of rational functions with integer coefficients" is a nonsensical statement (for example, 2x does not have an inverse in this ring). However, the field of rational functions with rational coefficient is. However, modifying this to the rationals makes no difference, as the rationals are countable. As for the rest of the argument, modifying "polynomial with integer coefficients" to "rational function with rational coefficients" and a slight modification to the following sentences in the proof will make it true, I believe. Not an entirely rigrous modification, but convincing enough.

[Ben]

Ben is from Great Britain (correct me if I'm wrong), so perhaps there (over in Europe) they refer to algebraic closure as algebraic completion, in much the same way that in French the term we use for a field is "body" (le corps).

Yes it's true, I am European. But no, it's not true that we use completion to mean algebraic closure, As I said earlier, I don't like it, but as long as we all know what we mean, WTF?

Not entirely sure what you mean by this? Are you asking under what conditions is an algebraically closed field isomorphic to C?

That's easy; an algebraically closed field [math]L:K[/math] is an extension of the field [math]K[/math] iff [math] K \subset L[/math] AND iff it is a splitting field for [math]K[/math], that is it contains all roots for any polynomial in [math]K[/math].

 

So it is not hard to show, I think, that if, [math]C= R(i),\,\,\, C' = R(j)[/math] are both algebraic extensions for the reals, then [math]C \simeq C'[/math]. I don't think countability/uncountability comes into it.

 

Theorem: Any uncountable set has a countable subset.

 

Proof; Let [math] S[/math] be uncountable, and start counting its elements. If, when Hell freezes over, OR you exhaust the counting numbers (i.e. [math]\mathbb{N}[/math]), whichever is sooner, you still haven't exhausted [math]S[/math], you will have found a countable subset.

 

Now every field has at least one subfield, that is itself. So let [math]L[/math] be an uncountable field, then, by the above, there must be some proper SUBSET of the SET, say, [math]K[/math] that contains 0 and 1 that is a countable subfield of the field [math]L[/math] i.e. [math]K\subsetneq L[/math].

 

But if any polynomial in [math] K[/math] has a root in [math]L:K[/math], then this latter is the algebraic closure of [math]K[/math]

 

So, for example, consider the field [math]Q[/math] of rational numbers which is obviously countable. Then any polynomial over this field will have roots in [math]Q(i),\,\,\,( i^2 \equiv -1)[/math], which is also countable.

[Qfwfq]

Well I was gonna give a more detailed reply yesterday but, as I was typing, I got a call about a new possible job that could be the best opportunity I've had ever, so far. So I just dropped my efforts, aborted everything and now, WTF about so much detail!!!:hihi: I'll just reply simply to state the matter better.

 

Theorem: Any uncountable set has a countable subset.

 

Proof; Let [math] S[/math] be uncountable, and start counting its elements. If, when Hell freezes over, OR you exhaust the counting numbers (i.e. [math]\mathbb{N}[/math]), whichever is sooner, you still haven't exhausted [math]S[/math], you will have found a countable subset.

Just love it, man!

 

However, you're off track about what I was looking for. Nootropic comes much closer.

 

I'm talking about abstractly defining [imath]\mathbb{C}[/imath] which is to say defining it as an abstract structure rather than a construct from [imath]\mathbb{R}[/imath]. Get it? We don't yet know what the heck [imath]\mathbb{R}[/imath] is, nor [imath]\mathbb{Q}[/imath] etc. We haven't cast Peano's axioms and constructed upward, we want to do the opposite. Obviously this also means: "Yes, up to isomorphism." So we have two requisite properties but they aren't enough to single out one and only one abstract structure.

 

[imath]\mathbb{C}[/imath] is a field which is:

1. of uncountable cardinality
   
2. such that any polynomial of degree [imath]n[/imath] has [imath]n[/imath] roots in it, counting multiplicities.
   
3. ??????
   
4. ??????????
   

 

Now every field has at least one subfield, that is itself. So let [math]L[/math] be an uncountable field, then, by the above, there must be some proper SUBSET of the SET, say, [math]K[/math] that contains 0 and 1 that is a countable subfield of the field [math]L[/math] i.e. [math]K\subsetneq L[/math].

There's a non sequitur here, now! :naughty:

[Ben]

Well I was gonna give a more detailed reply yesterday but, as I was typing, I got a call about a new possible job that could be the best opportunity I've had ever, so far.

Say what, I wish you all the best with that interview. Tie? No tie? Who knows

 

Busy getting drunk just now. Later for you and Nootropic

[Nootropic]

So it is not hard to show, I think, that if, C= R(i),,,, C' = R(j) are both algebraic extensions for the reals, then C simeq C'. I don't think countability/uncountability comes into it.

 

No. Two-dimensional extensions of the reals (of which are the only nontrivial extensions of the reals) are isomorphic, but this is true for any two finite dimensional extensions of a field.

 

That's easy; an algebraically closed field L:K is an extension of the field K iff K subset L AND iff it is a splitting field for K, that is it contains all roots for any polynomial in K.

 

Let me recast what you said in a more readable format for an algebraically closed field L and a subfield K:

 

1. L is an extension of K

2. K is a subset of L

3. L is a splitting field for K.

 

Theorem: Any uncountable set has a countable subset.

 

Proof; Let S be uncountable, and start counting its elements. If, when Hell freezes over, OR you exhaust the counting numbers (i.e. mathbb{N}), whichever is sooner, you still haven't exhausted S, you will have found a countable subset.

 

Be more precise. This is simple, map a function from the natural numbers that assigns to each natural number a distinct element of the set (easy to show it's bijective) This can be rephrased as saying that every infinite set contains a countable set.

 

So, for example, consider the field Q of rational numbers which is obviously countable. Then any polynomial over this field will have roots in Q(i),,,,( i^2 equiv -1), which is also countable.

 

No. The polynomial x^2-2 has no root in Q(i). The algebraic closure for the rational numbers is the field of algebraic numbers, this is however countable, but for a slightly less obvious reason.

 

mathbb{C} is a field which is:

 

1. of uncountable cardinality

2. such that any polynomial of degree n has n roots in it, counting multiplicities.

3. ??????

4. ??????????

 

1. Uncountably infinite 2. Characteristic zero (big one here) 3. Algebraically closed

4. Two-dimensional field extension of the real numbers (here 4 implies 1, thus 1 essentially not needed).

 

Clearly any field satisfying only all four is isomorphic to C (send a+bj to a+bi) and this might even be true for a field satisfying just 4 (need to double check...late and tired). If C is isomorphic to such a field, then all of these are preserved under any such isomorphism.

 

I'm fairly sure though if any field is a two-dimensional extension of the real numbers (now this seems plainly obvious, but I don't want to tread on any mistakes, as I'm a bit tired) then it is isomorphic to C (I'm sure this is correct, really).

 

I have more to say, but sleep beckons.

[Ben]

I'm talking about abstractly defining [imath]\mathbb{C}[/imath] which is to say defining it as an abstract structure rather than a construct from [imath]\mathbb{R}[/imath]. Get it?

Actually no, I don't. As far as I am aware, the field [math]\mathbb{C}[/math] is defined as the the algebraic closure of the real field, which implies that [math]\mathbb{C}[/math] is an algebraic extension of [math]\mathbb{R}[/math]

 

I have no idea how to argue this "backwards".

We don't yet know what the heck [imath]\mathbb{R}[/imath] is, nor [imath]\mathbb{Q}[/imath] etc.

Now I am lost - what can this possibly mean? If we don't know "what the heck" the real field is, how can we possibly define the complex field?

 

I'm fairly sure though if any field is a two-dimensional extension of the real numbers.....

Tsk, tsk!! Fields don't have "dimensions", as I am sure you know.

[Nootropic]

Tsk, tsk!! Fields don't have "dimensions", as I am sure you know.

 

Of course I know! :) My language was a little misleading, but to be more specific the complex numbers form a two-dimensional vector space over the real numbers, much like R^2 forms a two-dimensional vector space over R and hence, is isomorphic to C (as vector spaces!). However, if we have some algebraically closed field of characteristic zero properly containing the real numbers (note this is a bit redundant, because a field of characteristic p > 0 cannot contain the real numbers) that forms a two-dimensional vector space over the real numbers, then (I assert now this is true). All you really have to do is show that it's an algebraic closure for the real numbers and since algebraic closures are unique up to isomorphism (also nontrivial!), this is indeed isomorphic to the complex numbers. Actually, you may not even need to go that far. From the way the field forms a vector space, you can pretty much end up identifying (this is a bit vague, I know) some element in this field as behaving identically to i in C, and just extend the vector space isomorphism to a field isomorphism (you really need only show the map is a multiplicative homomorphism of the nonzero elements of the field into the nonzero elements of the complex numbers).

 

Now the question posed here is, are there any more relaxed conditions? So what we have so far is that F is a field properly containing the real numbers and it forms a two-dimensional vector space over the real numbers. Hm...this is much tougher to relax the vector space condition...if not impossible.