Algebraic completion (closure)

[Qfwfq]

Great, this is getting to be a real bun fight! Thanks Ben, and do your best at getting drunk. ;)

 

Neither of you are getting out of the groove:

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]

Say what? I'm not aware of that, at all! :hihi:

 

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

But it was you that started this whole thing by insisting that [imath]\mathbb{C}[/imath] is essentially [imath]\mathbb{R}^2[/imath]!

 

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.

Be more precise. It's easy to show it's injective. Of course it'll be surjective on its own image, but not on the uncountable codomain. Actually I think English language mathies more often say "one to one" and "onto" and perhaps you even mean bijective as being the former. Anyway Ben's proof was a lot better!

 

OK, enough tomfoolery, let's get down to it:

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).

Actually it's 4. that's superfluous! With a bit of googling I found that the answer seems to be it's enough to add 3. to the first two. The matter is however somewhat... complex (not so simple).

 

Can't stay here longer so I'll be back, we'll get to an abstract definition of [math]\mathbb{C}[/math] that doesn't need [math]\mathbb{R}[/math]!

[Nootropic]

Be more precise. It's easy to show it's injective. Of course it'll be surjective on its own image, but not on the uncountable codomain. Actually I think English language mathies more often say "one to one" and "onto" and perhaps you even mean bijective as being the former. Anyway Ben's proof was a lot better!

 

The point is that it's surjective on it's image and that the image is a subset of the set we're mapping into, hence the (infinite) set contains a countably infinite subset (be it the whole set or not).

 

Actually it's 4. that's superfluous! With a bit of googling I found that the answer seems to be it's enough to add 3. to the first two. The matter is however somewhat... complex (not so simple).

 

Can't stay here longer so I'll be back, we'll get to an abstract definition of mathbb{C} that doesn't need mathbb{R}!

 

Not so sure. My counterexample to the question that started satisfies 1-3, but not 4.

 

Really the definition of the complex numbers can be given quite simply (and abstractly) by the quotient ring R[x]/(x^2+1), since R is a field and x^2+1 is an irreducible polynomial (since in this principal ideal domain, x^2+1 is a nonzero prime element, hence the ideal it generates is a maximal ideal). It's quite easy to show this is isomorphic to C. This of course relies on the existence of the real numbers. Quite honestly, there is no way to "abstractly" definite C without assuming the existence of R in some way. The above construction is perhaps the best characterization of C you can get without having a mess of conditions.

[Nootropic]

ah, earlier I realize I wasn't clear when I mentioned the mapping being bijective. I assumed you would just take it to mean on the image, sorry for not clarifying!

[Ben]

Well, I'm confused.

 

Nootropic: In your post #17, you talk like you and I are in disagreement in this maatter - Qfwfq seems to think so too. But that's not true!

 

I earlier asserted that if any 2 fields are algebraic closures of the same field, then they are isomorphic as fields.

 

I would go further, and assert this is true for any 2 simple algebraic extensions of the same field, and further still, that this will still be true if I replace the words "same field" with the words "isomorphic fields".

 

Qfwfq: Regarding the remark you attribute to me, namely that "[math]\mathbb{C}[/math] is essentially [math]\mathbb{R}^2[/math]", what I said (I think) was that as REAL VECTOR SPACES [math]\mathbb{C} \simeq \mathbb{R \times R}[/math].

 

As you know, of course, vector spaces are isomorphic iff they have the same dimension. A suitable basis for [math]\mathbb{C}[/math] might be [math]\{1,\,i\}[/math] and a basis for [math]\mathbb{R}^2[/math] could be [math]\{(1,0),\,(0,1)\}[/math]. Then clearly they are isomorphic.

 

Finally, if anyone thinks I really don't how to define a countable set, they would do well to ignore all my posts from now on. It seems my attempt at a little levity bit me on the backside. I'll know better in future.

 

Sorry if I seem grumpy. Today is not being a good one for me.

[Qfwfq]

Quite honestly, there is no way to "abstractly" definite C without assuming the existence of R in some way.

But that's exactly what I'm after! Well, I said the matter isn't very clear, however I've seen stuff that suggests it is possible but it isn't so clear due to details. :confused:

 

ah, earlier I realize I wasn't clear when I mentioned the mapping being bijective. I assumed you would just take it to mean on the image, sorry for not clarifying!

Well, I only wasn't sure if that's what you meant. :)

 

Regarding the remark you attribute to me, namely that "[math]\mathbb{C}[/math] is essentially [math]\mathbb{R}^2[/math]", what I said (I think) was that as REAL VECTOR SPACES [math]\mathbb{C} \simeq \mathbb{R \times R}[/math].

That's what you started with, but then you disagreed with my subtleties.

 

The trouble is that we common mortals need to give the generic complex number a concretization in terms of real numbers (whether in rectangular, polar or matrix form). This makes us tend to think of [imath]\mathbb{C}[/imath] as being its usual well known concrete construction (as it is the most known). The effect is that folks therefore tend to think of it as being [imath]\mathbb{R}^2[/imath] which isn't true.

 

While the vector space of more than one dimension is, as such, isotropic, the extra operation which makes it a field clearly breaks isotropy. So, [imath]\mathbb{R}^2[/imath] when used as a concrete [imath]\mathbb{C}[/imath] isn't really and truly a vector space.

 

It seems my attempt at a little levity bit me on the backside. I'll know better in future.

It didn't bite my backside! :)

[Nootropic]

But that's exactly what I'm after! Well, I said the matter isn't very clear, however I've seen stuff that suggests it is possible but it isn't so clear due to details

 

I'd be curious in seeing this. So assuming nothing about the real numbers, we have C as an algebraically closed field of characteristic zero whose cardinality is uncountable. I mean, like I proved, this does not classify C by any means. I personally don't know any other condition that doesn't utilize the real numbers. Hm...I suppose a cute fact is that while both R and C are infinite dimensional extensions of Q, Aut(R/Q) consists only of the identity but Aut(C/Q) is infinite (Where Aut(K/F) denotes the subgroup of the group automorphisms on K such that any automorphism in Aut(K/F) fixes F), though that's not qite so abstract. Another more useful fact is that C cannot be totally ordered, as both R and Q can. I think that's worthy to add to the list...

 

Ah, well I found it!

 

Complex number - Wikipedia, the free encyclopedia

The field C is characterized up to field isomorphism by the following three properties:

 

* it has characteristic 0

* its transcendence degree over the prime field is the cardinality of the continuum

* it is algebraically closed

 

One consequence of this characterization is that C contains many proper subfields which are isomorphic to C (the same is true of R, which contains many subfields isomorphic to itself[citation needed]). As described below, topological considerations are needed to distinguish these subfields from the fields C and R themselves.

 

Hm, well I have to check out transcendence degree, but I'm studying out of Atiyah and Macdonald's commutative algebra textbook, so I should know that in the coming weeks.

[Qfwfq]

That's what I had found, but note that I had to fix the link (the wiki tag doesn't work like that) and I reduced the text to the essential. :)

 

Hm, well I have to check out transcendence degree, but I'm studying out of Atiyah and Macdonald's commutative algebra textbook, so I should know that in the coming weeks.

This is one of the reasons why I said the matter isn't so simple, also because of "which contains many subfields isomorphic to itself" and I hadn't had time to examine it properly.

 

In the meantime I remembered the other thing I had thought of and I now think it could be the answer. As we know, [imath]\mathbb{C}[/imath] has an involutive automorphism (one, apart from identity), it is what is vulgarly known as complex conjugation. It is quite closely and obviously linked to there being the subfield [imath]\mathbb{R}[/imath] which suggests it maybe being a good replacement for your axiom 4. but is of course cleaner. ;)

 

Now I went back to the wiki to have a better look and what do I find? Lo and behold, it mentions exactly that! It seems like we might also need to add a topological axiom:

 

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

1. is of uncountable cardinality
   
2. is such that any polynomial of degree [imath]n[/imath] has [imath]n[/imath] roots in it, counting multiplicities.
   
3. has one non trivial involutive automorphism.
   
4. is a connected and locally compact topological field.
   

Whaddya think?

[Ben]

Silly me! I finally see what you were driving at.

 

Suppose that [math]K[/math] is a field satisfying your 4 axioms.

 

Let [math]G[/math] be the group of all automorphisms [math]K \to K[/math]. Form the subgroup [math]H \subseteq G[/math] comprising just the identity [math]e[/math] and the automorphism [math]\alpha: K \to K [/math] such that [math] \alpha^2 = e[/math]

 

Then define [math]\alpha(k) = k',\,\,\, k \in K[/math] by [math]\alpha(kk') = \alpha(k)\alpha^2(k) = k'k = kk' \Rightarrow \alpha = e[/math] for any product [math]k_ik'_i \in K[/math]. Let these guys live in the subset [math]L\subsetneq K[/math].

 

Now [math]K[/math] is a field, so [math]0,\,\,1 \in K[/math], and, since [math]\alpha(1) = 1,\,\, \alpha(0) =0 \Rightarrow 0,\,\,1 \in L[/math] then [math]L[/math] is also a field. (Yeah, yeah, I know, I should really argue for additive and multiplicative inverses .... don't nag)

 

Now if it is the case that, for any [math]j \in K[/math], that [math]\alpha(j) = j[/math], let's send it to [math]L[/math] also.

 

Thus we have a subfield [math]L \subsetneq K[/math] whose elements are such that [math]\alpha(l) = l \Rightarrow l = l',\,\, \forall l \in L [/math]. (Galois theorists call this the "fixed field" for the automorphism group [math]H[/math])

 

We have assumed that [math]K[/math] is algebraically closed. What does this tell us about the closure of [math]L[/math]?

 

Nothing, as far as I can see. I'm pretty sure that the local compactness of [math]K[/math] isn't going to help us here...... am I wrong about this?

 

(Sorry chum, but you did ask for abstraction!! Of course you may make your own substitutions K = C, L = R)

[Qfwfq]

Well I had no doubt that complex conjugation can be seen as defining [imath]\mathbb{R}[/imath] in [imath]\mathbb{C}[/imath] (as I had said), just as much as vice versa. What I'm aiming at is not needing to "already have" [imath]\mathbb{R}[/imath] and I think the involutive automorphism can help circumvent Nootropic's objection. I'm not sure myself if axiom 4 is necessary but I threw it in just for the fun of it... I mean, cuzuv Complex number - Wikipedia, the free encyclopedia

 

So the trick is to see if any field having these properties (perhaps even without 4 and/or with some other axiom) is consequently isomorphic to [imath]\mathbb{C}[/imath]. :cup:

[maddog]

For the properties of Closure, I am interested in not just . I am interested in the

Quaternions, and the Octonians as well. I am aware that Algebraic Complete no longer holds though.

 

maddog