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Q: Do complex-numbers contain all other fields of size<=Aleph-1? ( Answered,   2 Comments )
Subject: Do complex-numbers contain all other fields of size<=Aleph-1?
Category: Science > Math
Asked by: mark1075-ga
List Price: $20.00
Posted: 26 Oct 2003 01:48 PDT
Expires: 25 Nov 2003 00:48 PST
Question ID: 269764
Do complex-numbers contain all other fields of size<=Aleph-1?

By "contain" I mean can all other fields be mapped into a
subset of the complex-numbers in such a way that all field-operations
are preserved under the mapping?  I believe there are "very large" 
fields (with Aleph-2 or more elements in it - so the size of the 
sets themselves are not 1-1 mappable onto the complex-numbers) that 
can obviously not be mapped onto a subset of the complex-numbers, but
I'm only interested in sets of size aleph-1 or smaller.  

(I think 'aleph-1' is standard terminology, but I mean there's a
1-to-1 mapping from the set into the continuum - e.g. real numbers.)

The fields I'm aware of (integers, rationals, reals, etc.) all fall
into this category, but I figure someone must have proved or disproved
more generally?

Because this may not be on the internet, full credit is given to a
reference to a valid proof in any media.  If you don't know but have
an interest comment on the matter, please speak up, too!

This is just for idle curiosity - something I wondered back when
taking modern-algebra in college.

Clarification of Question by mark1075-ga on 26 Oct 2003 10:11 PST
dannidin-ga (and mathtalk)

Thank you for your wonderful response!  First of all, I really
appreciate you addressing the "intended" question instead of just
giving a lame example of a finite-field-case I'd overlooked!  Your
explanation is quite nice and though not a solid proof, is sufficient
for my purposes.  If you could provide references to your favorite
reference-books where'd you'd go to looking for some fill-ins, that
would be a nice touch, but great job!
Subject: Re: Do complex-numbers contain all other fields of size<=Aleph-1?
Answered By: dannidin-ga on 28 Oct 2003 07:36 PST
Hi again mark1075-ga,

I am glad you find my answer satisfactory. (and thanks to mathtalk for
chipping in.) I finally had time to do some digging at our local
library. I couldn't find a book that contains the exact claim that
we're trying to prove. But I did find some useful references: The book
"Field and Galois Theory" by Patrick Morandi (Springer 1996, vol. 167
in the Graduate Texts in Mathematics series) contains an entire
chapter on transcendental field extensions, with a very clear
discussion on transcendence degree. The claim that C is of
transcendence degree aleph-1 over Q appears there as Example 19.20
(page 180). Some of the exercises on page 181 also seem relevant to
our discussion.

You may also wish to consult a textbook on set theory such as "Set
Theory for the Working Mathematician", by Krzysztof Ciesielski (London
Mathematical Society Student Texts 39, Cambridge University Press
1997), to learn more about transfinite induction, the well-ordering
principle, Zorn's lemma and other equivalent formulations of the Axiom
of Choice, which are used heavily in the context of properties of
transcendental field extensions (in Morandi's book mostly Zorn's lemma
is used). This is basic stuff, so you'll find it in any other set
theory textbook you are likely to lay your hands on.

Subject: Re: Do complex-numbers contain all other fields of size<=Aleph-1?
From: dannidin-ga on 26 Oct 2003 06:38 PST

surely when you say "field" you mean "field of characteristic 0"?
finite fields, such as the field Z_2 of two elements, cannot be mapped
isomorphically (i.e. in a way that preserves the field operations)
into any subset of the complex numbers. but if one considers only
fields of characteristic 0, i.e. fields where any finite multiple
x+x+...+x of any non-zero element x is itself non-zero, then i believe
the answer to your question is affirmative - the complex numbers
indeed contain isomorphic copies of any (zero-characteristic) field of
cardinality at most aleph-1. i'm not sure i remember exactly how to
prove this. i'm pretty sure it requires the use of some form of
transfinite induction such as zorn's lemma or the well-ordering
principle. roughly, it's related to the fact that the complex numbers
(henceforth: C) are an algebraically closed field that is of
"transcendence degree" aleph-1 over the rationals. algebraically
closed means that any polynomial has a root. transcendence degree
aleph-1 means that you can find aleph-1 distinct elements each of
which is transcendental over the rationals (i.e. satisfies no
polynomial equation with rational coefficients) and which furthermore
are algebraically independent over the rationals, i.e., no finite
subset of them satisfies a polynomial equation with rational
coefficients in several variables.

now, assuming that the above is true (this requires proof, of course.
i don't remember exactly how it's proved...), given any
zero-characteristic field F, we can try to imbed F isomorphically into
C, as follows: first, F contains a sub-field Q' isomorphic to the
rationals Q, obtained by taking the unit 1 of F, adding it to itself
any number of times (positive, negative or 0) to obtain an isomorphic
copy of the integers, and taking quotients of such to get Q'.

the next step is to build the isomorphism from F into C step by step,
by, at each step, taking an element of F which still hasn't been
mapped into C, and finding a suitable candidate to map it to. if the
element is algebraic over the sub-field of F composed of elements
which have already been mapped, that is to say, satisfies a polynomial
equation whose coefficients are in that sub-field, then one may map
the element into the element of C that is a root of the corresponding
polynomial equation in complex numbers obtained by translating the
above equation to complex numbers using the isomorphism (applied to
elements for which it is already defined). one should take care to use
the irreducible polynomial equation satisfied by our element - one of
minimal degree, that is.

another possibility is that our element is transcendental over the
sub-field of F composed of elements for which we have defined the
isomorphism. in this case, locate a complex number which is
transcendental over the sub-field of C which is the image (under the
isomorphism) of the above sub-field of F. map the element to this
complex number.

if you are familiar with transfinite induction, you will see that this
construction works, and will map F isomorphically into some subset of
C, provided only that we never run out of complex numbers which are
transcendental over the current sub-field where the isomorphism is
defined. because of the fact stated above without proof concerning the
degree of transcendence of C over Q, this will be the case when the
cardinality of F is no greater than aleph-1.

i know that this is not a complete proof. i probably missed a couple
of details, and do not have time right now to get more deeply involved
with your question, to find you references or to make the arguments
more precise, which is why i'm posting this as a comment. but if you
are satisfied with this as an answer to your question, let me know and
i will post it as such. if you require more detailed information, i
will try to help, or maybe some other researcher or interested party
could pick it up from here. (mathtalk? any ideas?)

Subject: Re: Do complex-numbers contain all other fields of size<=Aleph-1?
From: mathtalk-ga on 26 Oct 2003 08:33 PST
I think dannidin's discussion is right on target.  I seem to remember
from Galois theory that, at least for characteristic 0, one can always
"separate" a field extension into algebraic and "pure transcendental"
pieces.  I'll take a look at my copy of Kaplansky's Fields and Rings
when I get home.  If that's the case, the only question can be the
dimension of the purely transcendental component, i.e. showing that
within C we can find Aleph_1 algebraically independent complex numbers
(over Q).

regards, mathtalk-ga

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