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Q: Do complex-numbers contain all other fields of size<=Aleph-1? ( Answered,   2 Comments )
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 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 this 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! -mark```
 ```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. Regards, dannidin```
 ```mark1075-ga, 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?) regards, dannidin```
 ```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```