Construct a field with all positive intergers.

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Homework Statement



Is it possible to construct a field, that satisfies all basic propeties (http://en.wikipedia.org/wiki/Field_(mathematics)#Definition_and_illustration , the six bolded part ), with only and all positive integers?

Remeber, you don't have 0/negative numbers/fraction in positive integers, so you may have to redefine addition and multiplication to help.

And, quoted from my professor, "If you rephrase this question in terms of set theory, it would be rather easy then".

Homework Equations



None.

The Attempt at a Solution



OK first I tried finite field approach, and thought that I could be able to make a value table to enumerate all possible values of addition and multiplication, based on that we refer to Galois Extension to construct higher order of the table, until the p^n order of a finite field. But after speaking with professor, it was said to be wrong direction because I'm supposed to use all postive intergers, other than such special finite field case. Obviously what I'm trying to achieve is a infinite field, and I only know three of such, rational/real/complex.

And there I got a remark from him, that I wrote above, "If you rephrase this question in terms of set theory, it would be rather easy then". I'm only begnning in set theory as a junior math major, and I know I could probably be not able to answer it right now. But curiosity is killing me inside and I'd love to know even a little about what I should know for proving or disproving this postulate.

Thank you very much, and let me know if I do not make it clear somewhere.
 
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Rephrase the question to: You have a countable set. Can you turn it into a field?
 
Office_Shredder said:
Rephrase the question to: You have a countable set. Can you turn it into a field?

That's such amazing rephrase, I suddenly feel I have more for it.

Thanks for the tips, I'll see what I can do now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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