Talk:Formally real field

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[edit] Equivalence of the definitions

The text says: The equivalence of the first two properties is easy, and the third property easily implies the first two, but it is not easy to show either of the first two properties implies the third (that is, it is not evident how the assumption that a sum of squares being 0 forces each square to be 0 actually implies F has some ordering as a field)..

What is the meaning of not easy? I can imagine that it would require the axiom of choice; if so, this should be mentioned in the text. Albmont 17:35, 26 March 2007 (UTC)

It probably does require choice; certainly the proof I just found on Google books uses AC. I haven't got a reference that it's required though. In any case, the meaning of not easy is the usual meaning of that phrase in the English language: the equivalence of the first two properties, and the fact that the third implies them, can be proved in under a minute while drunk, while the proof of (iii) from the others requires a fair amount of algebraic work. Algebraist 14:00, 17 May 2008 (UTC)

OR:It does require AC. Take a set X. Adjoin the elements of X to Q as independent indeterminates. The field obtained clearly satisfies the first two properties; suppose the third follows and take an ordering. This restricts to a total ordering on X. It is well known that (assuming ZF is consistent) ZF does not prove every set can be totally ordered. Algebraist 14:10, 17 May 2008 (UTC)