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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 ofnot easy? I can imagine that it would require the axiom of choice; if so, this should be mentioned in the text.Albmont17:35, 26 March 2007 (UTC)[reply]

It probably does require choice; certainly theproof I just found on Google books uses AC. I haven't got a reference that it's required though. In any case, the meaning ofnot 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.Algebraist14:00, 17 May 2008 (UTC)[reply]

OR:It does require AC. Take a set X. Adjoin the elements of X toQ 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.Algebraist14:10, 17 May 2008 (UTC)[reply]

See[1]212.87.13.69 (talk)16:16, 5 May 2010 (UTC)[reply]

OK I may be thinking wrong here, but I really don't see how the second condition ("There exists an element ofF which is not a sum of squares inF, and the characteristic ofF is not 2") can be correct. I am not able to prove any of 1 and 3 from it. I also see it has been removed and reinstated previously, with an added condition of characteristic not equal to 2.

However, let's take the field of complex numbers, and extend it with a transcendental element x. Then I believe we have that x is not a sum of squares, unless we add even more elements to the field. And this field also has characteristic 0. But still i² = -1, so it clearly cannot be ordered. I think I'm going to remove condition 2 for now. --Ørjan (talk)17:58, 3 February 2010 (UTC)[reply]

OK, I'm stupid (or maybe I'd had too little beer, see above). Clearly if the characteristic is not 2, then

x = x*1 = ((x+1)/2 + (x-1)/2)*((x+1)/2 - (x-1)/2) = ((x+1)/2)2 - ((x-1)/2)2

which makes everything a sum of squares if -1 is. --Ørjan (talk)17:19, 4 February 2010 (UTC)[reply]

The given definition:"a formally real field is a field that admits an ordering which makes it an ordered field." provides no clue that there is any difference between the idea of "ordered field" and "formally real field". That is, is there an ordered field that is not a formally real field, or vice versa? From the definition, it appears not. If not, why does the term "formally real field" even exist? I think the definition should be improved or the whole article should be removed.Gsspradlin (talk)13:42, 2 May 2013 (UTC)[reply]

A formally real field is a field such that thereexists an order which makes it an ordered field. Such an order is not given as part of the field, nor is it necessarily unique if it exists. (E.g. the subfield of the reals consisting of elements of the form a+b*sqrt(2) where a,b are rationals can be ordered in two ways - the obvious one, and the one that puts an element in the order as if it were a-b*sqrt(2) instead. This is possible because there is no purely algebraic distinction between sqrt(2) and -sqrt(2).) I have attempted to clarify the definition. --Ørjan (talk)15:00, 2 May 2013 (UTC)[reply]

Orjan: that was fast. Thanks. Also, the phrase "The definition given above is not a first-order definition, as it requires quantifiers over sets." is opaque to me. I have a Ph.D. in mathematics. I am not an algebraist, but I have done a fair amount of reading on algebra, and I have no clue what this is supposed to mean. There should be at least a link to "first-order definition", if a page exists for this term.Gsspradlin (talk)15:30, 2 May 2013 (UTC)[reply]

I believe it refers tofirst-order logic, and have linked that. Many algebraic structures such as fields can be defined with statements only quantifying over elements of the algebra (varieties being one particularly simple form, although fields are more complicated than that) and some consider it important to know when this is the case. --Ørjan (talk)14:12, 4 May 2013 (UTC)[reply]

...Not that I see how to do the "can be coded as first-order sentences in the language of fields" part myself, as the alternative definitions also use sets for summing over. But I'm not an expert on this, so I just added a citation needed template. --Ørjan (talk)11:51, 5 May 2013 (UTC)[reply]

@Oerjan: In fact it can be coded asinfinitely many first-order sentences in the language of fields, as${\displaystyle \mathrm{\forall }{x}_1(-1\ne x_1^2)}$,${\displaystyle \mathrm{\forall }{x}_1{x}_2(-1\ne x_1^2+x_2^2)}$, etc., one sentence for each number of variables. This is allowed in first-order logic: as long as each sentence has finite length you're good to go. It is probably not easy to find a specific reference to this; thus I will remove the citation needed template and clarify this in the text.Tony Beta Lambda (talk)04:23, 2 August 2018 (UTC)[reply]

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