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In Van der Waerden B L. Algebra Vol.I[M]. Springer, 2003, Pro. Waerden announced in page 256 that if an element $\gamma$ of a formally real field K is not a sum of squares, there exist an ordering of K, in which $\gamma$ turns out to be negative. In the proof, he deduced the equation: $\gamma=\dfrac{1+\sum \beta_{v}^{2}}{\sum \alpha_{v}^{2}}$, and claimed that $\gamma$ would then be the sum of squares. I think that he mistakenly utilized a propoty of real close field proved in page 249: in a real closed field, every sum of squares is a square. Indeed, if K is a real closed field, then $\gamma$ is a sum of squares. But here K is only a formally real field, so a sum of squares might fail to be a square. For example, in the field Q of rational numbers, $\dfrac{1}{2}=\dfrac{1}{2^2}+\dfrac{1}{2^2}$ is a sum of squares, whereas $\dfrac{1}{\sqrt{2}}$ does not belong to Q. So the proof given by Waerden is not correct. Did Waerden make a mistake? Or do I make a mistake?

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up vote 3 down vote accepted

The product $ab$ of sums of squares $a$, $b$ is clearly a sum of squares. It follows that a quotient $a/b$ is also a sum of squares, since $a/b = ab(b^{-1})^2$.

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Thank you very much! –  Sun Mar 28 '13 at 7:52

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