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If R_1\subset R_2 are two real closed fields (R_2 is an extension of R_1), then is it always the case that R_1 contains {R_2}_alg; By the latter I mean algebraic elements of R_2.

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What do you mean by "algebraic elements of $R_2$"? Do you mean those elements that are algebraic over $R_1$? Then the answer is yes. There is a straightforward proof: the only algebraic extension of $R_1$ is $R_1[i]$, and $i \notin R_2$. (where $i^2 = -1$)

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As far as I know, "algebraic" without an "over $k$" clause, always means algebraic over the prime field, in this case $\mathbb Q$. –  Andreas Blass Dec 12 '11 at 19:25
    
That was my first thought too, but that makes the question nearly trivial. –  Hurkyl Dec 12 '11 at 22:22
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I think I've found it out. If R_1\subset R_2 are two real closed fields, then (R_1)_alg should be the same as (R_2)_alg, since otherwise (R_2)_alg would be a proper real extension of (R_1)_alg contradicting real-closedness of (R_1)_alg.

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And are both of these identified with the algebraic reals? –  Gerald Edgar Nov 28 '11 at 14:23
    
@Gerald Edgar: Yes, the algebraic elements of any real-closed field are (under the obvious identification) the algebraic reals. –  Andreas Blass Dec 12 '11 at 17:11
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