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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.

flag	asked Nov 28 2011 at 11:11 mosen 175●4

Hurkyl · Answer 1 · 2011-12-12 15:28:41Z UTC

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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 $i^2 = -1$)

answered Dec 12 2011 at 15:28

Hurkyl
1,163●3●15

	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 2011 at 19:25
	That was my first thought too, but that makes the question nearly trivial. – Hurkyl Dec 12 2011 at 22:22

mosen · Answer 2 · 2011-11-28 12:38:52Z UTC

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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.

answered Nov 28 2011 at 12:38

mosen
175●4

	And are both of these identified with the algebraic reals? – Gerald Edgar Nov 28 2011 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 2011 at 17:11

two real closed fields- algebraic elements

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