two real closed fields- algebraic elements - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:48:38Z http://mathoverflow.net/feeds/question/82071 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82071/two-real-closed-fields-algebraic-elements two real closed fields- algebraic elements mosen 2011-11-28T11:11:40Z 2011-12-12T15:28:41Z <p>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.</p> http://mathoverflow.net/questions/82071/two-real-closed-fields-algebraic-elements/82075#82075 Answer by mosen for two real closed fields- algebraic elements mosen 2011-11-28T12:38:52Z 2011-11-28T12:38:52Z <p>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.</p> http://mathoverflow.net/questions/82071/two-real-closed-fields-algebraic-elements/83247#83247 Answer by Hurkyl for two real closed fields- algebraic elements Hurkyl 2011-12-12T15:28:41Z 2011-12-12T15:28:41Z <p>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\$)</p>