Monday, 1 July 2013

Determine all algebraic extensions of BbbQ contained in BbbQ(sqrt2,pi)




Determine all algebraic extensions of Q contained in Q(2,π)




Assuming π to be transcendental over Q , it seems to me that the answer must be only Q(2) .



π transcendental πn transcendental nN , any linear combination of π with 2 and in fact, 2kπl transcendental k,lN, π is transcendental but do not know how to show (or whether it is at all true!) that π1n transcendental nN.




So my intuitive idea is that π should not come into the picture and hence, Q(2) should be only such extension, but how make my argument rigorous (if it's at all true!)



Thanks in advance for help!


Answer



Let F=Q(2).



If π was algebraic over F, [F(π):F] would be finite. Since [F:Q]=2, [F(π):Q] would be finite, thus π would be algebraic. A contradiction.



So Q(2,π)=F(π)F(X).




It is well-known that F(X) contains no nontrivial algebraic extension of F: thus neither does F(π).



So if G is any algebraic extension of Q in F(π), G(2) is an algebraic extension of F contained in F(π): it is F. Thus GF.


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