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 ∀n∈N , any linear combination of π with √2 and in fact, √2kπl transcendental ∀k,l∈N, √π is transcendental but do not know how to show (or whether it is at all true!) that π1n transcendental ∀n∈N.
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 G⊂F.
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