transcendental root theorem
Suppose a constant is transcendental over some field . Then is also transcendental over for any .
Proof.
Let denote an algebraic closure of . Assume for the sake of contradiction
that . Then since algebraic numbers
are closed under multiplication
(and thus exponentiation by positive integers), we have , so that is algebraic over , creating a contradiction.
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Title | transcendental root theorem |
---|---|
Canonical name | TranscendentalRootTheorem |
Date of creation | 2013-03-22 14:04:23 |
Last modified on | 2013-03-22 14:04:23 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 8 |
Author | mathcam (2727) |
Entry type | Theorem![]() |
Classification | msc 11R04 |