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When is an element of a finitely generated field extension algebraic?

  1. Jan 28, 2012 #1
    Hungerford says that
    But if we take K = ℝ and [itex]K(x_{1})[/itex] = ℝ(i) = ℂ, we have that i is not in ℝ yet is algebraic over ℝ. Guess I'm missing something here. Is it that this need not be true for simple extensions if the primitive element is algebraic over the field?
     
  2. jcsd
  3. Jan 28, 2012 #2

    morphism

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    Hungerford's x_i's are supposed to be indeterminates, i.e. they're transcendental over K by definition (essentially).

    When you write down "i" you're implicitly thinking of a complex number that satisfies the polynomial x^2+1 - i.e., you're not thinking of an indeterminate.
     
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