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

  • Thread starter TopCat
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  • #1
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Hungerford says that
In the field extension
[tex]K \subset K(x_{1},...,x_{n})[/tex]
each [itex]x_{i}[/itex] is easily seen to be transcendental over K. In fact, every element of [itex]K(x_{1},...,x_{n})[/itex] not in K itself is transcendental over K.

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?
 

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  • #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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