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TopCat
Jan28-12, 10:54 PM
Hungerford says that In the field extension
K \subset K(x_{1},...,x_{n})
each x_{i} is easily seen to be transcendental over K. In fact, every element of K(x_{1},...,x_{n}) not in K itself is transcendental over K.

But if we take K = ℝ and K(x_{1}) = 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?

morphism
Jan28-12, 11:13 PM
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.