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

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SUMMARY

The discussion centers on the nature of elements in finitely generated field extensions, specifically addressing the distinction between transcendental and algebraic elements. Hungerford asserts that in the field extension K ⊆ K(x₁,...,xₙ), each xᵢ is transcendental over K. However, the example of K = ℝ and K(x₁) = ℝ(i) = ℂ illustrates that the element i is algebraic over ℝ, contradicting the earlier assertion. This highlights the importance of distinguishing between indeterminates and specific algebraic elements within field extensions.

PREREQUISITES
  • Understanding of field extensions in abstract algebra
  • Familiarity with transcendental and algebraic elements
  • Knowledge of polynomial equations and their roots
  • Basic concepts of finitely generated fields
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  • Study the properties of algebraic and transcendental elements in field theory
  • Explore the concept of simple field extensions and their implications
  • Investigate the role of indeterminates in polynomial rings
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Mathematicians, particularly those specializing in abstract algebra, graduate students studying field theory, and anyone interested in the properties of algebraic structures and extensions.

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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?
 
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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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