Let \( E \subset F \) and \( F \subset K \) be field extensions which are not algebraic. Let \( \alpha \in K \) be algebraic over \( F \) and \( \alpha \notin F \). Let \( L \) be the subfield of \( K \) generated over \( E \) by the coefficients of the monic polynomial of minimal degree over \( F \) which has \( \alpha \) as a zero. Then, which of the following is/are TRUE?

Question

Consider the following subsets of the Euclidean space \( \mathbb{R}^4 \):
\( S = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = 0 \} \),
\( T = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = 1 \} \),
\( U = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = -1 \} \).
Then, which one of the following is TRUE?

Answer 1

The problem involves field extensions and requires us to determine the truth of certain statements about them. The essence of the question is about comparing the extension properties between various fields when given specific conditions.

Setup of the Problem: E \subset F \subset K are given as field extensions, where these are not necessarily algebraic. We have an element \alpha \in K that is algebraic over F but not a member of F itself. L is defined as the subfield of K generated over E by the coefficients of the monic polynomial of minimal degree over F for which \alpha is a root.
Analysis of Options:
- The correct answers given are:
  1. F(\alpha) \supset L(\alpha) is a finite extension if and only if F \supset L is a finite extension.
  2. F(\alpha) \supset L(\alpha) is an algebraic extension if and only if F \supset L is an algebraic extension.
Explanation of the First Correct Option:
- Finite Extensions and Algebraic Elements: If F(\alpha) \supset L(\alpha) is finite, it means that \alpha minimally satisfies a polynomial over F that constrains the extension's degree. Since \alpha's minimal polynomial over F has coefficients generating L, it implies F \supset L must also be finite. Conversely, if F \supset L is finite, it leads to F(\alpha) \supset L(\alpha) being finite because generating F(\alpha) involves finitely extending through \alpha. Therefore, the finiteness condition is equivalent in both scenarios.
Explanation of the Second Correct Option:
- Algebraic Extensions: When F(\alpha) \supset L(\alpha) is algebraic, every element within this extension satisfies a polynomial with coefficients in L. This condition reciprocates to F \supset L being algebraic due to \alpha's polynomial acting as a bridge for algebraicity. If F \supset L is algebraic, each element in F satisfies a polynomial within L, making F(\alpha) \supset L(\alpha) algebraic as well, showcasing a parallel in algebraic properties.
Elimination of Other Options:
- The claim that the dimension of L(\alpha) over L is greater or smaller than that of F(\alpha) over F lacks generality as L is specific to the coefficients and doesn't independently control such dimensional relations without additional constraints.

Thus, the confirmed true options correctly describe the finite or algebraic nature of the field extensions in relation to the subfield L.

Answer 2

The problem involves field extensions and requires us to determine the truth of certain statements about them. The essence of the question is about comparing the extension properties between various fields when given specific conditions.

Setup of the Problem: E \subset F \subset K are given as field extensions, where these are not necessarily algebraic. We have an element \alpha \in K that is algebraic over F but not a member of F itself. L is defined as the subfield of K generated over E by the coefficients of the monic polynomial of minimal degree over F for which \alpha is a root.
Analysis of Options:
- The correct answers given are:
  1. F(\alpha) \supset L(\alpha) is a finite extension if and only if F \supset L is a finite extension.
  2. F(\alpha) \supset L(\alpha) is an algebraic extension if and only if F \supset L is an algebraic extension.
Explanation of the First Correct Option:
- Finite Extensions and Algebraic Elements: If F(\alpha) \supset L(\alpha) is finite, it means that \alpha minimally satisfies a polynomial over F that constrains the extension's degree. Since \alpha's minimal polynomial over F has coefficients generating L, it implies F \supset L must also be finite. Conversely, if F \supset L is finite, it leads to F(\alpha) \supset L(\alpha) being finite because generating F(\alpha) involves finitely extending through \alpha. Therefore, the finiteness condition is equivalent in both scenarios.
Explanation of the Second Correct Option:
- Algebraic Extensions: When F(\alpha) \supset L(\alpha) is algebraic, every element within this extension satisfies a polynomial with coefficients in L. This condition reciprocates to F \supset L being algebraic due to \alpha's polynomial acting as a bridge for algebraicity. If F \supset L is algebraic, each element in F satisfies a polynomial within L, making F(\alpha) \supset L(\alpha) algebraic as well, showcasing a parallel in algebraic properties.
Elimination of Other Options:
- The claim that the dimension of L(\alpha) over L is greater or smaller than that of F(\alpha) over F lacks generality as L is specific to the coefficients and doesn't independently control such dimensional relations without additional constraints.

Thus, the confirmed true options correctly describe the finite or algebraic nature of the field extensions in relation to the subfield L.

Show Hint

The Correct Option is A, D

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