Question:
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?
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?
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When dealing with field extensions, remember that algebraic extensions preserve the degree of extension, and the field's dimension can be compared between subfields.
Updated On: Apr 9, 2025
- \( F(\alpha) \supset L(\alpha) \) is a finite extension if and only if \( F \supset L \) is a finite extension
- The dimension of \( L(\alpha) \) over \( L \) is greater than the dimension of \( F(\alpha) \) over \( F \)
- The dimension of \( L(\alpha) \) over \( L \) is smaller than the dimension of \( F(\alpha) \) over \( F \)
- \( F(\alpha) \supset L(\alpha) \) is an algebraic extension if and only if \( F \supset L \) is an algebraic extension
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The Correct Option is A, D
Solution and Explanation
Step 1: \( F(\alpha) \supset L(\alpha) \) is a finite extension if and only if \( F \supset L \) is a finite extension
Since \( \alpha \) is algebraic over \( F \), both \( F(\alpha) \) and \( L(\alpha) \) are algebraic extensions. Therefore, the finiteness of the extension is preserved between the fields, making (A) TRUE.
Step 2: The dimension of \( L(\alpha) \) over \( L \) is greater than the dimension of \( F(\alpha) \) over \( F \)
Since \( L \subset F \) and \( L(\alpha) \) is generated by a minimal polynomial over \( L \), the dimension of \( L(\alpha) \) over \( L \) cannot be greater than that of \( F(\alpha) \) over \( F \). Hence, (B) is FALSE.
Step 3: The dimension of \( L(\alpha) \) over \( L \) is smaller than the dimension of \( F(\alpha) \) over \( F \)
This is not necessarily true either, as \( \alpha \) might have the same minimal polynomial over both \( F \) and \( L \). Therefore, (C) is FALSE.
Step 4: \( F(\alpha) \supset L(\alpha) \) is an algebraic extension if and only if \( F \supset L \) is an algebraic extension
Since \( \alpha \) is algebraic over \( F \), both \( F(\alpha) \) and \( L(\alpha) \) are algebraic extensions. However, this doesn't imply that \( F \supset L \) is algebraic unless further conditions are given. So (D) is also NOT always TRUE. Hence, (D) is FALSE.
Final Answer
\[ \boxed{(A) \ F(\alpha) \supset L(\alpha) \text{ is a finite extension if and only if } F \supset L \text{ is a finite extension}} \]
Since \( \alpha \) is algebraic over \( F \), both \( F(\alpha) \) and \( L(\alpha) \) are algebraic extensions. Therefore, the finiteness of the extension is preserved between the fields, making (A) TRUE.
Step 2: The dimension of \( L(\alpha) \) over \( L \) is greater than the dimension of \( F(\alpha) \) over \( F \)
Since \( L \subset F \) and \( L(\alpha) \) is generated by a minimal polynomial over \( L \), the dimension of \( L(\alpha) \) over \( L \) cannot be greater than that of \( F(\alpha) \) over \( F \). Hence, (B) is FALSE.
Step 3: The dimension of \( L(\alpha) \) over \( L \) is smaller than the dimension of \( F(\alpha) \) over \( F \)
This is not necessarily true either, as \( \alpha \) might have the same minimal polynomial over both \( F \) and \( L \). Therefore, (C) is FALSE.
Step 4: \( F(\alpha) \supset L(\alpha) \) is an algebraic extension if and only if \( F \supset L \) is an algebraic extension
Since \( \alpha \) is algebraic over \( F \), both \( F(\alpha) \) and \( L(\alpha) \) are algebraic extensions. However, this doesn't imply that \( F \supset L \) is algebraic unless further conditions are given. So (D) is also NOT always TRUE. Hence, (D) is FALSE.
Final Answer
\[ \boxed{(A) \ F(\alpha) \supset L(\alpha) \text{ is a finite extension if and only if } F \supset L \text{ is a finite extension}} \]
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