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algebraic over k(α₁, ₂,..., α). Therefore, [k(α₁, ₂,..., an) (an+1): k(α₁, ₂,..., α₁)] is finite. By the induction hypothesis, [k(a₁, 0₂,..., an): k] is finite. Thus [F:k] [k(α₁, ₂,..., anan): k] = [k(α₁, ₂,..., α₁)(a): k(α₁, α₂,..., a)][k (α₁, ₂,..., α):k] is finite. Hence, F/k is a finite extension for any F generated by a finite number of U

algebraic over k(α₁, ₂,..., α). Therefore, [k(α₁, ₂,..., an) (an+1): k(α₁, ₂,..., α₁)] is finite. By the induction hypothesis, [k(a₁, 0₂,..., an): k] is finite. Thus [F:k] [k(α₁, ₂,..., anan): k] = [k(α₁, ₂,..., α₁)(a): k(α₁, α₂,..., a)][k (α₁, ₂,..., α):k] is finite. Hence, F/k is a finite extension for any F generated by a finite number of U

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.3: The Field Of Quotients Of An Integral Domain
Problem 11E
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Request explain these steps Theorem 9: Let F/k be an extension field generated by a finite number of algebraic elements over k. Then F/k is a finite extension. Proof: We prove the result by applying the principle of mathematical induction on the number of generators of F over k. Suppose first that F is generated over k by a single algebraic element say α₁, i.e., F=k(α₁). Since a, is algebraic over k, by Theorem 9, F/k is finite. Next, assume that if a field is generated over k by n algebraic elements, then it is a finite extension. We will now prove the result for extensions generated over k by n+1 algebraic elements. For this, let F=k(α₁, ₂,..., a, a), where *n+1 α; (1≤ i ≤n+1) are algebraic over k. Since is algebraic over k, it is also algebraic over k(α₁,₂,..., α). Therefore, [k(α₁, ₂,..., α) (an+1): k(α₁, α₂,..., a)] is finite. By the induction hypothesis, [k(α₁, ₂,..., a): k] is finite. Thus [F:k] [k(α₁, ₂,..., an, an+1): k] *n+l U = [k(α₁, ₂,..., α₁) (α₁+1): k(α₁, ₂,..., α)][k(α₁, ₂,..., an):k] is finite. Hence, F/k is a finite extension for any F generated by a finite number of algebraic elements over k.
Transcribed Image Text:Request explain these steps Theorem 9: Let F/k be an extension field generated by a finite number of algebraic elements over k. Then F/k is a finite extension. Proof: We prove the result by applying the principle of mathematical induction on the number of generators of F over k. Suppose first that F is generated over k by a single algebraic element say α₁, i.e., F=k(α₁). Since a, is algebraic over k, by Theorem 9, F/k is finite. Next, assume that if a field is generated over k by n algebraic elements, then it is a finite extension. We will now prove the result for extensions generated over k by n+1 algebraic elements. For this, let F=k(α₁, ₂,..., a, a), where *n+1 α; (1≤ i ≤n+1) are algebraic over k. Since is algebraic over k, it is also algebraic over k(α₁,₂,..., α). Therefore, [k(α₁, ₂,..., α) (an+1): k(α₁, α₂,..., a)] is finite. By the induction hypothesis, [k(α₁, ₂,..., a): k] is finite. Thus [F:k] [k(α₁, ₂,..., an, an+1): k] *n+l U = [k(α₁, ₂,..., α₁) (α₁+1): k(α₁, ₂,..., α)][k(α₁, ₂,..., an):k] is finite. Hence, F/k is a finite extension for any F generated by a finite number of algebraic elements over k.
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