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- 82

- Homework Statement
- 84i) If ##B## and ##C## be subfields of a field ##E##, then their ##\textbf{compositum}## ##B \lor C## is the intersection of all the subfields of ##E## containing ##B## and ##C##. Prove that if ##\alpha_1, \dots, \alpha_n \in E##, then ##F(\alpha_1) \lor \dots \lor F(\alpha_n) = F(\alpha_1, \dots, \alpha_n)##.

84ii) Prove that any splitting field ##K/F## containing ##B## (as in exercise 83) has the form ##K = B_1 \lor \dots \lor B_r##, where each ##B_i## is isomorphic to ##B## via an isomorphism that fixes ##F##. (Hint: If ##Gal(K/F) = \lbrace \sigma_1, \dots, \sigma_r \rbrace##, then define ##B_i = \sigma_i(B)##.

- Relevant Equations
- Definition: Let ##E/F## be a field extension and ##\alpha_1, \dots, \alpha_n \in E##. Then ##F(\alpha_1, \dots, \alpha_n)## is the intersection of all subfields containing ##F## and ##\alpha_1, \dots, \alpha_n##.

Definition: ##Gal(E/F)## is the group of automorphisms of ##E## that fix ##F##.

Exercise 83 (Since it is referred to in 84ii): Let ##B/F## be a finite extension. Prove that there is an extension ##K/B## so that ##K/F## is a splitting field for some polynomial ##f(x) \in F[x]##.

Proof of 84i): We assume that ##E/F## is a field extension. For each ##i##, ##F(\alpha_i)## is the smallest subfield of ##E## containing ##F## and ##\alpha_i##. Let ##F'## be a subfield of ##E## containing ##F## and ##\alpha_1, \dots, \alpha_n##. Then ##F'## containing ##F(\alpha_i)## for all ##i##. This implies ##F'## contains ##F(\alpha_1) \lor \dots \lor F(\alpha_n)##. This shows ##F(\alpha_1) \lor \dots \lor F(\alpha_n)## is the smallest subfield of ##E## containing ##F## and ##\alpha_1, \dots, \alpha_n##. We can conclude ##F(\alpha_1) \lor \dots \lor F(\alpha_n) = F(\alpha_1, \dots, \alpha_n)##. []

Proof of 84ii) Let ##f(x) \in F[x]## be a polynomial whose splitting field is ##K/F##. Denote the roots of ##f(x)## as ##\alpha_1, \dots, \alpha_n##. Let ##Gal(K/F) = \lbrace \sigma_1, \dots, \sigma_r \rbrace## and define ##B_i = \sigma_i(B)## for all ##i##. We know ##K = F(\alpha_1, \dots, \alpha_n)##. By part i), it is enough to show ##r = n## and ##\sigma_i(B) = F(\alpha_i)##. ... But I'm not sure how to proceed. I know the ##\sigma_i##'s permute the roots of ##f(x)## but I'm not sure if i can use that here.

Does this work as a counterexample? Consider ##B = F = \mathbb{Q}## and ##E = \mathbb{Q}(\sqrt{2})##. Then ##B/F## is a finite extension. We see ##E/F## is a splitting field for ##f(x) = x^2 - 2 \in \mathbb{Q}[x]## and ##Gal(E/F) = \lbrace id, \sigma \rbrace## where ##\sigma: \mathbb{Q}(\sqrt{2}) \mapsto \mathbb{Q}(\sqrt{2})## is defined by ##a\sqrt{2} + b \mapsto a(-\sqrt{2}) + b## for all ##a, b \in \mathbb{Q}##.

Then ##id(B) \lor \sigma(B) = B \lor B## since ##id## and ##\sigma## fix ##B##. This implies ##E \neq id(B) \lor \sigma(B)##.

Proof of 84ii) Let ##f(x) \in F[x]## be a polynomial whose splitting field is ##K/F##. Denote the roots of ##f(x)## as ##\alpha_1, \dots, \alpha_n##. Let ##Gal(K/F) = \lbrace \sigma_1, \dots, \sigma_r \rbrace## and define ##B_i = \sigma_i(B)## for all ##i##. We know ##K = F(\alpha_1, \dots, \alpha_n)##. By part i), it is enough to show ##r = n## and ##\sigma_i(B) = F(\alpha_i)##. ... But I'm not sure how to proceed. I know the ##\sigma_i##'s permute the roots of ##f(x)## but I'm not sure if i can use that here.

Does this work as a counterexample? Consider ##B = F = \mathbb{Q}## and ##E = \mathbb{Q}(\sqrt{2})##. Then ##B/F## is a finite extension. We see ##E/F## is a splitting field for ##f(x) = x^2 - 2 \in \mathbb{Q}[x]## and ##Gal(E/F) = \lbrace id, \sigma \rbrace## where ##\sigma: \mathbb{Q}(\sqrt{2}) \mapsto \mathbb{Q}(\sqrt{2})## is defined by ##a\sqrt{2} + b \mapsto a(-\sqrt{2}) + b## for all ##a, b \in \mathbb{Q}##.

Then ##id(B) \lor \sigma(B) = B \lor B## since ##id## and ##\sigma## fix ##B##. This implies ##E \neq id(B) \lor \sigma(B)##.

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