Lectures on the Algebraic Theory of Fields - Tata Institute of ...

4. Algebraic Closure 11¯σx=σ α xwhere x∈K α . (Note that every x∈K is in some K α in the simplyordered subfamily). It is easy to see that ¯σ is well - defined. Supposex∈K β and K β ⊂ K α thenσ α is an extension ofσ ρ and soσ α x=σ β x.This proves that ¯σ is an isomorphism of K on K ′ and extendsσ. Thusthe triplet (K, K ′ , ¯σ) is in M and is an upper bound of the subfamily.By Zorn’s lemma there exists a maximal triplet (Ω,Ω ′ ,τ). We assertthatΩis algebraically closed; for if not letρbe a root of an irreduciblepolynomial f (x)∈Ω[x]. Then f ζ (x)∈Ω ′ [x] is also irreducible. Letρ ′be a root of f τ (x). Thenτcan be extended to an isomorphism ¯τ ofΩ(ρ)onΩ ′ (ρ). Now (Ω(ρ), ¯τ is in M and hence leads to a contradiction. ThusΩ is an algebraic closure of k,Ω ′ of k ′ andτan isomorphism ofΩonΩ ′ extendingσ.In particular if k=k ′ andσthe identity isomorphism, thenΩandΩ ′ are two algebraic closures of k andτis then a k-isomorphism.Out theorem is completely demonstrated.Let f (X) be a polynomial in k[x] and K= k(α 1 ,...,α n , a splittingfield of f (x), so thatα 1 ,...,α n are the distinct roots of f (x) in K. Let K ′be any other splitting field andβ 1 ,...β m the distinct roots of f (x) in K ′ .LetΩbe an algebraic closure of K andΩ ′ of K ′ . ThenΩandΩ ′ are twoalgebraic closures of k. There exists therefore an isomorphismσofΩonΩ ′ which is identity on k. LetσK= K 1 . Then K 1 = k(σ α1 ,...,σ αn ).Sinceα 1 ,...α n are distinctσα 1 ...,σα n are distinct and are roots off (x). Thus K 1 is a splitting field of f (x) inΩ ′ . This proves that 13K ′ = K 1 .β 1 ,...,β m are distinct and are roots of f (x) inΩ ′ . We have m=nandβ i =σα 2 in some order. Therefore the restriction ofσto K is anisomorphism of K on K ′ . We haveTheorem 6. Any two splitting fields K, K ′ of a polynomial f (x) in k[x]are k− isomorphic.Let K be a finite field of q elements. Then q=P n where n is aninteger≥ 1 and p is the characteristic of K. Also n=(K :Γ),Γbeing