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By Wilkins D.R.

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K } of L such that αi is algebraic over K for i = 1, 2, . . , k and L = K(α1 , α2 , . . , αk ). Proof Suppose that the field extension L: K is a finite. 2). Thus if {α1 , α2 , . . , αk } is a basis for L, considered as a vector space over K, then each αi is algebraic and L = K(α1 , α2 , . . , αk ). Conversely suppose that L = K(α1 , α2 , . . , αk ), where αi is algebraic over K for i = 1, 2, . . , k. Let Ki = K(α1 , α2 , . . , αi ) for i = 1, 2, . . , k. Clearly Ki−1 (αi ) ⊂ Ki for all i > 1, since Ki−1 ⊂ Ki and αi ∈ Ki .

K. Moreover αi is clearly algebraic over Ki−1 since it is algebraic over K, and K ⊂ Ki−1 . 4 that the field extension Ki : Ki−1 is finite for each i. 1), we deduce that L: K is a finite extension, as required. 6 Let M : L and L: K be algebraic field extensions. Then M : K is an algebraic field extension. Proof Let α be an element of M . We must show that α is algebraic over K. Now there exists some non-zero polynomial f ∈ L[x] with coefficients in L such that f (α) = 0, since M : L is algebraic.

A K-homomorphism θ: L → M is a homomorphism of fields which satisfies θ(a) = a for all a ∈ K. A K-monomorphism is an injective K-homomorphism. A K-isomorphism is a bijective K-homomorphism. A K-automorphism of L is a K-isomorphism mapping L onto itself. Two extensions L1 : K and L2 : K of a field K are said to be K-isomorphic (or isomorphic) if there exists a K-isomorphism ϕ: L1 → L2 between L1 and L2 . If L: K is a field extension then we can regard L as a vector space over the field K. If L is a finite-dimensional vector space over K then we say that the extension L: K is finite.

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