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Math Help - algebraic over K

  1. #1
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    algebraic over K

    Perhaps this is not the proper place but, nonetheless, there it goes: E, K are fields, K included in E. A is a subset of E and all elements of A are algebraic over K. And the author says: "Clearly K(A) is the union of all subfields of the form K(M), where M ranges over finite subsets of A".

    It is this statement I cannot understand. I have proved that the union of the K(M) is included in K(A) but I cannot prove the inverse inclusion. Thank you for you time.

    Enrique.
    Last edited by Plato; October 13th 2009 at 01:20 PM.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Perhaps this is not the proper place but, nonetheless, there it goes: E, K are fields, K included in E. A is a subset of E and all elements of A are algebraic over K. And the author says: "Clearly K(A) is the union of all subfields of the form K(M), where M ranges over finite subsets of A".

    It is this statement I cannot understand. I have proved that the union of the K(M) is included in K(A) but I cannot prove the inverse inclusion. Thank you for you time.

    Enrique.

    Well, take any element x in K(A): since this is a vector space over K generated by A (perhaps not finite dimensional, though) , there exist elements y_1,...,y_n in K(A) s.t. x is K-linear combination of y_1,...,y_n.
    Now, each element y_i is a K-lin. comb. of a finite number of elements from A since, as pointed out above, K(A) is generated as K-vec. space by A ==> in y_1,...,y_n intervene only a finite number of elements of A ==>
    the element x is a K-lin. com. of a finite number of elements of A, say a set M ==> x belongs to K(M), for M a finites subset of A.

    Tonio
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  3. #3
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    I see. But why not say directly: take any element x in K(A). Since this
    is a vector space over K generated by A, x is a K-lin. comb. of a
    finite number of elements from A, say a set M. Therefor, x belongs to
    K(M)?

    After all, afterwards you take y_i belonging to K(A) and say that y_i
    is a lin.comb. of a finite number of elems. of A. But x was also in
    K(A). Regards,

    Enrique.
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  4. #4
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    Hi

    Take care, there can be more elements in K(A) than just K-linear combinations of elements of A:

    for instance, with A=\{\sqrt[3]{2}\} and K=\mathbb{Q},
    \sqrt[3]{2}^2\in\mathbb{Q}(\sqrt[3]{2}) but \sqrt[3]{2}^2 can't be written as a \mathbb{Q}-lin. comb with \sqrt[3]{2}, it would implie that \sqrt[3]{2} minimal polynomial has degree at most 2
    (while X^3-2 is irreducible and has \sqrt[3]{2} as a root, therefore it is its minimal polynomial)

    But there is a similar argument as what tonio proposed, using that:

    K(A)=\{\frac{P(a_1,...,a_n)}{Q(b_1,...,b_m)}\ ;\ n,m\in\mathbb{N}, (P,Q)\in K[X_1,...,X_n]\times(K[X_1,...,X_m]-\{0\}),\ a_1,...,a_n,b_1,...,b_m\in A\}
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  5. #5
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    Alright. Then let x belong to K(A). Then x = P(a_1,...,a_m)/Q(b_1,...,b_n) for some m, n, P, Q and a_1,...,a_m,b_1,...,b_n belonging to A. Now, if I put M= {a_1,...,a_m,b_1,...,b_n} then M is a subset of A and, using the same definition for K(M) as that used for K(A), x belongs to K(M) and all the more so to the union of all the K(M) s.t. M is a finite subset of A.

    I also see that the notion of a vector space has not been used here. Well,
    thank you very much.
    Enrique.
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  6. #6
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    I see. But why not say directly: take any element x in K(A). Since this
    is a vector space over K generated by A, x is a K-lin. comb. of a
    finite number of elements from A, say a set M. Therefor, x belongs to
    K(M)?

    After all, afterwards you take y_i belonging to K(A) and say that y_i
    is a lin.comb. of a finite number of elems. of A. But x was also in
    K(A). Regards,

    Enrique.
    .
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