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Math Help - A-algebra, element that is integral and finite A-module, explanation a proof

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    An A-algebra, element that is integral and finite A-module, proof explanation

    let \varphi A\to B be an A-algebra and let y\in B then there are 3 equivalent conditions:
    i)  y is integral over A
    ii)the subring A^{\prime}[y]\subset A generated by A^{\prime}=\varphi(A) is finite over A (as an A-module?)
    ii) \exists an A-algebra C\subset B such that A^{\prime}[y]\subset C and C is finite over A.

    Proof:
    i\Rightarrow ii since y is integral then we know that there exists a monic polynomial f(Y)=Y^n+a_{n-1}Y^{n-1}+...+aY+a_{0}\in A^{\prime}[Y] s.t f(y)=0(*) but then the proof says A^{\prime}[y] is generated by 1,y,y^2,....,y^{n-1} , as an A-module.
    To show this let
    b_{m}y^{m}+....+b_{1}y+b_{0}\in A^{\prime}[y] if m\ge n then substracting b_{m}y^{n-m} times the relation (*) kills the leading term......I cannot get to this.... please help

    how to show (ii)\Rightarrow (i)?

    ii\Rightarrow iii can I just set C=A^{\prime}[y]?? then A^{\prime}[y]\subset B and A^{\prime}[y]\subset A^{\prime}[y] (every ring is a subring of itself right?)
    Last edited by rayman; December 12th 2012 at 05:56 AM.
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    MHF Contributor Drexel28's Avatar
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    Re: A-algebra, element that is integral and finite A-module, explanation a proof

    These are classic proofs.

    i implies ii is correct.


    ii implies iii Correct, you can just set C=A'[y].


    iii implies i This is annoying, but should be able to be found in any book on algebraic number theory. You basically construct the characteristic polynomial of a matrix from y which will annihilate it. See Milne's notes on alg. number theory for a full proof.
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