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Math Help - every finite field is noetherian

  1. #1
    cig
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    every finite field is noetherian

    i really need some help in proofing that Every Finite Field is Noetherian .
    if anyone could help me i would be greatfull
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cig View Post
    i really need some help in proofing that Every Finite Field is Noetherian .
    if anyone could help me i would be greatfull
    Every field has only two ideals the trivial and the non-proper. Is this really what you meant?
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    Quote Originally Posted by Drexel28 View Post
    Every field has only two ideals the trivial and the non-proper. Is this really what you meant?
    Actually,the question is : "Is it true that every finite ring is noetherian ? " I have to JUSTIFY my answer.. Nothing else is specified . Thank you in advance .
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cig View Post
    Actually,the question is : "Is it true that every finite ring is noetherian ? " I have to JUSTIFY my answer.. Nothing else is specified . Thank you in advance .
    What would the existence of a non-terminating AC imply about the cardinality? Think about it..
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    Quote Originally Posted by Drexel28 View Post
    What would the existence of a non-terminating AC imply about the cardinality? Think about it..

    I haven't understand the AC condition. In a finite ring it means that i have specific number of Ideals and that somehow it ends the series of ideals that are subsets. so by this condition it means that it is true ? I am really confused !
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    Senior Member Tinyboss's Avatar
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    The problem is that you don't know what Noetherian means. Take care of that, and the answer to this problem will become obvious.
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    Quote Originally Posted by Tinyboss View Post
    The problem is that you don't know what Noetherian means. Take care of that, and the answer to this problem will become obvious.

    I know, as i said i am confused. Can you give me a hint or help me somehow ??
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cig View Post
    I know, as i said i am confused. Can you give me a hint or help me somehow ??
    Suppose that \#(R)=n then what if for any \left\{A_1,\cdots,A_{n+1}\right\}\subseteq 2^R one has A_1\subsetneq\cdots\subsetneq A_{n+1}?
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    Quote Originally Posted by Drexel28 View Post
    Suppose that \#(R)=n then what if for any \left\{A_1,\cdots,A_{n+1}\right\}\subseteq 2^R one has A_1\subsetneq\cdots\subsetneq A_{n+1}?

    This gives us a finite strincly increasing chain,contradicting the ascending chain condition that stabilises for Io=In for n>n0 ?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cig View Post
    This gives us a finite strincly increasing chain,contradicting the ascending chain condition that stabilises for Io=In for n>n0 ?
    You're thinking about this way too hard. What you want to prove is that any ascending chain terminates, right? But for any (non-empty) subsets A_1,\cdots,A_{n+1} of your ring R with n elements if A_1\subsetneq\cdots\subsetneq A_{n+1} then this implies that \#(A_1)<\cdots\#(A_n) and so evidently \#(A_{n+1})\geqslant n+\#(A_1)\geqslant n+1...but this isn't possible since \#(A_{n+1})\leqslant \#(R)=n. Get the point?
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  11. #11
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    Quote Originally Posted by Drexel28 View Post
    You're thinking about this way too hard. What you want to prove is that any ascending chain terminates, right? But for any (non-empty) subsets A_1,\cdots,A_{n+1} of your ring R with n elements if A_1\subsetneq\cdots\subsetneq A_{n+1} then this implies that \#(A_1)<\cdots\#(A_n) and so evidently \#(A_{n+1})\geqslant n+\#(A_1)\geqslant n+1...but this isn't possible since \#(A_{n+1})\leqslant \#(R)=n. Get the point?
    So since that this isnt possible for the cardinality of our #R=n ,then all A's are subsets and satisfies the AC condition and thus it is true that every finite ring is noetherian ?
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