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Math Help - rings and ideals

  1. #1
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    rings and ideals

    Let  I_{1} \subseteq I_{2} \subseteq I_{3} \ldots be an increasing sequence of ideals in a ring R

    1.) Prove that  \bigcup_{i\in\aleph} I_{i} is an ideal.

    2.) Prove that if every ideals of R is finitely generated, then exist a number  n \in \aleph  such that
    I_{1} \subseteq I_{2} \ldots \subseteq I_{n-1}\subseteq I_{n} = I_{n} = I_{n+1} =I_{n+2} = \ldots <br />
    Last edited by Magics6; November 29th 2009 at 01:08 PM.
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  2. #2
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    Quote Originally Posted by Magics6 View Post
    Let  I_{1} \subseteq I_{2} \subseteq I_{3} \ldots be an increasing sequence of ideals in a ring R

    1.) Prove that  \bigcup_{i\in\aleph} I_{i} is an ideal.

    2.) Prove that if every ideals of R is finite, then exist a number  n \in \aleph such that
    I_{1} \subseteq I_{2} \ldots \subseteq I_{n-1}\subseteq I_{n} = I_{n} = I_{n+1} =I_{n+2} = \ldots <br />

    Use definitions, as simple as that...and in (2), shouldn't it be "finitely generated ideals" instead of "finite ideals"? Because if it is finite then it is a very trivial exercise.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    ...and in (2), shouldn't it be "finitely generated ideals" instead of "finite ideals"?
    confirm... finitely generated ideals
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  4. #4
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    Quote Originally Posted by Magics6 View Post
    Let  I_{1} \subseteq I_{2} \subseteq I_{3} \ldots be an increasing sequence of ideals in a ring R

    1.) Prove that  \bigcup_{i\in\aleph} I_{i} is an ideal.
    Let I= \bigcup_{i\in\aleph} I_{i} ; let x, y \in I and r \in R. Without loss of generality, x \in I_n and y \in I_m. Let N=max{m,n}.

    Since I_N is an ideal in R, we see that x + y \in I_N \subset I, rx \in I_N \subset I and xr \in I_N \subset I. Thus I is an ideal in R.

    2.) Prove that if every ideals of R is finitely generated, then exist a number  n \in \aleph such that
    I_{1} \subseteq I_{2} \ldots \subseteq I_{n-1}\subseteq I_{n} = I_{n} = I_{n+1} =I_{n+2} = \ldots <br />
    Assume every ideal in R is finitely generated and consider I in (1). It follows that I is finitely generated by a_1, a_2, \cdots, a_m. Each a_i for i=1,2,...,m lies in one of the chains of I, say I_{k_i}. Let n= \text{max}\{k_1, k_2, \cdots, k_m\}. Then a_i \in I_n for all i. This implies I \subseteq I_n. Thus, I_n =I and I_k = I_n for all k \geq n.
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