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Thread: rings and ideals

  1. #1
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    rings and ideals

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

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

    2.) Prove that if every ideals of R is finitely generated, then exist a number $\displaystyle n \in \aleph $ such that
    $\displaystyle I_{1} \subseteq I_{2} \ldots \subseteq I_{n-1}\subseteq I_{n} = I_{n} = I_{n+1} =I_{n+2} = \ldots
    $
    Last edited by Magics6; Nov 29th 2009 at 12:08 PM.
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  2. #2
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    Quote Originally Posted by Magics6 View Post
    Let $\displaystyle I_{1} \subseteq I_{2} \subseteq I_{3} \ldots$ be an increasing sequence of ideals in a ring R

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

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

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

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

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

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