# rings and ideals

• Nov 29th 2009, 04:57 AM
Magics6
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
$
• Nov 29th 2009, 06:13 AM
tonio
Quote:

Originally Posted by Magics6
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
$

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
• Nov 29th 2009, 07:08 AM
Magics6
Quote:

Originally Posted by tonio
...and in (2), shouldn't it be "finitely generated ideals" instead of "finite ideals"?

confirm... finitely generated ideals
• Nov 29th 2009, 07:52 PM
aliceinwonderland
Quote:

Originally Posted by Magics6
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.

Quote:

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
$

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 $k \geq n$.