1. ## 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
$

2. 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

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

4. 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.

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 $k \geq n$.