Let $R$ be any ring.
I need to prove that $R$ satisfies the ascending chain condition on ideals iff every ideal of $R$ is finitely generated.
Here's what I have so far:
$(\Rightarrow)$ We prove this implication using the contrapositive. Assume $R$ has an ideal $I$ which is not finitely generated. Let $x_1\in I$. Now $(x_1)\neq I$ so there exists $x_2\in I $ which is not in $(x_1)$.
Similarly, since $(x_1, x_2)\neq I $, there exists $x_3\in I $ which is not in $(x_1,x_2) $. Continuing this process indefinitely, since no finite set of $x_i$'s will generate $I$, we have the following strictly increasing chain:
$$(x_1)\subsetneqq (x_1,x_2)\subsetneqq (x_1,x_2,x_3)\subsetneqq\cdots$$
This proves one direction of the theorem.
$(\Leftarrow)$ We prove this implication by the contrapositive, as well. Assume $R$ has a strictly increasing chain of ideals $$I_1\subsetneqq I_2\subsetneqq \cdots$$
Let $I=\bigcup\limits_{k=1}^\infty I_k$. (I'm able to prove $I$ is an ideal easily, but I omit it here).
Now I claim that $I$ is not finitely generated by showing that an arbitrary set of generators $x_1,\dots,x_\ell$ does not generate $I$.
I thought I could do this by choosing the minimum $n$ such that $I_n$ contains all the $x_i$, but I don't think this step is valid if $R$ does not have an identity element. Without $1_R$, I'm unsure whether the ideals of the chain necessarily contain the generators of their union. Suggestions on how to get around this difficulty would be appreciated.
Best Answer
Whether $R$ has a unit is totally irrelevant here. By definition, a set $S$ generates an ideal $I$ if $I$ is the smallest ideal containing every element of $S$. In particular, $I$ does contain every element of $S$. So in your situation, if $x_1,\dots,x_\ell$ are generators of $I$, then they are all elements of $I$, and so some $I_n$ contains them all.