# Rings: Ascending Chain Condition iff Finitely Generated Ideals

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.

• $x\in I$ implies $x\in I_k$ for some $k$ and we have finite set of generators. So we can choose largest $n$ that $I_n$ contains a generator and $I_n$ contains all of generators. Commented Feb 4, 2017 at 6:21
• Thanks @Tyler! for asking this question. In it you answered my own question regarding the equivalency, which I was searching on the web. Commented Jan 6 at 9:15

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.
• My understanding was that $(x_1,\dots,x_\ell)=\lbrace \sum_{i=1}^\ell r_ix_i \vert r_i\in R\rbrace$. Is this not correct? Commented Feb 4, 2017 at 6:43
• That is not correct in a ring that does not have unit. Or at least, it is not correct if $(x_1,\dots,x_\ell)$ means "the ideal generated by $x_1,\dots,x_\ell$", because $x_1,\dots,x_\ell$ may not be elements of that set. Commented Feb 4, 2017 at 6:45