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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.

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    $\begingroup$ $x\in I$ 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. $\endgroup$
    – Hanul Jeon
    Commented Feb 4, 2017 at 6:21
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    $\begingroup$ Thanks @Tyler! for asking this question. In it you answered my own question regarding the equivalency, which I was searching on the web. $\endgroup$
    – bogec
    Commented Jan 6 at 9:15

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Tyler
    Commented Feb 4, 2017 at 6:43
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    $\begingroup$ 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. $\endgroup$
    – Eric Wofsey
    Commented Feb 4, 2017 at 6:45
  • $\begingroup$ Oh dear. I thought that the ideal generated by a set was defined as the combinations of ring operations on the set, and that when the ring has a unit, such an ideal is the smallest ideal containing the generators. But apparently I've got the definition and the special case backwards. Thanks for the clarification, it looks like I've pretty much completed the original proof, then. $\endgroup$
    – Tyler
    Commented Feb 4, 2017 at 6:49

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