# quick question on ascending chain condition for rings

I know that if $R$ is a commutative ring with an identity in which every ideal if finitely generated then it satisfies the ascending chain condition. Just wondering if the converse is also true?

Proof of "If $R$ satisfies the ascending chain condition then every ideal is finitely generated".
Take an ideal $I\subset R$. Suppose $R$ has ascending chain condition but $I$ is not finitely generated. Choose $f_0\in I$ and look at the ideal $(f_0)$. Now $I\neq (f_0)$ (because $I$ is not finitely generated) so there exists $f_1\in I$ but $f_1\notin I_0$. The consider the ideal $(f_0, f_1)$ and repeat the same argument to find $f_2$. Therefore we have an ascending chain of ideals $$(f_0)\subsetneq (f_0, f_1)\subsetneq (f_0, f_1,f_2)\subsetneq \cdots$$ such that $f_n\in I$ but $f_{n}\notin (f_0, \cdots, f_{n-1})$. This is a contradiction because each ascending chain of ideals should stabilize but the above chain doesn't. This means our original assumption, that $I$ is not finitely generated, is wrong.