# Every subset of ideals of $R$ has maximal element $\implies$ Every ideal of $R$ is finitely generated, if $R$ is Noetherian.

We know that a commutative ring $$R$$ is Noetherian, if for the increasing sequence of ideals $$I_1\subseteq I_2 \subseteq...,\ \exists m\in \mathbb{Z}^+,\forall k\in \mathbb{Z}_{\geq m}:\ I_m=I_k$$.

Question. If $$R$$ is Noehterian and every non-empty subset of ideals of $$R$$ has maximal element, then every ideal of $$R$$ is finitely generated.

Proof. We take an ideal $$I$$ of $$R$$ and we define the set $$F:=\{J: J\text{ finitely generated ideal of } R \text{ and } J\subseteq I\}.$$ We can see that $$F\neq \emptyset$$ (because $$\langle 0_R \rangle \in F$$). So, from hypothesis, there exists a maximal element $$M\in F$$.

We want to show that $$I=M$$.

Obviously, $$M\subseteq I$$, because $$M\in F$$. We suppose now that $$M\subsetneq I$$. Then, there exists an element $$a\in I\backslash M\iff a\in I$$, with $$a\notin M$$.

We consider the ideal $$J:=M+\langle a\rangle.$$ Then, if $$M=\langle m_1,...,m_n \rangle$$, we have $$J=M+\langle a\rangle=\langle m_1,..., m_n,a \rangle \implies J\in F$$

Question: Why $$M\neq M+\langle a \rangle$$?

If $$M= M+\langle a \rangle$$ $$\underline{ \text{and}}$$ $$R$$ has $$1_R$$, then $$a=1_R\cdot a \in M+\langle a \rangle =M$$, contradiction.

But what happens if $$R$$ hasn't unity?

The argument you are giving in no way requires $$R$$ to be unital. Note that if $$R$$ is not unital, then $$\langle a\rangle$$ still refers to the ideal generated by $$a$$. That means it is the smallest ideal which contains $$a$$, which is not necessarily the same as the set $$Ra$$. So, by definition, $$a\in\langle a\rangle.$$
(Note also that your statement of what you are proving seems to be mangled. You have never used the assumption that $$R$$ is Noetherian, nor do you need to. The assumption that every nonempty set of ideals has a maximal element is actually equivalent to assuming that $$R$$ is Noetherian.)