I am confused about the equivalence of the different definitions of Noetherian Rings. The confusion essentially stems from the fact that a Noetherian ring need not be finitely generated. Essentially the proof goes like this: If $$I_1 \subseteq I_2 \subseteq \cdots \subseteq I_k \subseteq \cdots $$ is an ascending chain of finitely generated ideals, then $\bigcup_{j \in \mathbb{N}}I_j$ is an ideal which is then finitely generated. Then, all the generators are in some $I_N$ for $N$ sufficiently large, proving the result.
This is clear if $\bigcup_{j \in \mathbb{N}} I_j$ is a strict ideal of $R$. I'm not sure if this proof works when $\bigcup_{j \in \mathbb{N}} I_j = R$, though. Given a situation like this, where all $I_j$ are strict ideals of $R$, can the union be $R$?
If $R$ has a 1, then definitely not since if $\bigcup_{j} I_j = R$, then $1 \in I_n$ for some $n$ which implies that $I_k = R$ for all $k \geq n$. I'm having trouble seeing why this would be true if $R$ didn't have a 1, though.
Thanks for the help!