This question is not about rings, but groups.
Somewhere I saw a claim that:
Let $H$ and $K$ be groups satisfying ACC (the ascending chain condition) on their normal subgroups. Then the product $H\times K$ also satisfies ACC on its normal subgroups.
The person who said this gave me a sketch of the proof:
Sketch of the proof. Let $\{G_i\}$ be an increasing sequence of normal subgroups of $H\times K$. Let $\pi_H$ and $\pi_K$ be the projection maps. Then the increasing sequences $\{\pi_H(G_i)\}$ and $\{\pi_K(G_i)\}$ consist of normal subgroups of $H$ and $K$, respectively. Then by the ACC of $H$ and $K$, we must have fixed natural numbers $n$ and $m$ such that $\pi_H(G_n)=\pi_H(G_{n+1})=\pi_H(G_{n+2})=\cdots$ and $\pi_K(G_m)=\pi_K(G_{m+1})=\pi_K(G_{m+2})=\cdots$. Hence for every $i$ larger than both $n$ and $m$, we must have a fixed group $G_i$, being a subgroup of the fixed $\pi_H(G_i)\times \pi_K(G_i)$. The end.
But this is so strange. What if the group $\pi_H(G_i)\times \pi_K(G_i)$ doesn't have ACC? I think there still is a possibility for $G_i$ to grow strictly. Is this proof correct? If not, let me know the correct proof of the proposition.