TITLE: Direct product of finitely many Noetherian non-unital rings is Noetherian
QUESTION [9 upvotes]: Let $A_1, A_2,...,A_n$ be Noetherian rings (not necessarily unital). Is the direct product $A:=A_1×A_2×⋯×A_n$ necessarily a Noetherian ring?

If $A_1, A_2,...,A_n$ are unital, then one can prove that the ideals of $A$ have the form $I_1 \times I_2 \times \dotsc \times I_n$, where $I_k$ is an ideal of $A_k$, and then show that indeed $A$ is Noetherian. But what about the general question?

REPLY [6 votes]: I think some proofs for unital rings carry through without change, but maybe I'm being stupid? For example:
As Martin noted in comments, we can assume $n=2$, so consider an ascending chain
$$I_1\leq I_2\leq \dots$$ 
of ideals of $A\times B$, where $A$ and $B$ are Noetherian. I'll identify $A$ with the ideal $A\times\{0\}$ of $A\times B$, and let $\pi:A\times B\to B$ be the projection map.
Then
$$I_1\cap A\leq I_2\cap A\leq\dots$$
is an ascending chain of ideals of $A$, and
$$\pi(I_1)\leq\pi(I_2)\leq\dots$$
is an ascending chain of ideals of $B$.
Since $A$ and $B$ are Noetherian, there is some $t$ such that $I_i\cap A=I_t\cap A$ and $\pi(I_i)=\pi(I_t)$ for all $i\geq t$.
But if $(a,b)\in I_i$ for $i\geq t$, then $b\in\pi(I_i)=\pi(I_t)$, so 
$(a',b)\in I_t$ 
for some $a'\in A$. So 
$$(a-a',0)=(a,b)-(a',b)\in I_i\cap A=I_t\cap A,$$
and so 
$$(a,b)=(a-a',0)+(a',b)\in I_t.$$ 
So $I_i=I_t$.