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?

  • $\begingroup$ Take an ideal $I$ of $A$ and consider its projection onto $A_k$. You know that $I$ is closed under addition and under multiplication by arbitrary elements of $A$, as it is an ideal of $A$. What does this tell you about the projection? $\endgroup$ – qaphla Jul 15 '14 at 23:48
  • $\begingroup$ Am I completely missing something here? Surely the fact that $I$ is multiplicatively closed gives that its projection onto $A_k$ is also multiplicatively closed, as if some $a_k$ is in this projection, coming from $a \in I$, then for any $r \in A_k$, we can take some $r' \in A$ such that the $k$th coordinate of $r'$ is $r$, and it must be that $r'a \in I$, and thus that $ra_k$ is in the projection. $\endgroup$ – qaphla Jul 16 '14 at 0:20
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    $\begingroup$ Of course it suffices to consider $n=2$. $\endgroup$ – Martin Brandenburg Jul 20 '14 at 21:17
  • $\begingroup$ Consider noetherian non-unital rings $A,B$. Notice that $A$ is a module over its unitalization $A^+$. Ideals of $A$ are $A^+$-submodules of $A$. We have $(A \times B)^+ \cong A^+ \times_{\mathbb{Z}} B^+$ (fiber product w.r.t the augmentation). Thus, the question is a special case of the following one, which has the advantage that the rings are unital: Let $A^+,B^+$ be augmented unital rings and let $M$ (resp. $N$) be a noetherian $A^+$-module (resp. $B^+$-module). Clearly $M \times N$ is a noetherian $A^+ \times B^+$-module. Does it stay noetherian over the subring $A^+ \times_\mathbb{Z} B^+$? $\endgroup$ – Martin Brandenburg Jul 21 '14 at 5:33

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 $I_i=I_t$.


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