Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – qaphla Jul 15 '14 at 23:48
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. – qaphla Jul 16 '14 at 0:20
Of course it suffices to consider $n=2$. – Martin Brandenburg Jul 20 '14 at 21:17
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 $A^+ \times_\mathbb{Z} B^+$? – Martin Brandenburg Jul 21 '14 at 5:33
up vote 4 down vote accepted

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 $I_i=I_t$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.