# If $A/I_i$ is noetherian for all $i$, and $\cap I_j=\{0\}$, is $A$ noetherian?

I want to show the following:

Let $I_{1},...,I_{n}$ be ideals of a commutative ring (with 1) $A$ such that $\bigcap_{i=1}^{n} I_{i}$ is the zero ideal. If each quotient $A/I_{i}$ is a Notherian ring show that $A$ is a Notherian ring.

My work:

Let $f: A \rightarrow A/I_{1} \times A/I_{2} \times .... \times A/I_{n}$ be the map defined by $f(a)=(a+I_{1},a+I_{2},...,a+I_{n})$. Then $A$ embeds in $A/I_{1} \times A/I_{2} \times .... \times A/I_{n}$. Since the direct sum of Noetherian modules is Noetherian and every submodule of a Noetherian module is Noetherian the result follows.

Is the above OK?

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Yes. ${}{}{}{}$ – Mariano Suárez-Alvarez Mar 23 '11 at 2:44
@Mariano: Dear Mariano, You should make this an answer! (And also, how did you get around the character limit? I know there are ways, but I always forget them!) Cheers, – Matt E Mar 23 '11 at 4:22
@Matt, I usually write ${}{}{}{}{}{}$... It is very silly to have to do that! – Mariano Suárez-Alvarez Mar 23 '11 at 4:28
a sub module of a Noetherian module is also Noetherian, but this is not true in general for rings. For example, take any integral domain which is not Noetherian (can take the polynomial ring with infinite indeterminates) and take its field of fractions which is always Noetherian. – Prometheus Mar 23 '11 at 7:13

Yes. ${}{}{}{}{}{}{}{}{}{}{}{}$