# Is the direct product of quotient rings left Noetherian?

I have to figure out how to show the following ring is not left Noetherian:

$\prod _{1<n\in \mathbb{N}}\mathbb{Z}_{n}$

It is the direct product of all quotient rings.

I have been given the hint that if:

$A_{i+1}\setminus A_{i}\neq \varnothing$

That is, if the elements of an ideal's predecessor are removed from it, you are not given the empty set and that is enough of a basis to show that the ring you've been given is not left Noetherian.

I'm not really sure how that works, so any help given would be greatly appreciated.

• Hint : there is an easy infinite increasing chain of ideal which does not station. – user171326 Nov 10 '16 at 19:45
• Would it be something along the lines of: $A_{n}={(x_{1},x_{2},x_{3},...,0,0,0)}$ – Dazzler95 Nov 10 '16 at 19:51
• Yes, that's exactly it ! – user171326 Nov 10 '16 at 19:54
• Why are you saying "left Noetherian", since it is a commutative ring? – Watson Nov 10 '16 at 19:58
• @Watson: I didn't realise it was a commutative ring. Thanks for pointing that out. I fail to see how $A_{n}$ above does not become stationary, when it looks like it does exactly that when you get 0,0,0. – Dazzler95 Nov 10 '16 at 20:05