# Are all commutative rings with identity Noetherian?

According to the definition of Noetherian ring, a ring is said to be Noetherian if all increasing chains of ideals eventually stabilize.

Now, let $$R$$ be a commutative ring with identity and we take a collection $$C$$ of proper ideals of $$R$$. Consider an increasing chain of ideals $$I_1\subseteq I_2\subseteq I_3\subseteq ....$$ Now, $$I=\cup_{i}I_i$$ is a maximal ideal, as any proper ideal containing $$I$$, is in $$C$$ and falls in the increasing chain, hence it falls inside $$I$$ and if, $$I=R$$, then $$1\in I$$, therefore, $$1\in I_i$$ for any $$i$$, which again contradicts the definition of $$I_i$$.

So, does all commutative rings with identity falls inside Noetherian rings?

• Your $I$ need not be maximal – Hagen von Eitzen Nov 17 '20 at 14:23
• Why do you think that?@HagenvonEitzen – roydiptajit Nov 17 '20 at 14:29
• Why do you think it is maximal? Why would any proper ideal containing $I$ have to be in $C$? – Daniel Hast Nov 17 '20 at 14:30
• As, $C$ is the collection of proper ideals. Also, if $I\subseteq J\ne R$, then $J$ falls in the increasing chain of $I_i$ as it should also contain any $I_i$. – roydiptajit Nov 17 '20 at 14:35
• This maybe wrong, but I cannot get how... – roydiptajit Nov 17 '20 at 14:42

No: for example $$F[x_1,x_2,\ldots]$$ in countably many variables is commutative, has identity, and is not Noetherian.

There are holes in the reasoning you've given. The first one is

Now, $$I=\cup_{i}I_i$$ is a maximal ideal

This does not follow, even if the ring was Noetherian. In a Noetherian ring, that chain is always going to stabilize, and so the union will be the place where it stabilizes. But there's no reason to expect it will be a maximal ideal. The reason you gave: "Any ideal containing $$\cup I_i$$ falls in the increasing chain" does not hold. Consider the ring I gave above and the chain of ideals $$(x_2)\subseteq (x_2,x_4)\subseteq (x_2,x_4,x_6)\subseteq \cdots$$. The ideal $$(x_1, x_2,x_3,x_4,\ldots)$$ contains the union of the "even variable chain" but it is not one of the members of the chain, and it is not even equal to the union of the chain.

Secondly, even if the union is a maximal ideal, it does not help determine if the ring is Noetherian or not. One can construct non Noetherian valuation rings for which the union of all proper ideals is indeed the maximal ideal, but the ring is not Noetherian.

• Which one of my reasons are failing to make $I$ maximal? Can you point out.. – roydiptajit Nov 17 '20 at 14:32
• @DiptajitRoy You say "Any ideal containing $\cup C$ is in $C$" which is just non-sequitur. Consider the ring I gave in my solution, and consider the chain of ideals $(x_2)\subseteq (x_2,x_4)\subseteq (x_2,x_4, x_6)\subseteq \cdots$. Why would $(x_1,x_2,x_3,...)$ , which contains the union of the chain, be in the chain? – rschwieb Nov 17 '20 at 14:55
• I stated $C$ to be the set of all proper ideals. I claim, $\cup_i I_i\in C$. If not then $\cup _iI_i= R$ as it is not $(0)$. That means, $1\in I\Longrightarrow 1\in I_i$ for any $i$, which is a contradiction to $C$'s definition. – roydiptajit Nov 17 '20 at 15:34
• @DiptajitRoy Sorry, yes I misinterpreted "a collection" and $C$ as the chain. I will refer to it the way you wanted from now on. The union of the chain is indeed a proper ideal if the things in the chain are proper. But it still does not follow that the proper ideal containing $I$ "falls inside the chain". Did you consider the example I gave in the comment above? – rschwieb Nov 17 '20 at 15:53
• That is actually my confusion. Your example and one another example I saw is contradicting my argument. I am not sure why is that.. Regarding whether $I$ is not in the chain, I think $I$ is in the chain as it contains all the $I_i$s and the $I_i$, makes a chain so I sits at the top. Why do you think $I$ is not in the chain? – roydiptajit Nov 17 '20 at 16:46