Are all commutative rings with identity Noetherian?

Question

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?

Answer 1

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.