infinite strictly increasing chain of ideals $\{0\}\subsetneq I_1\subsetneq I_2\subsetneq\dots\subsetneq I_n\subsetneq\dots R$

Question

Does $\mathbb{Z}$ have an infinite strictly increasing chain of ideals

$\{0\}\subsetneq I_1\subsetneq I_2\subsetneq\dots\subsetneq I_n\subsetneq\dots\mathbb{Z}$?

If so, exhibit such a chain, and if not, give an example of a commutative ring $R$ with such a chain.

My attempt: I do not believe this statement holds for $\mathbb{Z}$, as any element in $\mathbb{Z}$ has only finitely many divisors, so the chain of ideals must be finite.

To find a commutative ring $R$ with such a chain, I was thinking of using some sort of polynomial ring, because $\mathbb{Z}[x]\subsetneq\mathbb{Z}[x,y]$, and so on, but defining the actual ring $R$ is a problem.

Any help appreciated!

Answer 1

Any ascending chain of ideals must end at the ideal $p \mathbb{Z}$ for some prime $p.$

For an example your intuition is correct. Try $R[x_1,x_2,\ldots]$ where its elements are polynomials in a finite number of indeterminates $x_i$ and the ideal $\langle x_1,x_2,\ldots \rangle.$