Example of a ring with an infinite inclusion chain of ideals

By definition, such a ring is non-Noetherian. A good example of a non-Noetherian ring is $F[X_1, X_2, X_3,...]$, the ring of polynomials over a field F in countably infinite indeterminates. In this ring, we have the infinite chain of generated ideals $(X_1) \subsetneq (X_1, X_2) \subsetneq (X_1, X_2, X_3) \subsetneq...$.

Claim: $X_{n+1} \notin (X_1, ..., X_n)$

Proof: Suppose for contradiction that $X_{n+1} \in (X_1, ..., X_n)$. Then, we would be able to write $X_{n+1} = a_1 \cdot X_1 \,+ ... + \,a_n \cdot X_n$ for some $a_1, ..., a_n \in F[X_1, X_2, X_3,...]$. If we take all the indeterminate arguments of each $a_i$ to be $0$ (there are finitely many arguments for each), it follows that $X_{n+1} = a_k \cdot X_k\, + ... + \,a_l\cdot X_l$, where $a_k,...,a_l$ are the constant functions among these $a_1, ..., a_n$. This is a contradiction because $X_{n+1}$ can now be written as a linear combination of certain other indeterminates, when its choice should be unconstrained.

To see this last step more clearly, it can help to actually consider $f(X_{n+1}) = X_{n+1}$ and $g(X_k,...,X_l) = a_k \cdot X_k\, + ... + \,a_l\cdot X_l$. Note that both of these polynomials reside in $F[X_1, X_2, X_3,...]$. Moreover, by the results above, $f(X_{n+1}) \equiv g(X_k,...,X_l)$. Now take $X_{n+1} = 1$ and $X_k= ...=X_l=0$. Then, it follows that

\begin{equation} 1 = f(1) = g(0,...,0) = a_k \cdot 0\, + ... + \,a_l\cdot 0 = 0 \end{equation} Hence, $1=0$, which is a contradiction because F is a field and taken to have nontrivial multiplication. The claim follows.

For more info on Noetherian rings, check out the wikipedia page.

The classical example is the ring of polynomials in a countable number of indeterminates: $k[x_1,x_2,\dots,x_n,\dotsc]$ and the chain is $$ 0\subsetneq (x_1)\subsetneq(x_1,x_2)\subsetneq\dotsb $$

Note that $x_{n+1}\notin(x_1,\dots,x_n)$ by a standard argument: suppose $$ x_{n+1}=f_1 x_1 + f_2x_2 + \dots + f_nx_n $$ where $f_i\in k[x_1,x_2,\dots,x_n,\dotsc]$ (for $i=1,2,\dots,n$). Substitute $x_i=0$ for $i=1,2,\dots,n$; this gives $x_{n+1}=0$, a contradiction.

Edit: By "ring" I mean commutative ring with identity.

Every ring is a quotient of an infinite polynomial ring. Let $R$ be any ring, and let $\{r_i\}_{i \in I}$ be a set of generators (under the ring operations). For example, we could simply take every single element of the ring $R$. Then define a ring homomorphism $\varphi : \mathbb{Z}[x_i : i \in I] \rightarrow R$ by sending $x_i \rightarrow r_i$. This is a surjective homomorphism, and thus $$R \cong \mathbb{Z}[x_i : i \in I]/\ker \varphi.$$

In particular, any non-Notherian integral domain is a quotient of an infinite polynomial ring. So in some sense, your example is a fundamental one.