The subring $k[x,y,x\mathrm{/}y,x\mathrm{/}{y}^2,\dots ]$ of $k(x,y)$ has a strictly ascending chain of principal ideals

A field is an integral domain: it is a commutative ring with $1\mathrm{\ne }0$, and if $ab=0$ with $a\mathrm{\ne }0$ then $b=a^{-1}ab=0$. Hence $k[x,y]$ is an integral domain, its field of fractions $k(x,y)$ is a field, and $k[x,y]$ embeds in it.

Let $R$ be the intersection of all subrings of $k(x,y)$ containing $k$, $x$, $y$ and every $x\mathrm{/}{y}^i$ with $i\ge 1$; this family is nonempty because $k(x,y)$ itself belongs to it, and an intersection of subrings is a subring, so $R$ is the smallest such subring.

$R$ is the set of $k$-linear combinations of the elements $x^c{y}^d$ with $c\ge 1$ and $d\in \mathbb{Z}$, together with the elements $y^d$ with $d\ge 0$. Call that set $S$. It is a subring: it contains $1=y^0$, is visibly closed under addition and additive inverses, and is closed under multiplication because $(x^c{y}^d)(x^{c^{\mathrm{\prime }}}{y}^{d^{\mathrm{\prime }}})=x^{c+c^{\mathrm{\prime }}}{y}^{d+d^{\mathrm{\prime }}}$, where $c+c^{\mathrm{\prime }}\ge 1$ as soon as one of $c,c^{\mathrm{\prime }}$ is, and where $d+d^{\mathrm{\prime }}\ge 0$ when $c=c^{\mathrm{\prime }}=0$. It contains the listed generators, since $x=x^1{y}^0$, $y=y^1$ and $x\mathrm{/}{y}^i=x^1{y}^{-i}$. Conversely every element of $S$ lies in any subring containing the generators, because $x^c{y}^d$ with $c\ge 1$ is $x^{c-1}$ times $x{y}^d$, and $x{y}^d$ is $x\cdot y^d$ when $d\ge 0$ and is the generator $x\mathrm{/}{y}^{-d}$ when $d<0$. So $S=R$.

The elements $x^c{y}^d$ of $k(x,y)$, for $c\ge 0$ and $d\in \mathbb{Z}$, are $k$-linearly independent, and consequently $1\mathrm{/}y\mathrm{\notin }R$. For the independence, a finite relation $\sum {\lambda }_{c,d}{x}^c{y}^d=0$ becomes, after multiplication by $y^M$ for $M$ large enough that every $d+M$ occurring is at least $0$, a relation $\sum {\lambda }_{c,d}{x}^c{y}^{d+M}=0$ among distinct monomials of $k[x,y]$; a polynomial is a finitely supported family of coefficients, so it vanishes exactly when every coefficient does, and the embedding of $k[x,y]$ is injective, so all ${\lambda }_{c,d}=0$. Now $1\mathrm{/}y=x^0{y}^{-1}$, whose index pair has $c=0$ and $d=-1<0$; by step 3.1 and the independence just proved, $1\mathrm{/}y$ is not among the $k$-linear combinations making up $R$.

For every $n\in \mathbb{N}$, $(x\mathrm{/}{y}^n)\subseteq (x\mathrm{/}{y}^{n+1})$ and the inclusion is strict. The inclusion holds because $y\in R$ and $x\mathrm{/}{y}^n=y\cdot (x\mathrm{/}{y}^{n+1})$, so the generator of the left ideal lies in the right one. If the two were equal then $x\mathrm{/}{y}^{n+1}=r\cdot x\mathrm{/}{y}^n$ for some $r\in R$; multiplying by $y^n\mathrm{/}x$ in the field $k(x,y)$, which is legitimate because $x\mathrm{\ne }0$ and $y\mathrm{\ne }0$ there, gives $r=1\mathrm{/}y$, contradicting step 4.1. At $n=0$ this reads $(x)⊊(x\mathrm{/}y)$.

The chain $(x)\subseteq (x\mathrm{/}y)\subseteq (x\mathrm{/}{y}^2)\subseteq \cdots$ is an ascending chain of ideals of $R$ indexed by $\mathbb{N}$ with every inclusion strict, so for no index $N$ is it constant from $N$ onwards. The ascending chain condition fails and $R$ is not Noetherian.