HINT:

$k[x,y]/(xy-1)$ is naturally isomorphic to the ring of fractions $k[x][\frac{1}{x}] = S^{-1}\ k[x]$, where $S= \{1,x,x^2, \ldots\}$.

I found on the Web and, also on mathstack, several examples and I decided to write an answer.

**Example of ring where the gcd of two elements doesn't exist:**

Consider the ring $\Bbb Z[\sqrt{-d}]=\{a+bi\sqrt{-d} : a,b\in \Bbb Z\}$, $d\ge 3$ ($d$ free-square). Then in this paper D. Khurana has proved that:

*Let $a$ be any rational integer such that $a\equiv d\quad (mod\quad 2)$ and let $a^2 + d = 2q$. Then the elements $2q$ and $(a+ i\sqrt{d})q$ do not have a $GCD$.*

**Two examples of ring where the lcm of two elements doesnâ€™t exist:**

I use the following *theorem*:
Let $D$ be a domain and $a,b\in D$.
Then, $\text{lcm}(a,b)$ exists if and only if for all $r\in D\setminus\{0\}$, $\gcd(ra,rb)$ exists.

$1)$Consider the ring $\Bbb Z[\sqrt{-d}]=\{a+bi\sqrt{-d} : a,b\in \Bbb Z\}$, $d\ge 3$ ($d$ free-square); and a rational integer $a$ such that $a\equiv d$ $(mod\quad 2)$, then $lcm(2,a+i\sqrt{d})$ doesn't exist. Indeed this follows from the previous theorem. Note that $GCD(2,a+i\sqrt{d})=1$.

$2)$ let $R$ be the subring of $\Bbb Z[x]$ consisting of the polynomials $\sum_i c_ix^i$ such that $c1$ is an even number. If we consider $p_1(x)=2$ and $p_2(x)=2x$, $lcm(p_1, p_2)$ doesn't exist.

## Best Answer

Sure. The trivial example is letting $I = (0)$ and taking the quotient map. But there are also plenty of examples with $I \neq (0)$.

The first example to come to me would be the following: Let $A = k[x_1, x_2, \cdots]$ be the polynomial ring with infinitely many variables and let $I = (x_1)$, so $A/I \cong k[x_2, x_3, \cdots]$. The map $$k[x_1, x_2, \cdots] \to k[x_2, x_3, \cdots], \ x_i \mapsto x_{i+1}$$ is an isomorphism (and in particular injective).