Is every simple extension of a local domain inside its fractions again local?

Question

Let $A$ be a local domain with fractional field $K$. Let $x\in K\setminus A$. My question is that, is the ring $A[x]$ also local?

(It is easy to see that the anwser is yes, if $A$ is further a valuation ring.)

Answer 1

The answer is no.

Let $A$ be the localization of $k[x, y]$ at the maximal ideal $(x, y)$, so that $K = \operatorname{Frac}(A) = k(x, y)$.

Let $t$ be the element $\frac 1 y \in K$.

In the ring $B = A[t]$, we have elements $\frac x y$ and $\frac{y - x}y$ which sum up to $1$. However neither of them is a unit in $B$, as shown below.

Every element in $B$ can be written as $\frac f{y^d}$ for some $f \in A$ and some $d \geq 0$, which again can be written as $\frac h{gy^d}$ for some $g, h \in k[x, y]$ with $g(0, 0) \neq 0$.

Suppose that $\frac y x = \frac h{gy^d}$. This leads to $xh = gy^{d + 1}$. As $k[x, y]$ is a UFD and $x$ is an irreducible polynomial, it follows that $x$ divides $g$, which is impossible because $g(0, 0) \neq 0$.

In the same way, $\frac y{y - x} = \frac h{gy^d}$ implies that $y - x$ divides $g$, which is impossible because $g(0, 0) \neq 0$.