Is localization of a henselian ring still henselian?

If $(A,m,k)$ is a henselian local ring and $P$ is a prime ideal of $A$, does it follow that also $A_P$ is henselian?

Solution 1:

No.

Let $k$ be a field of characteristic not $2$. Let $A = k [[ x,y ]]$, the ring of formal power series in $x$ and $y$. This is a complete local ring, with maximal ideal $\langle x,y \rangle$, and hence henselian.

Let $P \subset A$ be the ideal $\langle x \rangle$. So elements of $A_P$ is ratios of formal power series $f(x,y)/g(x,y)$ where $g$ is not divisible by $x$. The equation $u^2 = 1-x/y$ obviously has two distinct roots, $\pm 1$, modulo $P_P$. I claim that it does not have roots in $A_P$; in fact, it doesn't even have roots in $\mathrm{Frac}(A_P) = \mathrm{Frac}(A)$. Suppose to the contrary that $(f(x,y)/g(x,y))^2 = 1-x/y$. So $y f(x,y)^2 = g(x,y)^2 (y-x)$ for $f$ and $g$ in $A$. But $A$ is a UFD where $y$ and $y-x$ are relatively prime irreducibles, so $y$ divides $y f^2$ an odd number of times and $(y-x) g^2$ an even number of times, contradiction.