Henselian ring

Remarks

An equivalent condition: for every monic $f \in R[x]$ and every $a_0 \in k$ with $\bar f(a_0) = 0$ and $\bar f'(a_0) \ne 0$, there is $a \in R$ with $f(a) = 0$ and $a \equiv a_0 \pmod{\mathfrak m}$. That is, simple roots modulo $\mathfrak m$ lift to roots.

Example: every complete Noetherian local ring is Henselian, by the general form of Hensel's lemma; this includes the $p$-adic integers $\Z_p$ and $k[\![t]\!]$ for a field $k$. Every field is Henselian, since $\mathfrak m = 0$ and $\bar f = f$.

Non-example: $R = \Z_{(5)}$ is not Henselian. For $f = x^2 + 1$, one has $\bar f = (x - 2)(x - 3)$ in $\Z/5\Z[x]$, and the factors $x - 2$ and $x - 3$ are coprime because $2 \ne 3$. A lift would give a root $a \in \Z_{(5)} \subseteq \Q$ of $x^2 + 1$, but $-1$ has no rational square root.