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Let $R$ be a henselian ring. Is $R[[x]]$ also a henselian ring?

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Yes, it is. More generally, the following result holds.

Proposition. If $R$ is henselian at the maximal ideal $\mathfrak{m}$, then $R[[x_1, \ldots, x_n]]$ is henselian at the maximal ideal lying over $\mathfrak{m}$.

A reference is the paper by N. Sankharan A Theorem on Henselian Rings, Canad. Math. Bull. 11 (1968), 275-277. See in particular Corollary 2.

Remark. The Proposition above is no longer valid if one takes the polynomial ring instead of the power series ring. For instance, if $K$ is a field than $K$ is henselian but $K[x]$ is not.

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    $\begingroup$ More generally: if $A$ is a local ring and $I$ an ideal such that $A$ is $I$-adically complete and separated, then $A$ is henselian if (and only if!) $A/I$ is henselian. This is easy by observing that each $A_n:=A/I^n$ is henselian (because $A_{n,\mathrm{red}}$ is) and then inductively lifting roots to $A$. See Raynaud, LNM 169, I, §2 where this result is given as an exercise. $\endgroup$ – Laurent Moret-Bailly Feb 18 '13 at 14:46

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