This might not be a question appropriate for this forum, I apologize in this case... Is it true that any element of the completion of a valued ring $R$ that is algebraic over the field of fractions of $R$ is in the henselization of $R$ ?

$\begingroup$ @Sell: What do you mean by the Henselization of a valued <I>field</I> $R$? Do you mean to ask whether every element in the completion of some valuation <I>ring</I> $R$ that is algebraic over $R$ is in the Henselization of $R$? $\endgroup$ – Jason Starr Aug 24 '12 at 12:26

$\begingroup$ Yes, that's exactly what I meant. Thank you for having reformulated it in a better way. Is it possible for me to edit my question so I can reformulate it ? $\endgroup$ – seli Aug 24 '12 at 12:59

$\begingroup$ This is true if $R$ is a discrete valuation ring (I don't know whether it is true in general). See Ex. 8.3.34 (p360 first ed.) of Liu's book "Algebraic geometry and arithmetic curves". In fact in that case, the henselization is actually equal to the set of elements of $\widehat{R}$, which are algebraic over $R$. $\endgroup$ – Damian Rössler Aug 24 '12 at 14:22

9$\begingroup$ It is false: if $R$ is a nonexcellent dvr (so of equichar. $p > 0$) there's $a \in R$ not a $p$th power that is a $p$th power in $\widehat{R}$, so $a^{1/p} \in \widehat{R}  R^h$. By ArtinPopescu approximation, for any excellent normal local noetherian ring, the "algebraic closure" in the completion is the henselization. (The henselization is also excellent.) This includes any excellent dvr. The omission of excellence is an error in Liu's book, at least in the 1st edition; I don't know if this was corrected in later editions. $\endgroup$ – user22479 Aug 24 '12 at 17:43

$\begingroup$ @quasicoherent: you are quite right, one has to suppose that $R$ is excellent. Otherwise, the algebraic closure in the completion might not be separable (in fact this is what you point out). Sorry for that. $\endgroup$ – Damian Rössler Aug 24 '12 at 17:51
Edit Add noetherian hypothesis because $R$ might no be a subring of $\hat{R}$ otherwise.
The answer is no as pointed out by quasicoherent in the comments. But suppose $R$ is a discrete valuation ring and denote $K=\mathrm{Frac}(R)$, then
$R^h=\hat{R}\cap K^{sep}$.
There are several equivalent properties defining henselien local rings. I will use two of them: let $(A, m)$ be a local ring and $k=A/m$. Then $A$ is henselian iff
(a) for any monic polynomial $f(X)\in A[X]$, any simple root of $\bar{f}(X)\in k[X]$ lifts to root of $f(X)$ in $A$.
or equivalently
(b) If $A\to A'$ is a local homomorphism, étale and with trivial residue extension, then $A'=A$.
One can find a proof in Raynaud: Anneaux locaux henséliens, IV, §3. For discrete valuation rings, one can certainly find easier references.
Now let us prove the claim above $R^h$. As $\hat{R}$ is henselian (Hensel Lemma), we have $R\subseteq \hat{R}$.
(1) For any algebraic separable extension $F/K$ contained in $\hat{K}$, $B_F=\hat{R}\cap F$ is a DVR with ramification index $1$ over $R$ and trivial residue extension. This is easy.
(2) Let $B=B_L$ where $L$ is the separable closure of $K$ in $\hat{K}$. Let us prove that $B$ is henselian using Property (a) above. Let $f(X)\in B[X]$ as in (a). Then any simple root of $\bar{f}(X)\in k[X]$ lifts to a root $\lambda\in \hat{K}$. It is automatically a simple root, so $\lambda$ is separable over $L$, hence $\lambda\in L\cap \hat{R}=B$.
(3) Let $R\to R'$ be an extension to a henselian DVR. Let's us prove it factorizes throught $R\to B$, which will show that $B$ is a henselization of $R$. It is enough to prove this factorization for $B_F$ for any finite separable exension $F/K$ contained in $\hat{K}$. As $B_F$ is an étale extension of $R$, $B_F\otimes_R R'$ is étale over $R'$, with a unique maximal ideal above the maximal ideal of $R'$ and trivial residue extension at this maximal ideal (because the quotient $B_F\otimes_R R'/(m')\simeq R'/m'$). By Property (b), this implies that $F\otimes_K K'$ has a direct factor equal to $K'$. Hence $F\subseteq K'$ (the minimal polynomial of a primitive element of $F$ has a root in $K'$) and $B_F\subseteq R'$.
Remark Parts (1)(2) work for any local ring $R$ such that $\hat{R}$ is a domain and such that $R\to \hat{R}$ is injective. In particular $\hat{R}\cap K^{sep}$ is henselian. (3) works for any valuation ring But in general, I don't think (3) holds because $B_F$ has no reason to be étale over $R$.

1$\begingroup$ Warning: this is OK only if by "valuation" you mean "valuation of rank one", i.e. the value group is a subgroup of $\mathbb{R}$. There are valuation rings (of rank $>1$) which are complete but not henselian! $\endgroup$ – Laurent MoretBailly Aug 30 '12 at 6:23

$\begingroup$ Laurent, yes thanks ! I had in mind the $m$adic completion, but a nonnoetherian valuation ring is not separated in general and so doesn't embedd in its $m$adic completion (can be equal to its residue field). And you remark makes the situation even worse. I will edit the answer. $\endgroup$ – Qing Liu Aug 30 '12 at 8:54

$\begingroup$ Sorry if I write so many years later but why is $B_F$ etale over $R$? $\endgroup$ – Tommaso Scognamiglio Sep 27 '19 at 11:01