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First I would like to be clear about the definition, which I am having trouble finding.

What does: The local ring $A$ is algebraically closed in $B\supset A$. (e.g. for $B:=\hat{A}$, the completion of $A$ at its maximal ideal).

1- Does it mean that every element of $B$ satisfying a polynomial with coefficients in $A$, belongs to $A$?

2- Does it mean the same as above but assuming $b$ is an element of some commutative finite extension of $B$?

Afterwards I am trying to understand why if $A$ is local, noetherian, Henselian, and $\hat{A}$ is normal, characteristic is zero, then $A$ is algebraically closed in $\hat{A}$.

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  • $\begingroup$ Crossposted from math.SE: math.stackexchange.com/q/425539/264 $\endgroup$
    – Zev Chonoles
    Commented Jun 20, 2013 at 20:55
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    $\begingroup$ I suppose $A$ is noetherian. If $\widehat{A}$ is normal then it is a domain, so $A$ is a domain. Normality descends through a flat local map of local noetherian rings (Cor. to Theorem 23.9 in Matsumura "Commutative ring theory"), so $A$ is normal. Hence, we can think inside the fraction fields of $A$ and $\widehat{A}$, so we could take the condition to mean the fraction field of $A$ is algebraically closed in that of $\widehat{A}$: it is the same as your suggestion #1 (demanding the polynomial be nonzero!). If $A$ is excellent then we can apply Artin approximation, without assuming char. 0. $\endgroup$
    – user29720
    Commented Jun 20, 2013 at 23:35
  • $\begingroup$ I think I am having problems understanding two basic things. 1) How Artin's approximation becomes an 'exact' solution. 2) How Solutions of monic polynomials (what is involved in the notions of integral, and norma) gives solutions of arbitrary polynomials (algebraically closed). Maybe 1 comes from convergence, but it seems it is being needed in $A$ instead of the (complete) $\hat{A}$. Maybe 2 is done in a field, so that you can get rid of the leading coefficient. I will 'do my homework'. Let me think it by myself to see if I get it. Thanks for your help. $\endgroup$
    – O.R.
    Commented Jun 21, 2013 at 11:56
  • $\begingroup$ No, was not able to understand it. Let $P(x)=a_nx^n+...+a_0\in A[x]$, with solution $s\in \hat{A}$. If I need a monic polynomial I could work with $a_ns\in \hat{A}$ which is a solution of $q(x)=x^n+a_na_{n-1}x^{n-1}+...+a_{n}^{n-1}a_0\in A[x]$ Now I am completely blocked. How do I get that $s$ is in $A$? $\endgroup$
    – O.R.
    Commented Jun 22, 2013 at 22:56
  • $\begingroup$ Never mind. I can't believe I have been misreading Artin's approximation. The theorem says that there is a solution, and the approximation part is about that solution being close to a formal one, not that the solution is approximate in the sense of f(a) being close to 0. I should really read things carefully. $\endgroup$
    – O.R.
    Commented Jun 23, 2013 at 0:30

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