9
$\begingroup$

Let $A$ be an henselian local noetherian ring. There is an old result of Lafon ("Anneaux henséliens et théorème de préparation" (1967)), which says that if $A$ is analytically normal and of characteristic zero, then it verifies Weierstrass preparation theorem.

Is there any progress on this? Is the analytic normality essential and what happens in characteristic p?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ A lovely question. Are there no examples known of Henselian non-Weierstrassable rings? $\endgroup$
    – Lubin
    Commented Feb 23, 2015 at 14:06
  • $\begingroup$ What is the statement of the Weierstrass preparation here? That any $f \in R[[x]]$, nonzero mod $m$, is equal to a unit of $R[[x]}$? times a monic polynomial in $R[x}$? It seems to me that, if this holds in a local ring $R$, then it holds in any quotient of $R$ with the same maximal ideal. So it could be possible to prove this in Noetherian non-analytically normal rings by viewing them as quotients of analytically normal rings. $\endgroup$
    – Will Sawin
    Commented Feb 27, 2017 at 22:28
  • $\begingroup$ I cannot find the paper you are referring to. Could you please explain briefly how that works? For example, when I have a Henselian pair $(A,I)$, how can I do a Weierstrass division of some formal power series $f\in R[[x]]$ by the simplest "Weierstrass polynomial" $x-a$ where $a\in I$? Seemingly this will force me to "evaluate" $f(a)$ in $R$, which seems impossible without any kind of completeness of $R$? $\endgroup$
    – user20948
    Commented Feb 7, 2021 at 11:25

0

Reset to default

You must log in to answer this question.