8
$\begingroup$

Elkik in Solutions d'equations a coefficients dans un anneu Henselian, Theorem 7 proves that:

Let $A$ be a Noetherian ring that is Henselian with respect to a principal ideal $(a)$.

That is, if $f(x)$ is a monic polynomial with a root $\alpha \in A/(a)$ such that $df/dx (\alpha)$ is a unit in $A/(a)$, then $\alpha$ lifts to a root of $f$ in $A$.

Let $\hat{A}$ be its $a$-adic completion, and let $\hat{B}$ be a formally finitely generated algebra over $\hat{a}$ that is formally smooth over $\hat{A}[a^{-1}]$. Then there exists a finitely generated algebra $B$ over $A$ that is smooth over $A[a^{-1}]$ and such that its $a$-adic completion is isomorphic to $\hat{B}$.

Is this theorem true in the non-Noetherian case? If not, what is a counterexample?

I would ideally prefer counterexamples of relative dimension $0$ such as finite etale covers.

$\endgroup$
  • $\begingroup$ $B$ is only smooth over $A[a^{-1}]$. $\endgroup$ – Laurent Moret-Bailly Aug 28 '14 at 6:59
3
$\begingroup$

If I understand the notation of Gabber and Romero corectly, the theorem is true in the non-Noetherian case for finite etale covers. Finite etale torsors on $X$ of degree $n$ are classified by elements of $H^1(X,S_n)$.

Thus we apply Theorem 5.8.14 of Almost Ring Theory by Gabber and Ramero.

To convert from my notation to theirs, take $R=A$, $t=a$, $I=(1)$, so that $R^ = \hat{A}$. This satisfies the conditions of Proposition 4.21.

Let $G$ be $S_n$. Then $G$ satisfies the conditions of 5.8.4 by explicit construction, or because of Lemma 5.8.5 and the fact that it is defined over the Dedekind domain $\mathbb Z$.

So we may apply the theorem and get an equality

$$H^1( \operatorname{Spec} A[a^{-1}], S_n) = H^1( \operatorname{Spec} \hat{A}[a^{-1}],S_n)$$

giving what I asked for (in the finite etale case).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.