The Stacks project

Lemma 15.11.4. Let $(A, I)$ be a pair. If $A$ is $I$-adically complete, then the pair is henselian.

Proof. By Algebra, Lemma 10.96.6 the ideal $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma 15.11.2 we can successively lift this factorization to $f \bmod I^ n = g_ n h_ n$ with $g_ n, h_ n$ monic in $A/I^ n[T]$ for all $n \geq 1$. As $A = \mathop{\mathrm{lim}}\nolimits A/I^ n$ this finishes the proof. $\square$

Comments (9)

Comment #5370 by Kestutis Cesnavicius on

It suffices to assume that be derived -adically complete, it would be nice to include this stronger version as another lemma (one may need to rearrange the order of the sections to be able to use some results from For the proof, reduces to the case when is principal, and then one concludes by noting that is a square-zero extension of its -adic completion.

Comment #5371 by on

Yes, this is nice and answers my question in

Comment #5607 by on

Thanks again! I have added this result in the section on derived completion by 1 element. See this commit.

Comment #5764 by Kestutis Cesnavicius on

Thanks. In that commit "Let be an ideal derived complete with respect to an ideal ." should be "Let be a ring derived complete with respect to an ideal ."

Comment #5767 by Kestutis Cesnavicius on

In that commit, one could also insert an additional corollary: if a ring is reduced and derived -adically complete for a finitely generated ideal , then is actually -adically complete (use

Comment #5771 by on

@#5767: This is immediate from Proposition 15.91.5 and Lemma 15.93.11. So I guess I will leave it unstated for now, unless you wanted a direct reference for it.

Comment #5775 by Kestutis Cesnavicius on

I agree with what you say, but a direct reference would nevertheless be nice.

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  • 5 comment(s) on Section 15.11: Henselian pairs

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