The Stacks project

Lemma 15.11.12. The property of being Henselian is preserved under limits of pairs. More precisely, let $J$ be a preordered set and let $(A_ j, I_ j)$ be an inverse system of henselian pairs over $J$. Then $A = \mathop{\mathrm{lim}}\nolimits A_ j$ equipped with the ideal $I = \mathop{\mathrm{lim}}\nolimits I_ j$ is a henselian pair $(A, I)$.

Proof. By Categories, Lemma 4.14.11, we only need to consider products and equalizers. For products, the claim follows from Lemma 15.11.11. Thus, consider an equalizer diagram

\[ \xymatrix{ (A, I) \ar[r] & (A', I') \ar@<1ex>[r]^{\varphi } \ar@<-1ex>[r]_{\psi } & (A'', I'') } \]

in which the pairs $(A', I')$ and $(A'', I'')$ are henselian. To check that the pair $(A, I)$ is also henselian, we will use the Gabber's criterion in Lemma 15.11.6. Every element of $1 + I$ is a unit in $A$ because, due to the uniqueness of the inverses of units, this may be checked in $(A', I')$. Thus $I$ is contained in the Jacobson radical of $A$, see Algebra, Lemma 10.19.1. Thus, let

\[ f(T) = T^{N - 1}(T - 1) + a_{N - 1} T^{N - 1} + \dotsb + a_1 T + a_0 \]

be a polynomial in $A[T]$ with $a_{N - 1}, \dotsc , a_0 \in I$ and $N \ge 1$. The image of $f(T)$ in $A'[T]$ has a unique root $\alpha ' \in 1 + I'$ and likewise for the further image in $A''[T]$. Thus, due to the uniqueness, $\varphi (\alpha ') = \psi (\alpha ')$, to the effect that $\alpha '$ defines a root of $f(T)$ in $1 + I$, as desired. $\square$

Comments (2)

Comment #3637 by Brian Conrad on

In the proof, replace "radical" with "Jacobson radical".

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

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