Let $\pi: R\to S$ be a local morphism of Henselian local rings. Let $f: R \to \hat{R}$ and $g: S \to \hat{S}$ be their completions. Let $\mathcal F$ be a constructible $l$-adic sheaf on $\operatorname {Spec} S$ for $l$ invertible in $R$. Is the natural base change map $f^*R^i\pi_* \mathcal F \to R^i \hat{\pi}_* g^* \mathcal F$ an isomorphism?

Motivation: For most purposes, Henselian local rings are as good as complete ones, and completions are as good as Henselizations. But are they cohomologically the same? $H^i$ is the same because global sections are just the stalk at the closed point in both cases, but the behavior of pushforwards and higher pushforwards are not so clear.

EDIT: Thanks to user61789, I realize my assumptions are much too general! Thus I will ask about a much more specific case. Let $R$ be the etale local ring of $\mathbb A^n_k$ at the origin, let $S$ be the etale local ring of $\mathbb A^m_k$ at the origin for $m\geq n$, and let $\pi$ be induced by the natural projection $\mathbb A^m_k \to \mathbb A^n_k$. Then is the map an isomorphism?


1 Answer 1


You probably meant to assume $R$ and $S$ are noetherian. The answer is "no" to the initial hypergeneral part of the question. EDIT: In the 2nd half (below the long line), I now give a proof of an affirmative answer to the added part involving maps of affine spaces.

Counterexamples to the initial hypergeneral part can be made using 2-dimensional regular excellent local rings built from henselization and completion of local rings at $k$-points on smooth schemes over any field $k$.

Let $R$ be a noetherian henselian local ring, and $S = \widehat{R}$, so $\widehat{S}=S$. Let $\pi:{\rm{Spec}}(\widehat{R}) \rightarrow {\rm{Spec}}(R)$ be the natural map. Then in this case an affirmative answer to your question says exactly that $\pi_{\ast}$ is exact and $\pi^{\ast} \circ \pi_{\ast} \rightarrow {\rm{id}}$ is an isomorphism of functors (all on the category of constructible abelian etale sheaves on ${\rm{Spec}}(\widehat{R})$). In particular, every such sheaf on ${\rm{Spec}}(\widehat{R})$ would be the $\pi$-pullback of one on ${\rm{Spec}}(R)$. Clearly this cannot be true in general, so it is just a game to find a suitable counterexample to that.

Let $Z' \subset {\rm{Spec}}(\widehat{R})$ is a Zariski-closed set and $j':U' \hookrightarrow {\rm{Spec}}(\widehat{R})$ the open subscheme complementary to $Z'$. Consider the extension-by-zero $F' := j'_{!}(\mathbf{Z}/(n))$ for an integer $n > 1$. Suppose $F' = \pi^{\ast}(F)$ for an abelian etale sheaf $F$ on ${\rm{Spec}}(R)$.

Let $Z$ be the set of points $x \in {\rm{Spec}}(R)$ such that $F_x = 0$. This satisfies $Z' = \pi^{-1}(Z)$, so $Z$ is closed since $\pi$ is topologically a quotient map (as it is fpqc). Hence, if there is a closed set $Z'$ that is not the preimage of a closed set in ${\rm{Spec}}(R)$ then we have a counterexample.

Suppose $R$ is excellent, so the fpqc map $\pi$ is a "regular morphism" (flat and its fiber algebras that are regular and remain so after finite extension of the base field). Thus, if $Z$ is a reduced closed set in ${\rm{Spec}}(R)$ then the scheme-theoretic preimage $\pi^{-1}(Z)$ is reduced. Consider any radical ideal $J'$ in $\widehat{R}$ and let $Z' := {\rm{Spec}}(\widehat{R}/J')$. If $Z' = \pi^{-1}(Z)$ topologically for a closed set $Z$ in ${\rm{Spec}}(R)$ then by using the reduced scheme structure on $Z$ we would have $\pi^{-1}(Z) = Z'$ as schemes. In other words, for the radical ideal $J$ in $R$ corresponding to such a $Z$ we would necessarily have $J' = J\widehat{R}$.

Note that $J \otimes_R \widehat{R} \rightarrow J\widehat{R}$ is an isomorphism, so if $J'$ is invertible then necessarily $J$ is invertible. In particular, if $J' = r'\widehat{R}$ for $r'$ that is not a zero-divisor then necessarily $J = rR$ for some $r \in R$ that is not a zero-divisor. In such a situation, $r'$ would have to be a unit multiple of $r$ in $\widehat{R}$. Thus, if we can find $r' \in \widehat{R}$ that isn't a zero-divisor such that no unit multiple of $r'$ lies in the subring $R$ then we have our counterexample (using $Z' = {\rm{Spec}}(\widehat{R}/(r'))$ and the associated $F'$ as above).

In the special case that $R$ is regular, so $\widehat{R}$ is also regular, we're just looking for a nonzero $r' \in \widehat{R}$ having no unit multiple in $R$. Obviously in the dvr case this cannot be arranged, so we go on to dimension 2.

Take $R = k[x,y]_{(x,y)}^{\rm{h}}$ for a field $k$, so $\widehat{R} = k[\![x,y]\!]$. The ring $R$ is excellent, since passage to henselization preserves excellence (18.7.6, EGA IV$_4$). Since $k(\!(x)\!)$ is not algebraic over $k(x)$ (lazy way is to use $e^x$ and assume characteristic is 0), we can find $h \in x k[\![x]\!]$ that is not algebraic over $k(x)$. For such $h$, the irreducible element $r' = y - h(x)$ in $\widehat{R}$ will do the job.

Indeed, suppose that $r'$ admits a unit multiple $r \in R$. We will show that $h$ must be algebraic over $k(x)$. Clearly $R/(r) \rightarrow \widehat{R}/(r') = k[\![x]\!]$ is a completion map, so $R/(r)$ is a dvr. Since henselization is compatible with quotients, it follows by the link between transcendence degree and dimension for finitely generated domains over $k$ that $R/(r)$ is a direct limit of a directed system of local-etale extensions of local rings at $k$-points on regular curves over $k$, and these curves must all be quasi-finite over the affine $x$-line over $k$ (since $x \in k[\![x]\!]$ is transcendental over $k$). Thus, everything in $R/(r)$ is algebraic over $k(x)$, including the class of $y$. But mapping that into $k[\![x]\!]$ with $y \mapsto h(x)$ in there, we conclude that $h$ is algebraic over $k(x)$.

Now let's give an affirmative proof under some additional "smoothness" hypotheses as follows. The key point is to use Artin-Popescu approximation and to work at the level of derived categories. Also, one has to be extremely careful because certain schemes will intervene that (as far as I know) might fail to be noetherian or excellent, even though our original setup will involve only excellent schemes and smooth maps. (Maybe for some reasons which escape me, the non-noetherian concerns in the argument below cannot really happen?)

Setup: Let $\mathcal{X} \rightarrow \mathcal{Y}$ be a smooth map of noetherian schemes with $\mathcal{Y}$ excellent. Choose $y \in Y$ and $x \in X_y$ a $k(y)$-rational point. Let $X = {\rm{Spec}}(O_{\mathcal{X},x}^{\rm{h}})$, $Y = {\rm{Spec}}(O_{\mathcal{Y},y}^{\rm{h}})$. Let $X'$ and $Y'$ denote Spec of the corresponding completed local rings, and let $\pi:X \rightarrow Y$, and $\pi':X' \rightarrow Y'$ be the natural maps.

Claim: Let $F$ be a torsion etale abelian sheaf on $X$ whose torsion-orders are invertible on $Y$. Let $G \rightsquigarrow G'$ be shorthand for pullback on derived categories (of abelian etale sheaves) from $Y$ to $Y'$ or from $X$ to $X'$. Then the natural base change morphism $$R\pi_{\ast}(F)' \rightarrow R\pi'_{\ast}(F')$$ is an isomorphism for any torsion $F$ on $X$ whose torsion-orders are invertible on $Y$.

Proof: By Artin-Popescu approximation, since $\mathcal{Y}$ is excellent we know that the map $Y' \rightarrow Y$ is a limit of smooth maps. Hence, if we let $T = X \times_Y Y'$ and $p:T \rightarrow Y'$ be the projection then by the smooth base change theorem and standard limit stuff with etale sheaf theory, the natural map $R\pi_{\ast}(F)' \rightarrow Rp_{\ast}(G)$ is an isomorphism, where $G$ is the pullback of $F$ along the first projection.

I don't know if $T$ is noetherian, by the way, so I am tacitly using the good behavior of the limit formalism in etale sheaf theory without noetherian hypotheses (but even the proof of the smooth base change theorem in SGA4.5 leaves the noetherian framework due to such kind of fiber products intervening, so I suppose you don't mind).

The $y$-fiber of $p$ is identified with $X_y$, so naturally $x \in T$, and $O_{T,x}$ is a "partial henselization" of the local ring at a rational point on the special fiber of a finite type (even smooth) $Y'$-scheme. So even though I/we do not know if $T$ is noetherian, we do know that $O_{T,x}$ is noetherian, by the same reasoning which proves that passage to henselization preserves the noetherian property (namely, EGA 0$_{\rm{III}}$,

Note that the henselization of $O_{T,x}$ coincides with that of a local ring $R$ on a finite type (even smooth) $Y'$-scheme, and $Y'$ is excellent (as for any complete local noetherian ring: IV$_2$, 7.8.3(iii)), so $R$ is excellent. Consequently $R^{\rm{h}}$ is excellent (IV$_4$, 18.7.6), so $O_{T,x}$ has excellent henselization. Now it is not true that excellence descends from the henselization (counterexample in IV$_4$, 18.7.7), but this is for geometric reasons related to failure of being universally catenary, and we don't actually care about that.

Indeed, what matters for the following is geometric regularity of formal fibers of $O_{T,x}$, which is to say that the flat map of noetherian schemes ${\rm{Spec}}(O_{T,x}^{\wedge}) \rightarrow {\rm{Spec}}(O_{T,x})$ is a regular morphism (since Artin-Popescu approximation is about regular morphisms being a limit of smooth morphisms, which in practice is easiest to remember under the banner of "excellence" but is not strictly a necessary condition). So all we care is to know that geometric regularity of formal fibers descends from the henselization, and that is fine; see IV$_4$, 18.7.4.

The upshot is that $O_{T,x}$ is noetherian and its completion morphism is regular.

Now comes the crux of the matter: I claim that the map $h:X' \rightarrow T$ (or rather, its factorization through ${\rm{Spec}}(O_{T,x})$) is computing the completion of the noetherian local ring $O_{T,x}$. (This is the step at which our argument breaks down in the setting of the counterexamples above! Indeed, in that setting the map in the role of $h$ would be akin to a graph morphism, super-far from an isomorphism.)

In view of the local structure theorem for smooth morphisms (applied to a smooth $Y'$-scheme whose "partial henselization" at a suitable point computes $T$), our hypothesis that $k(x) = k(y)$, and the fact that completion of a local noetherian ring is insensitive to "partial henselization", justifying our assertion about $h$ amounts to the following down-to-earth observation: if $A$ is a henselian (e.g., complete) local noetherian ring, $B = A\{x_1,\dots,x_N\}$ is the henselization at the origin of the special fiber of an affine space over $A$, and $B'$ is the local ring at the origin on the special fiber of the ring $\widehat{A} \otimes_A B$ (a ring I/we do not know to be noetherian, but the local ring $B'$ certainly is, as explained above), then the natural map $B' \rightarrow \widehat{A}[\![x_1,\dots,x_N]\!]$ is the completion of $B'$. This verification is a simple exercise using the compatibility of henselization and quotients.

From our description of $h$ via completion of a local noetherian ring having geometrically regular formal fibers, $h$ is a regular morphism. Thus, by Popescu's theorem, $h$ is a limit of smooth maps. Hence, by the acyclicity theorem for smooth maps and the usual limit games (which do not require knowing $T$ to be noetherian), it follows that the natural map $G\rightarrow Rh_{\ast}(h^{\ast}G)$ is an isomorphism. But $h^{\ast}(G)=F'$, so we have a natural isomorphism $$R\pi_{\ast}(F)' \simeq Rp_{\ast}(G)=Rp_{\ast}Rh_{\ast}(F')=R(p \circ h)_{\ast}(F').$$ But $p\circ h=\pi'$, and as such it is a standard exercise to check that the composite isomorphism we have made is the base change morphism of initial interest. QED

  • $\begingroup$ Thanks, this was helpful! However, what I need is a special case where this sort of counterexample won't apply. I have edited the question to reflect this. Do you know the answer in that situation? $\endgroup$
    – Will Sawin
    Jun 29, 2013 at 18:31

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