question on a concrete example of etale cohomology groups

Question

I don't understand something basic about the following answer to this question on mathoverflow:

Angelo Vistoli writes: Let $X = \mathop{\rm Spec}R$, where $R := k[x,y]/(y^2 - x^3 + x^2)$. Consider the finite morphism $\pi\colon \mathbb A^1 \to X$, which yields an exact sequence $$ 0 \to \mathbb Z_X \to \pi_*\mathbb Z_{\mathbb A^1} \to i_*\mathbb Z_x \to 0, $$ where $x$ is the singular point of $X$. Since both $\pi_*\mathbb Z_{\mathbb A^1}$ and $i_*\mathbb Z_x$ have trivial étale cohomology , by taking global sections we see that $\mathrm H^1(X, \mathbb Z) = \mathbb Z$.

Okay, my basic question is why $\pi_*\mathbb Z_{\mathbb A^1}$ and $i_*\mathbb Z_x$ have trivial étale cohomology? Also, even a more basic question just to check my understanding - $\pi_*\mathbb Z_{\mathbb A^1}$ and $i_*\mathbb Z_x$ are not coherent sheaves, right? $H^1_{et}(X, \pi_*\mathbb{Z}_{\mathbb{A}^1}) = R^1\pi_* \mathbb{Z}_{\mathbb A^1}$ ... not sure if that's true or useful.

Answer 1

First, your even more basic question. Indeed $\pi_*\mathbb{Z}_{\mathbb{A}^1}$ and $i_*\mathbb{Z}_x$ are not coherent sheaves. In fact, they are not even sheaves of $\mathcal{O}_X$ modules. And as Remy points out, $H^1_{ét}(X,\pi_*\mathbb{Z}_{\mathbb{A}^1})$ and $R^1\pi_*\mathbb{Z}_{\mathbb{A}^1}$ are not the same kind of objects, the first being a group, the latter being a sheaf on $X$.

For your second question, Angelo uses the well known fact that finite morphism have vanishing direct images. In general, I see this as a consequence of the proper base change theorem, but in fact there is a direct proof which work with every sheaves of sets, see here : http://stacks.math.columbia.edu/tag/03QN.

It follows that $H^i_{ét}(X,\pi_*\mathbb{Z}_{\mathbb{A}^1})=\mathbb{H}^i_{ét}(X,R\pi_*\mathbb{Z}_{\mathbb{A}^1})=H^i(\mathbb{A}^1,\mathbb{Z}_{\mathbb{A}^1})$. And similarly, you have $H^i_{ét}(X,i_*\mathbb{Z}_x)=\mathbb{H}^i_{ét}(X,Ri_*\mathbb{Z}_x)=H^i(x,\mathbb{Z}_x)$. So it remains to show that the two groups $H^1(\mathbb{A}^1,\mathbb{Z}_{\mathbb{A}^1})$ and $H^1(x,\mathbb{Z}_x)$ are zero. This follows from the canonical isomorphism $H^1(Y,A)=\operatorname{Hom}_{cont}(\pi_1(Y),A)$ and the fact that $\pi_1(Y)$ being profinite for $Y$ normal noetherian, there is no continuous group homomorphism $\pi_1(Y)\rightarrow\mathbb{Z}$.

Answer 2

Lemma. Let $f \colon X \to Y$ be a finite morphism, and let $\mathcal F$ be any sheaf on $X_{\operatorname{ét}}$. Then $R^if_* \mathcal F = 0$ for all $i > 0$.

Proof. It suffices to show that $(R^if_* \mathcal F)_{\bar y} = 0$ for all geometric points $\bar y \to Y$. But the stalk is computed by $H^i_{\operatorname{ét}}(X', \mathcal \pi^* F)$, where $X' = X \times_Y \operatorname{Spec} \mathcal O_{Y, \bar y}^{\operatorname{sh}}$, and $$\pi \colon X \times_Y \operatorname{Spec} \mathcal O_{Y, \bar y}^{\operatorname{sh}} \to X$$ is the first projection. But $X' \to \operatorname{Spec} \mathcal O_{Y, \bar y}^{\operatorname{sh}}$ is a finite morphism, and $\mathcal O_{Y, \bar y}^{\operatorname{sh}}$ is strictly Henselian. Thus, $X'$ is a finite union of spectra of strictly Henselian local rings (use the equivalent conditions of $\mathcal O_{Y, \bar y}^{\operatorname{sh}}$ being Henselian). Thus, $$H^i_{\operatorname{ét}}(X', \mathcal G) = 0$$ for any sheaf $\mathcal G$ and any $i > 0$, since the étale site of a strictly Henselian local ring has a cofinal system of coverings given by disjoint unions. $\square$

For more details, see Tag 03QP.

Corollary. Let $f \colon X \to Y$ be finite, and let $\mathcal F$ be a sheaf on $X_{\operatorname{ét}}$. Then $$H^i_{\operatorname{ét}}(X, \mathcal F) = H^i_{\operatorname{ét}}(Y, f_*\mathcal F)$$ for all $i$.

Proof. Immediate from the Leray spectral sequence. $\square$

Thus, it suffices to show that $\mathbb Z$ has no first cohomology on a point and on $\mathbb A^1$. For a point, this is trivial (let's assume that $k$ is algebraically closed). We will now consider $\mathbb A^1$.

Remark. Recall that for any field $K$, we have $H^1_{\operatorname{Gal}}(K,\mathbb Z) = 0$. Indeed, for all $i > 0$ the group $H^i_{\operatorname{Gal}}(K,\mathbb Q)$ is zero, since it is torsion (since $\operatorname{Gal}(K^{\operatorname{sep}}/K)$ is profinite) and divisible (since $\mathbb Q$ is divisible). Now use the short exact sequence $$0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.$$

Lemma. Let $X$ be any scheme, and let $j \colon x \to X$ be a point. We have $H^1_{\operatorname{ét}}(X, j_* \mathbb Z) = 0$.

Proof. We have a Leray spectral sequence $$E_2^{pq} = H^p_{\operatorname{ét}}(X, R^qj_* \mathbb Z) \Longrightarrow H^{p+q}_{\operatorname{ét}}(x, \mathbb Z),$$ whose low-degree sequence reads $$0 \to H^1_{\operatorname{ét}}(X, j_* \mathbb Z) \to H^1_{\operatorname{ét}}(x, \mathbb Z) \to H^0_{\operatorname{ét}}(X, R^1j_* \mathbb Z) \to \ldots.$$ By the remark above, $H^1_{\operatorname{ét}}(x, \mathbb Z) = 0$. $\square$

Now let $j \colon \eta \to \mathbb A^1$ be the generic point.

Lemma. The canonical morphism $\mathbb Z \to j_* \mathbb Z$ on $(\mathbb A^1)_{\operatorname{ét}}$ (coming from the adjunction $j^* \dashv j_*$ and the isomorphism $j^* \mathbb Z \cong \mathbb Z$) is an isomorphism.

Proof. It suffices to prove this after taking the stalk at any geometric point $\bar x \to \mathbb A^1$ corresponding to a closed point $x$. Now $\mathcal O_{\mathbb A^1, \bar x}^{\operatorname{sh}}$ is a DVR, and $K_{\bar x}^{\operatorname{sh}} := k(t) \otimes_{k[t]} \mathcal O_{\mathbb A^1, \bar x}^{\operatorname{sh}}$ is its field of fractions (since we are inverting a uniformiser). Thus, if $$j' \colon \operatorname{Spec} K_{\bar x}^{\operatorname{sh}} \to \operatorname{Spec} \mathcal O_{\mathbb A^1, \bar x}^{\operatorname{sh}}$$ denotes the base change of $j$ along $\operatorname{Spec} \mathcal O_{\mathbb A^1, \bar x}^{\operatorname{sh}} \to \mathbb A^1$, we have to show that $$\mathbb Z \to j'_*\mathbb Z$$ is an isomorphism on $(\operatorname{Spec} \mathcal O_{\mathbb A^1, \bar x}^{\operatorname{sh}})_{\operatorname{ét}}$. But global sections give an equivalence $\underline{\operatorname{Sh}}(X_{\operatorname{ét}}) \to \underline{\operatorname{Ab}}$ for $X$ the spectrum of a strictly Henselian local ring. Thus, we are done by the simple observation that for a DVR $R$ with fraction field $K$, the map $$\mathbb Z \to j'_* \mathbb Z$$ induces an isomorphism on global sections, since both $\operatorname{Spec} R$ and $\operatorname{Spec} K$ are connected. $\square$

Corollary. We have $H^1_{\operatorname{ét}}(\mathbb A^1, \mathbb Z) = 0$.

Proof. Combine the two lemmata. $\square$

Remark. In general, it turns out that for an integral scheme $X$ with generic point $j \colon \eta \to X$, the map $$\mathbb Z \to j_* \mathbb Z$$ is an isomorphism if and only if $X$ is geometrically unibranched (see SGA 4, Theorem IX.3.6(ii)). Every smooth variety over any field satisfies this condition, and more generally every normal scheme does. Thus, we get $H^1_{\operatorname{ét}}(X, \mathbb Z) = 0$ for any such scheme $X$.