# Comparison between etale and singular cohomology for a singular variety

In his Lectures on Etale Cohomology Milne proves in Theorem 21.1, that for all $r\geq 0$ $$H^r_{\acute{e}t}(X,\Lambda)\cong H^r(X_{cx},\Lambda)$$ with $X$ a nonsingular $\mathbb{C}$-variety and $\Lambda$ a finite abelian group. The right-hand-side is singular cohomology and $X_{cx}$ are the $\mathbb{C}$-points of $X$ with the complex topology.

Here a ''$\mathbb{C}$-variety'' is a sepatated and reduced scheme of finite type over $\operatorname{Spec}(\mathbb{C})$ and ''nonsingular'' means ''smooth over $\operatorname{Spec}(\mathbb{C})$''.

As suggested in Remark 21.2,the conditions of this theorem can be relaxed in the way

• If $X$ is connected, $\Lambda$ can be a locally constant sheaf with finite stalks, or even a little less (Milne,Theorem 21.5 and SGA IV,Theoreme 4.4),

and apparently also in the way

• $X$ can be singular.

I have two questions:

1. What is a reference for this second relaxation to a singular $X$?

2. This theorem can probably not be generalized to the case $\Lambda=\mathbb{Z}$. What is an example for $H^r_{\acute{e}t}(X,\mathbb{Z})\not\cong H^r(X_{cx},\mathbb{Z})$? Edit: Is there an affine example?

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In section 6 here, it is asserted that $H^2_{\textrm{ét}}(\mathbb{P}^1_{\mathbb{C}}, \mathbb{Z}) = 0$. Of course, we know that $H^2(\mathbb{P}^1 (\mathbb{C}), \mathbb{Z}) \cong \mathbb{Z}$. –  Zhen Lin Mar 19 '13 at 12:48
@ZhenLin: Yes, this answers 2., although I haven't understood how to calculate the $q$-th etale cohomology of $\mathbb{A}^1\setminus\{0\}_{\mathbb{C}}$ yet. According to the reference you gave, this is nonzero for $q=0,1$ only. This is like $S^1$ in topology! Does somebody know an affine example where $H^r_{\acute{e}t}(X,\mathbb{Z})\not\cong H^r(X_{cx},\mathbb{Z})$? –  Ronald Bernard Mar 19 '13 at 13:13

If $X$ is a smooth connected affine curve whose completion has genus $> 0$, then $H^1(X_{cx}, \mathbb Z)$ will be non-zero. (It will have rank $\geq 2g$.) On the other hand $H^1_{et}(X,\mathbb Z)$ will vanish. (This latter claim is not immediately obvious, but true. As you would have learned from the links in your MO crosspost, it's not true in certain singular cases, e.g. if $X$ is a nodal curve.)
Basically the point is that in algebraic geometry we can see connected components, and finite covering spaces, but not irreducible, infinite degree, covering spaces. If you look at my posts in the SBS thread I linked to on your MO post, you will see that the non-vanishing $H^1(X,\mathbb Z)$ for a nodal curve is related to the fact that the resulting copy of $\mathbb Z$ is not really coming from an infinite degree irreducible cover, but is rather counting the components of a reducible cover (a chain of lines covering the nodal curve).
Dear Matt E, thank you for your answer. I will try to understand your post in the SBS thread. This is perhaps worth a new question but since ''etale $=$ formally etale $+$ of finite presentation'', the problem for integer coefficients seems to be the property ''of finite presentation'', why doesn't one consider a Grothendieck topology constituted by formally-etale instead of etale morphisms? Why is ''etale'' the ''right'' thing? I apologize for the many parentheses. –  Ronald Bernard Mar 23 '13 at 23:28
@RonaldBernard: Dear Ronald, The problem is more than just the issue of finite presentation. Think about why $H^1(\mathbb C^{\times}, \mathbb Z)$ is free of rank one: it is because $\mathbb C^{\times}$ has a cover which is a $\mathbb Z$-torsor: namely $\mathbb C \to \mathbb C^{\times}$ by $z \mapsto e^z$. This cover is not an algebraic map, and you can't make it one by removing finite presentation hypotheses. Regards, –  Matt E Mar 24 '13 at 2:39