I was reading Milne's book "Arithmetic Duality Theorems". On page 166 there are a lot of useful lemmas on the etale cohomology with compact support on S-integers. However, I get confused when I tried to look at Prop 2.3 (a) and (d) at once.

Suppose $X$ is the ring of integers of a totally complex number field $K$. $U\hookrightarrow X$ is an open subscheme, $S=X\setminus U=\{p\}$ has only one finite prime. Consider the sheaf $\mathbb{G}_m$ and its pull-back on the various schemes involved, by (d) of Prop 2.3, we have

$$ H^0(\mathbb{F}_p, \mathbb{G}_m)\to H^1_c(U, \mathbb{G}_m)\to H^1_c(X, \mathbb{G}_m) $$

Here $H^1_c(X, \mathbb{G}_m)=H^1(X, \mathbb{G}_m)$ since $K$ is a totally complex field, but anyway it is finite. Also $H^0(\mathbb{F}_p, \mathbb{G}_m)=\mathbb{F}_p^*$ is finite, therefore $H^1_c(U, \mathbb{G}_m)$ is finite.

If we use (a) of Prop 2.3, we have

$$ H^0(U, \mathbb{G}_m)\to H^0(K_p, \mathbb{G}_m)\to H^1_c(U,\mathbb{G}_m)\to H^1(U,\mathbb{G}_m)$$

Here $K_p$ is the Henselization of $K$ at $p$. Since we have shown $H^1_c(U,\mathbb{G}_m)$ is finite, and since we know $H^0(U, \mathbb{G}_m)$, which is the $S$-units on $X$, is a finite rank abelian group, we should have $H^0(K_p, \mathbb{G}_m)=K_p^*$ is also a finite rank abelian group. This seems absurd to me, since the subfields of $K_p$ which are number fields has infinite rank multiplicative groups.

So I wonder, did I miss anything? Thanks very much in advance!

  • $\begingroup$ I don't have Milne on me, but I'd be careful of the first term in your first exact sequence. I don't think the coefficient sheaf there should be the $\mathbb{G}_m$ of $\mathbb{F}_p$, but rather the pull-back to $Spec(\mathbb{F}_p)$ of $\mathbb{G}_m$ on $X$. Go through the definition of pull-back in the etale topology carefully and work out the $H^0$, and see if this helps with your question. $\endgroup$ May 23, 2012 at 21:44
  • $\begingroup$ That could the case. But since $\mathbb{G}_m$ is represented by $Hom(-,Spec \mathbb{Z}(t,\frac{1}{t}))$, I thought the pull-back of $\mathbb{G}_m$ on any scheme is always $\mathbb{G}_m$ on that scheme, isn't it? The push-forward, on the other hand, could be something else. $\endgroup$
    – Ying Zhang
    May 23, 2012 at 22:44
  • 3
    $\begingroup$ I don't think so. Go through the definition as I suggested. The relevant $H^0$ should become the units in the ring of integers of $K_p$. This should then be consistent with the second exact sequence. Working out such things is very important in learning etale cohomology, by the way. Usual geometric intuition takes you part of the way, but these arithmetic examples, even basic ones, convey some of the subtleties. $\endgroup$ May 23, 2012 at 23:06
  • $\begingroup$ I added the tag "nt.number-theory" -- questions will be seen by more people if they have at least one Arxiv subject class tag. $\endgroup$ May 24, 2012 at 9:36
  • $\begingroup$ Thanks Minhyong, sometimes it helps me a lot to be pointed out by others you are wrong about this and this..." and I'll check again the assumptions I implicitly assumed to be true. I am definitely wrong in saying that the pull-back of a sheaf defined by a group scheme is still the sheaf defined by the same group scheme: we see counter examples here, and also Milne has it on page 68 of his etale cohomology'' book (It is a different example there). $\endgroup$
    – Ying Zhang
    May 24, 2012 at 20:02

1 Answer 1


Apologies: (as pointed out by Minhyong Kim), the answer below does not address the OP's question

$K_p$ is not a number field. It contains a $p$-adic field $Q_p$ and, as such, '$K_p^*$' is not a finitely generated abelian group.

  • $\begingroup$ Sorry about the notation, $K_p$ here is the Henselization, therefore it is the algebraic elements in the usual p-adic field. But you are right, it is not a finitely generated abelian group, which is why I am confused. I thought either I misinterpret something or one of the lemmas is wrong? $\endgroup$
    – Ying Zhang
    May 23, 2012 at 22:47
  • $\begingroup$ Even if $K_p$ is the Henselization, $K_p$ is not a field of finite type over $\Q$; it is not a number field. Any equation (with mild restrictions coming from.. um.. Hensel's lemma!) with integer coefficients that has a solution mod $p$ will have a solution in $K_p$; so the degree of $K_p$ over $Q$ is infinite. So there is no reason to expect the units to be finitely generated. $\endgroup$
    – SGP
    May 24, 2012 at 11:52
  • $\begingroup$ Dear SGP: It seems to me that Ying Zhang's question is not what you are addressing at all. He is asking why the (obvious) lack of finite generation of $H^1_c$ implied by the second exact sequence does not contradict the apparent finite generation obtained through the first. $\endgroup$ May 24, 2012 at 13:45
  • $\begingroup$ @Minhyong Kim: You are indeed right! I did not read the question carefully. Thanks for pointing this. I have modified my answer. $\endgroup$
    – SGP
    May 24, 2012 at 20:37

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