[I already posted this question on stackexchange a while ago, but did not get any response: http://math.stackexchange.com/questions/93437/ideal-class-groups-and-extension-of-number-fields]

Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be schemes and $ f:X \to Y$ be a morphism. Suppose $f^\#:\mathcal{O}_Y \to f_*\mathcal{O}_X $is injective. Then so is the restriction $\mathcal{O}_Y^* \to f_*\mathcal{O}_X^*$ and we can complete to an exact sequence $1 \to \mathcal{O}_Y^* \to f_*\mathcal{O}_X^* \to \mathcal{Q} \to 1$. From this we get an exact sequence in cohomology starting as $ 1 \to H^0(\mathcal{O}_Y^*) \to H^0(f_*\mathcal{O}_X^*) \to H^0(\mathcal{Q}) \to H^1(\mathcal{O}_Y^*) \to H^1(f_*\mathcal{O}_X^*) \to H^1(\mathcal{Q}) $.


Consider the case where $L/K$ is an extension of number fields with rings of integers $\mathcal{O}_L$, $\mathcal{O}_K$, $X = Spec(\mathcal{O}_L)$, $Y = Spec(\mathcal{O}_K)$ and $f$ induced from the inclusion $\mathcal{O}_K \to \mathcal{O}_L$.

  1. How do $H^1(f_*\mathcal{O}_X^*)$ and $H^1(\mathcal{O}_X^*) = Pic(X)$ relate? (In particular, are they equal?)
  2. Can we describe $\mathcal{Q}$ explicitely? In particular, is there a good expression for $Q = H^0(\mathcal{Q})$? Does $H^1(\mathcal{Q}) = 0$?.


The motivation for considering the above is to end up with an interesting exact sequence relating the Picard and unit groups of $X$ and $Y$ via $Q$. Nice results are obtainable for example if $Y$ is the spectrum of a Dedekind domain and $X$ is an open subset or if $Y$ is the spectrum of a one-dimensional integral domain, $X$ its normalisation, and some finiteness conditions hold. These two cases are worked out "by hand" in Neukirch, algebraic number theory, sections 1.11 and 1.12.

A start on question 1

This is probably not the most elegant method, but the five-term exact sequence from the Leray spectral sequence starts as $ 1 \to H^1(f_*\mathcal{O}_X^*) \to H^1(\mathcal{O}_X) \to \Gamma(Y, R^1f_*\mathcal{O}_X^*) $. A sufficient condition for $ Pic(X) = H^1(f_*\mathcal{O}_X^*) $ is thus that the stalks of the first higher direct image are trivial, which is easily seen to be equivalent to $Pic((f_*\mathcal{O}_X)_p) = 0$ for all $p \in Y$. This does hold for an open immersion of number rings, but it is not clear to me if it applies to a normalisation or to an extension of number fields.

It is also easy to see that $H^1(\mathcal{F}) = H^1(f_* \mathcal{F})$ for all $\mathcal{F}$ if $f_*$ is exact, for example a closed immersion. This does not seem applicable here either, though.

  • 2
    $\begingroup$ Pushforwards are exact along a finite morphism if you work in the etale topology (or Nisnevich, but not Zariski). So if you're working in the etale topology, the answer to (1) is always "yes," without much thought. In fact, even in the Zariski topology, in the special case at hand, one is in good shape: the stalks of R^1 f_* O_X* are given by Picard groups of semilocal rings of X; such rings are always PIDs (as X is Dedekind), so R^1 f_* O_X* does indeed vanish. $\endgroup$
    – Bhargav
    Dec 27, 2011 at 16:46

1 Answer 1


if your aim is really to relate $Pic(Y)$ to $Pic (X)$ it is probably a good idea to pybass the use of $\mathcal Q$ and to consider instead the more powerful technique of introducing the relative Picard scheme (functor) $Pic_{X/Y}$. There are plenty of references, among them :

Néron models.

Bosch; Lütkebohmert ; Raynaud,

Ergebnisse der Mathematik und ihrer Grenzgebiete Springer-Verlag, Berlin, 1990.

for instance :

chapter 8 page 204 proposition 4 if $f:X\to Y$ is quasi-compact and quasi-separated and $f_*(\mathcal O_X)=\mathcal O_Y$ then there is an exact sequence :

$$ 0 \to Pic (Y) \to Pic (X) \to Pic_{X/Y}(Y) \to Br(Y) \to Br(X)$$

If you are interested, the standard is FGA (Grothendieck's Fondations de la Géométrie Algébrique), and the modern version in

The Picard scheme

Steven L. Kleiman


  • $\begingroup$ Well my aim is "really" to learn about algebraic geometry and algebraic number theory; so I just "follow all leads". I'll look at the reference(s) soon. $\endgroup$ Dec 28, 2011 at 15:44
  • $\begingroup$ In fact the exact sequence you quote is the five-term exact sequence of the Leray spectral sequence for the Etale topology. Reading Kleiman now; this material seems very interesting and a good way to learn about lots of stuff. I gladly accept your answer. $\endgroup$ Dec 30, 2011 at 16:18

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