[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}) $.

**Question(s)**

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 $\mathcal{O}_K \to \mathcal{O}_L$.

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

**Examples/Motivation**

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.