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!