Endomorphisms of the multiplicative group

Question

Let $U$ be an open subscheme of the spectrum $X=\mathrm{Spec}(O_K)$ of the ring of integers of a number field. Let $\mathbb{G}_m$ be the étale sheaf represented by the multiplicative group. Do we know how to compute $$ \mathrm{Hom}_U(\mathbb{G}_m,\mathbb{G}_m) $$ and $$ \mathrm{Ext}^1_U(\mathbb{G}_m,\mathbb{G}_m) $$ ? I'm a bit at a loss here. It would be nice if the first ext group was finite or at least torsion.

Answer 1

EDIT: Now that I have more time I'll give a more precise answer.

Claim: Let $S$ be a normal integral scheme, then $\mathrm{Ext}^1_{S_\mathrm{et}}(\mathbb{G}_m,\mathbb{G}_m)=0$.

Proof: Let $\mathcal{F}$ be an etale sheaf given as an extension

$$0\to \mathbb{G}_m\to\mathcal{F}\to\mathbb{G}_m\to 0$$

Note then that $\mathcal{F}$ is a $\mathbb{G}_m$-torsor over $\mathbb{G}_m$ and so by etale descent for affine morphisms of schemes we know that $\mathcal{F}$ is representable by some affine group scheme $T$. We claim that $T$ is a torus (in the sense of Definition 3.1.1 of [Con1]). But, by Theorem B.4.1 of [Con1] it suffices to show that $T_{\overline{s}}$ is a torus for every geometric point $\overline{s}$ of $S$. But, this reduces us to the case when $S$ is the spectrum of an algebraically closed field $k$. Note then that $T$ is a smooth affine group scheme and so by Corollary 12.21 of [Mil] it suffices to show that $T(k)$ consists only of semisimple elements. But this is immediately from the fact that $T$ is an extension of $\mathbb{G}_m$ by itself and the fact that Jordan decomposition is functorial.

So, now that we know that $T$ is a torus it suffices by Corollary B.3.6 of [Con1] to show that the continuous representation $\pi_1^\mathrm{et}(S,\overline{s})\to \mathrm{GL}(X^\ast(T))$ is trivial. But, note that by the fact that $\mathcal{F}$ is an extension of $\mathbb{G}_m$ we know that $\pi_1^\mathrm{et}(S,\overline{s})$ takes values in $\left\{\begin{pmatrix}1 & \ast\\ 0 & 1\end{pmatrix}\right\}\cong\mathbb{Z}$. In other words, $T$ corresponds to a continuous homomomorphism $\pi_1^\mathrm{et}(S,\overline{s})\to \mathbb{Z}$. But, there is no non-trivial such homomorphism since $\pi_1^\mathrm{et}(S,\overline{s})$ is profinite. $\blacksquare$

TL;DR: There are no non-trivial continuous homomorphisms $\pi_1^\mathrm{et}(S,\overline{s})\to \mathbb{Z}$.

Note that this TL;DR also indicates what could go wrong over a non-normal base. Namely, while there are no non-continuous homomorphisms $\pi_1^\mathrm{et}(S,\overline{s})\to \mathbb{Z}$ for an arbitrary irreducible scheme $S$, there are non-trivial homomorphisms $\pi_1^{\mathrm{proet}}(S,\overline{s})\to \mathbb{Z}$ or $\pi_1^{\mathrm{SGA}3}(S,\overline{s})\to \mathbb{Z}$. The point is that B.3.6 of [Con1] can be stated for arbitrarly irreducible $S$ if one replaces $\pi_1^{\mathrm{et}}(S,\overline{z})$ with $\pi_1^{\mathrm{proet}}(S,\overline{s})$ or $\pi_1^{\mathrm{SGA}3}(S,\overline{s})$. So, for example the claim is false since, for example, if $S$ is the projective nodal curve then $\pi^{\mathrm{proet}}(S,\overline{s})\cong \pi_1^{\mathrm{SGA}3}(S,\overline{s})\cong \mathbb{Z}$ as topological groups.

Your first question is more elementary--the answer is $\mathbb{Z}$. See Exercise 1 here for a guided solution.

[Con1] Reductive group schemes, Brian Conrad, -- found here.

[Con2] http://virtualmath1.stanford.edu/~conrad/252Page/homework/hmwk1.pdf

[Mil] Algebraic groups: the theory of finite type group schemes over a field, Milne, published by Cambridge press.