Direct image for standart affine cover of projective line Let $U_1 \cong Spec(K[t])$ and $U_2\cong Spec(K[t])$ be a standard affine cover of a projective line $\mathbb{P^1}(K)$, where $K$ is some field. Let us denote open embedding as $j_k$
$$
j_k : U_k \to \mathbb{P^1}(K), k=1,2.
$$
Is it correct that $R^i(j_k)_*=0$ for $i > 0$ (because $j_k$ is an affine map)? How one can explititly describe $(j_k)_* \mathcal{O}_{U_k}$? Am I right that $(j_k)_* \mathcal{O}_{U_k}$ is a quasi-coherent sheaf, but not coherent?
 A: Yes, $j_k$ is an affine morphism (as every inclusion of an affine open into a separated scheme), and for an affine morphism $f$ it is clear from the local description that $f_*$ is exact, so that $R^i f_* = 0$ for $i>0$.
For the explicit description in the special case you have mentioned, let $R$ be an arbitrary ring (actually the same works for an arbitrary base scheme) and write $U_1 = \mathrm{Spec~} R[t]$ and $U_2 = \mathrm{Spec~} R[t^{-1}]$ for usual affine open cover of $\mathbb{P}^1_R$. Then there is an equivalence of categories between $\mathsf{Qcoh}(\mathbb{P}^1_R)$ and the category of triples $(M,N,\alpha)$, where $M$ is a $R[t]$-module, $N$ is a $R[t^{-1}]$-module and $\alpha$ is an isomorphism of $R[t,t^{-1}]$-modules $M_t \cong N_{t^{-1}}$. Under this equivalence, quasi-coherent modules of finite type correspond to those triples $(M,N,\alpha)$ for which $M,N$ are finitely generated.
The pullback functor $j_1^* : \mathsf{Qcoh}(\mathbb{P}^1_R) \to \mathsf{Qcoh}(U_1) \cong \mathsf{Mod}(R[t])$ corresponds to $(M,N,\alpha) \mapsto M$, similarily $j_2^*$ corresponds to $(M,N,\alpha) \to N$.
The direct image functor ${j_1}_* : \mathsf{Qcoh}(U_1) \to \mathsf{Qcoh}(\mathbb{P}^1_R)$ corresponds to $M \mapsto (M,M_t|_{R[t^{-1}]},\alpha)$. Here we restrict scalars from $R[t,t^{-1}]$ to $R[t^{-1}]$, and $\alpha$ is the obvious isomorphism. From here we see that ${j_1}_*$ doesn't preserve the property of finite type: If $M$ is a finitely generated $R[t]$-module, then $M_t$ is a finitely generated $R[t,t^{-1}]$-module, but it doesn't have to be a finitely generated $R[t^{-1}]$-module. Already $M=R[t]$ provides a counterexample when $R \neq 0$.
