Explicit basis for the space of global sections of a twisted arithmetic ideal sheaf Assume $x\in X=\mathbb{P}^1_{\mathbb{Z}}$ is a closed point with $f(x)=p\in Y$ where $f:X\rightarrow Y$, here $Y=Spec(\mathbb{Z})$. Assume $k(x)=\mathbb{F}_p$ and denote by $I_x$ the ideal sheaf of $x$.
Can we write down a basis for the free $\mathbb{Z}$-module $M(n)=H^0(X,I_x\otimes\mathcal{O}_X(n))$?
I think for n=0, we have $M(0)=p\mathbb{Z}$, so that $p$ is a basis.
This is because $H^0(X,I_x)=H^0(Y,f_{*}I_x)$ and we have $f_{*}I_x=I_{f(x)}$, which comes from $0\rightarrow I_x\rightarrow \mathcal{O}_X\rightarrow k(x)\rightarrow 0$ and the facts: $f_{*}\mathcal{O}_X=\mathcal{O}_Y$,  $f_{*}k(x)=k(f(x))$ and $H^1(X,I_x)=0$. But $I_{f(x)}$ is just the ideal sheaf of $p \in Y$, which is the associated sheaf to the ideal $(p)$.
But what happens for $n>0$?
We have $0\rightarrow I_x(n)\rightarrow \mathcal{O}_X(n)\rightarrow k(x)\rightarrow 0$ and $H^0(X,\mathcal{O}_X(n))$ is the free $\mathbb{Z}$-module generated by the homogeneous polynomials in two variables of degree $n$. We also have $H^1(X,I_x(n))=0$ so that $M(n)$ is a free submodule of $H^0(X,\mathcal{O}_X(n))$ with quotient $\mathbb{F}_p$, hence it has the same rank as $H^0(X,\mathcal{O}_X(n))$.
Can we write down a basis in terms of the homogeneous polynomials? It seems that in one basis vector there should appear a $"p"$. It may be obvious, but i don't see how to write it down explicitely in terms of these polynomials. Maybe we have to restrict to a special point like $x=(0:1)\in \mathbb{P}^1_{\mathbb{F}_p}$?
Can we more generally say something about an explicit basis for $M_Z(n)=H^0(X,I_Z(n))$, where $Z\subset X$ is a l.c.i. of codimension two? 
Edit: $H^1(X,I_x(n))=0$ because we have $R^1f_{*}(I_x(n))=0$. This is because we have a map $R^1f_{*}(I_x(n))\otimes\mathbb{F}_q\rightarrow H^1(X_q,I_x(n)\otimes\mathbb{F}_q)$. Here $X_q=\mathbb{P}^1_{\mathbb{F}_q}$. 
Now using $0\rightarrow I_x(n)\rightarrow O_X(n)\rightarrow \mathbb{F}_p\rightarrow 0$, we have $I_x(n)\otimes\mathbb{F}_q \cong \mathcal{O}_{X_q}(n)$ if $q\neq p$. If $q=p$ have have an exact sequence $0\rightarrow \mathbb{F}_p\rightarrow I_x(n)\otimes\mathbb{F}_p\rightarrow \mathcal{O}_{X_q}(n-1)\rightarrow 0$. This shows that for all $q$ and all $n\geq 0$ we have $H^1(X_q,I_x(n)\otimes\mathbb{F}_q)=0$.  Using the base change theorem we see that $R^1f_{*}(I_x(n))=0$. But since $Y$ is affine $R^1f_{*}(I_x(n))=\widetilde{H^1(X,I_x(n))}$. So we must have $H^1(X,I_x(n))=0$.
 A: For the sake of simplicity I will tackle the case that $x$ is the closed point $V_+(p,x_0)$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like
$$ 0 \to (x_0,p)_{x_0}(n) \times (x_0,p)_{x_1}(n) \stackrel{d}{\to}  (x_0,p)_{x_0x_1}(n) \to 0$$
where the notation $(x_0,p)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0,p)$ where $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.
Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume


*

*$f,g$ are respectively not multiples of $x_0,x_1$

*$\deg f - k = n$ and $\deg g - l = n$.


If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.
Conclusion
The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0,p)$. This is free of rank $n + 1$ and the good news is it agrees with what you found above! Hopefully from this you can see how to generalize to other closed points.
