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 $H^1(X,I\_x(n))=0$.



Explicit basis for the space of global sections of a twisted arithmetic ideal sheaf
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


1. $f,g$ are respectively not multiples of $x\_0,x\_1$
2. $\deg f - k = n$ 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.


