My question refers to a step in AriyanJavanpeykar's argumentation in following former thread of mine:
Assume we have a proper and smooth morphism of schemes $f:X \to Spec(\mathbb{Z})$. Assume, that the set of $\mathbb{F}_p$-valued points $X(\mathbb{F}_p)= Hom(Spec(\mathbb{F}_p),X)$ isn't empty.
How to conclude that $X(\mathbb{Q}_p)$ also isn't empty?
My considerations:
Up to now I tried two ideas without success:
since $f$ proper (therefore of finite type) we know $\Gamma(X, \mathcal{O}_X) = \mathbb{Z}[X_1,...,X_n] / I $ for appropriate ideal $I$.
The point $x \in X(\mathbb{F}_p)$ gives a map $ \mathbb{Z}[X_1,...,X_n] / I \to \mathbb{F}_p$. Can it in this situation be lifted to $\mathbb{Z_p}$? I don't see why. Universal properties for completions wrt p seem not to help here.
Other approach was to observe that since $x$ is a morphism $\mathbb{Z}$-schemes that it can lifted to $X_{\mathbb{Z}_p}(\mathbb{F}_p)$ where $X_{\mathbb{Z}_p}= X \times_{Spec\mathbb{Z}} Spec(\mathbb{Z_p})$. Maybe that could simplify the problem...
But I'm unsure if one of these ideas would lead me to desired result.
Does anybody see how it can be deduced that $X(\mathbb{Q}_p)$ is not empty?
I think here is a solution. Let $R$ be a Henselian DVR with fraction field $K$ and residue field $k$. Let $f:Y\rightarrow \mathrm{Spec}\,R$ be a morphism of schemes. If $f$ is smooth, then there is a surjective map of sets $Y(R)\rightarrow Y_k(k)$ (see "Neron models", Ch. 2.3, Prop. 5). If $f$ is proper, then there is a bijective map $Y(R)\rightarrow Y_K(K)$ (valuative criterion of properness). So if $f$ is smooth and proper and $Y_k(k)$ is non-empty, then $Y_K(K)$ is non-empty.
Now apply the above to $R=\mathbb{Z}_p$ and $Y=X\times_{\mathrm{Spec}\,\mathbb{Z}}\mathrm{Spec}\,\mathbb{Z}_p$ (note that if $X\rightarrow \mathrm{Spec}\, \mathbb{Z}$ is smooth and proper, then $Y\rightarrow \mathrm{Spec}\,\mathbb{Z}_p$ is going to be smooth and proper as well).