I am looking at the following part:
$$$$
$$$$
$$$$
$$$$
$$$$
I haven't really understood the proof...
We suppose that Grothendieck's problem stand and that almost all prime ideals of the algebraic number field $K$ are of degree one.
We want to show that $K=\mathbb{Q}$.
Why do we take $K=\mathbb{Q}(a)$ ?
Why does it stand that for each prime ideal $\mathfrak{p}$ of degree one there exists an integer $\alpha (\mathfrak{p})$ such that $\alpha=\alpha (\frak{p}) \pmod {\mathfrak{p}}$ ?
We can assume $K=\mathbb{Q}(\alpha)$ for some $\alpha$ by the primitive element theorem.
I'm not sure what degree means in this context; I'm guessing it means the degree of the residue field $\mathcal{O}_K/\mathfrak{p}$ over the subfield $\mathbb{Z}/(\mathfrak{p}\cap\mathbb{Z})$? Assuming that, then for the degree to be $1$ means that $\mathcal{O}_K/\mathfrak{p}=\mathbb{Z}/(\mathfrak{p}\cap\mathbb{Z})$, and any element of the latter has an integral representative. So if we consider $\alpha$ mod $\mathfrak{p}$, we can find some integer $\alpha(\mathfrak{p})$ that represents it modulo $\mathfrak{p}$.