Every irreducible representation comes via base change from a finite extension of the prime field I asked (see Representation on a splitting field and base change) about the meaning of Proposition 4.19 in ''introductory course on l-adic sheaves and their ramification theory on curves'' by Kindler and Rülling (see http://www.mi.fu-berlin.de/users/kindler/documents/madrid.pdf).
Now, the meaning is almost clear, but I don't understand the proof.
Proposition 4.19. Let $G$ be a finite group, $E$ a field and $F$ the prime field of $E$. If $E$ is a splitting field for $G$, then every irreducible representation of $G$ on $E$ comes via base change from a finite extension $K$ of $F$ ie
there exists an finite extension $K$ of $F$ and an $K$-representation $W$ of $G$ such that
    $$ V \simeq W \otimes_K E. $$
The proof goes as follows.
Let $\overline{E}$ be an algebraic closure of $E$ and $\overline{F} \subset \overline{E}$ be an algebraic closure of $F$.
Let $V$ be an irreducible representation of $G$ over $E$.
Since $E$ is a splitting field for $G$, $V \otimes_E \overline{E}$ is also irreducible and the map
    $$ R_{\overline{F}}(G) \longrightarrow R_{\overline{E}}(G) $$
is an isomorphism (where $R_{\overline{E}}(G)$ is the Grothendieck ring of the category of representations of $G$ on finite dimensional $\overline{E}$-vectors spaces).
So, there exists an representation $W'$ of $G$ over $\overline{F}$ such that
  $$ W' \otimes_{\overline{F}} \overline{E} \simeq V \otimes_E \overline{E}. $$
The representation $W'$ comes via base change from $\overline{F} \subset E_V$ contained in $\overline{F}$.
Let $E_0$ be the compositum of the field $E_V$ in $\overline{F}$.
As there is only finitely many isomorphism classes of irreducible representations of $G$, $E_0$ is finitely generated over $F$.
The proof ends here.
My questions are
1) In the proposition, is $E$ an extension of $K$?
2) I guess that we must choose $K = E_0$.
Is $E$ an extension of $E_0$?
3) How to find the representation $W$ of $G$?
 A: It's a good question, and I hadn't appreciated the subtleties until I thought about it.
The answer to your first two questions is "no": a splitting field $E$ doesn't have to contain a splitting field $E_0$ that is a finite extension of the prime subfield.
For example, let $G=Q_8$, the quaternion group of order $8$. Over an algebraically closed field of characteristic zero, $G$ has four $1$-dimensional representations that are defined over $\mathbb{Q}$, and a two-dimensional irreducible representation for which $\mathbb{R}$ is not a splitting field, but any field containing elements $a,b$ with $a^2+b^2+1=0$ is (in fact that's "if and only if", but we don't need that).
Let $\mathbb{R}(t)$ be the field of rational functions in an indeterminate $t$, and let $E=\mathbb{R}(t)(\alpha)$ be the quadratic extension where $\alpha^2+t^2+1=0$. Then $E$ is a splitting field for $G$, but the only algebraic numbers that it contains are real, so it doesn't contain any splitting field that is a finite extension of $\mathbb{Q}$.
This isn't a problem if $E$ is algebraically closed, and a correct version of the statement is that there is a splitting field $E_0\subseteq \bar{E}$ that is a finite extension of the prime subfield.
Or alternatively, for every irreducible $E$-representation $V$, there is a finite extension $E_0$ of the prime subfield with $E_0\subseteq \bar{E}$, and an $E_0$-representation $W$ such that
$$V\otimes_E\bar{E}\cong W\otimes_{E_0}\bar{E}.$$
