My question refers to a step in the proof of Prop. 3.3.5 Szamuely and Tamásin's "Galois groups and fundamental groups":
Here the statement and Thm 3.3.3 & lemma 3.3.6: The main ingredients for the proof:
AND here the proof of 3.3.5 with red tagged unclear argument:
My question is why $f$ satisfies an irreducible polynomial equation $(\phi^*a_n)f^n + ... +(\phi^*a_0)=0$? Why irreducible?
Lemma 3.3.6 says that the polynomial equation is not neccessary irreducible.
Intuitively I guess that should have something to do with the assumption that all values $f(y_i)$ are distinct. Why does it suffice? I don't find a clear argument.
I haven't looked into the whole question in detail, but this is just a general algebraic principle. The Lemma tells you that there is some non-zero polynomial $P$ with coefficients in $M(Y)$ that kills $f$, i.e. satisfies $P(f) =0$. Now take such a non-ozero polynomial $P_0$ of smallest degree $n \leq d$. If $P_0$ were not irreducible, you could factor $P_0$ as $P_0 = RS$ for some non-zero polynoimals $R,S$, both of strictly smaller degree than $P_0$. But then $0=P_0(f) = R(f) S(f)$ implies that either $R(f) =0$ or $S(f) = 0$, contradicting minimiality of the degree.