Let $L=ℚ(X)$ and $M=ℚ(Y)$ where $ℚ(X)$ (resp. $ℚ(Y)$ ) is the fraction field of $ℚ[X]$ (resp. $ℚ[Y])$. Let $ϕ: L ↪ M$ be a field extension. Suppose that $ϕ(X)=P(Y) / Q(Y)$, where $P(Y), Q(Y) ∈ ℚ[Y]$ are coprime polynomials.

Let $σ: ℚ(X) → ℚ(X)$ be an automorphism of fields. Let \[ ℚ(X)^σ:=\{s ∈ ℚ(X) ∣ σ(s)=s\} . \] Show that the field extension $ℚ(X) ∣ ℚ(X)^σ$ is algebraic if and only $σ$ is of finite order in $\operatorname{Aut}_{ℚ}(ℚ(X))$.
[Hint: do not use (b); derive the result from standard results.]
[The following result may be used without proof: Let $R$ be a PID and let $K$ be its fraction field. Suppose that $A(X) ∈ R[X]$ is irreducible in $R[X]$ and is of positive degree. Then $A(X)$ is irreducible in $K[X]$.

Let $G$ be the subgroup of $\operatorname{Aut}(ℚ(X))$ generated by $σ$. Suppose that $G$ is finite. Then the extension $ℚ(X) ∣ ℚ(X)^σ=ℚ(X)^G$ is finite by Artin's lemma.
Without Artin's lemma Suppose that $σ$ has finite order $n$. Then $λ=X+σ X+⋯+σ^{n-1}X=P(X)/Q(X)$ is in $ℚ(X)^σ$.
Then $X$ satisfies $λ⋅ Q(t)-P(t)$ in $ℚ(X)^σ[t]$.
Hence, $ℚ(X)|ℚ(X)^σ$ is finite.
Conversely, suppose the extension $ℚ(X) ∣ ℚ(X)^σ$ is algebraic, then it is finite, because $ℚ(X)$ is generated by $X$ over $ℚ(X)^σ$ and thus $ℚ(X) ∣ ℚ(X)^σ$ is simple and algebraic, and thus finite.
Now we have by construction $$G ⊆ \operatorname{Aut}_{ℚ(X)^G}(ℚ(X)) ⊆ \operatorname{Aut}(ℚ(X))$$ The group $\operatorname{Aut}_{ℚ(X)^G}(ℚ(X))$ embeds as a subgroup of the group of permutations of the roots of the minimal polynomial of $X$ over $ℚ(X)^G$ and is thus finite (one may also apply Th. 3.18). Thus $G$ is finite.