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.
- Find a non-zero polynomial $T(t) ∈ L[t]$ which vanishes at $t=Y$, that is, $T(Y)=0$ in $M$.
- Show that the extension $M ∣ L$ is simple and finite.
- We have $Q(Y) ⋅ ϕ(X)-P(Y)=0$ so we may choose $T(t)=X ⋅ Q(t)-P(t)$
- The element $Y$ generates $M$ over $ℚ$ as a field and hence it generates $M$ over $L$. By (i), $Y$ is algebraic over $L$ and thus $M$ is a simple and finite extension of $L$ (by Prop. 3.9).
- Compute the degree of the extension $M ∣ L$ in terms of the degrees of $P(Y)$ and $Q(Y)$.
- Let $σ: ℚ(X) → ℚ(X)$ be an automorphism of fields. Show that
\[
σ(X)=(a X+b) /(c X+d)
\]
where $a d-b c ≠ 0$.
- We need to compute the degree of the minimal polynomial of $Y$ (use Prop. 3.9). We show that $T(t)=X ⋅ Q(t)-P(t)$ is irreducible over $L$. Note that
\[
L[t]=(ℚ[X])[t] ≃ ℚ[X, t]
\]
and write $\tilde{T}(X, t) ∈ ℚ[X, t]$ for $T(t)$ viewed as an element of $ℚ[X, t]$. We first prove that $\tilde{T}(X, t)$ is irreducible. Suppose for contradiction that $\tilde{T}(X, t)=T_1(X, t) T_2(X, t)$, where $T_i(X, t) ∈ ℚ[X, t]$. Then either $T_1(X, t) ∈ ℚ[t]$ or $T_2(X, t) ∈ ℚ[t]$ for otherwise $\tilde{T}(X, t)$ would contain a term $X^k$ with $k ≥ 2$. Suppose that $T_1(X, t) ∈ ℚ[t]$. Write $T_2(X, t)=X H_2(X, t)-H_1(t)$, where $H_1(t) ∈ ℚ[t]$ and $H_2(X, t) ∈ ℚ[X, t]$. Then $T_1(X, t) H_1(t)=P(t)$ and $T_1(X, t) H_2(X, t)=Q(t)$. In particular, $H_2(X, t) ∈ ℚ[t]$ and $T_1(X, t)$ is a common factor of $P(t)$ and $Q(t)$. Hence $T_1(X, t) ∈ ℚ^*$, since $P(t)$ and $Q(t)$ are coprime. Hence $\tilde{T}(X, t)$ is irreducible. By the generalised Gauss lemma, we conclude that $T(t)$ is irreducible as an element of $L[t]$. Hence, by (a), $T(t)$ is the minimal polynomial of $Y$, up to multiplication by an element of $L^*$. Hence $[M: L]$ is the degree of $T(t)$, which is $\max (\deg(Q(t)), \deg(P(t)))$.
- Suppose that $σ(X)=R(X) / U(X)$, where $R(X), U(X) ∈ ℚ[X]$ are coprime. By (i), $R(X)$ and $U(X)$ have at most degree one and hence $σ(X)=(a X+b) /(c X+d)$ for some $a, b, c, d ∈ ℚ$. Also, $a X+b$ cannot be a $ℚ$-multiple of $c X+d$ for otherwise, $σ(X) ∈ ℚ$, which is not possible. Hence $a d-b c ≠ 0$.
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.
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.