Q. Let $F$ be any field and $K=F(X)$. Prove that any subextension $F\subset L\subset K$ either has $K\vert L$ is finite, or $L=F$.
Find $\operatorname{Aut}_F(K)$.
Proof. The first part amounts to proving that for any $f\in K\setminus F$,
\[[K:F(f)]=[F(X):F(f)]<\infty.\]We can in fact find the minimal polynomial of $X$ over $F(f)$. Let $f=g/h$ where $g$, $h\in F[X]$ are coprime. Clearly $fh(T)-g(T)\in F(f)[T]$ has $X$ as a root. Therefore $[K:F(f)]<\infty$.
Note that this polynomial is actually the minimal polynomial of $X$. In view of the Gauss Lemma we reviewed last time,
\[\text{irreducible in }R[T]\wedge\text{nonconstant}\Longrightarrow\text{irreducible in }\operatorname{Frac}(R)[T].\]So it suffices to check that $fh(T)-g(T)$ is irreducible in $F[f][T]=F[f,T]$. This is easy:
\[\begin{align*} &fh(T)-g(T)=(fa(T)+b(T))c(T)\\ &\implies ac=h\wedge bc=g\\ &\implies c\in F. \end{align*}\]In particular, $[F(X):F(f)]=\max(\deg h,\deg g).$
As a consequence, we also get that $f$ generates $F(X)$ over $F$ if and only if $f$ has the form
\[f(X)=\frac{aX+b}{cX+d}\notin F.\]That $f\notin F$ can be stated as $ad-bc\neq 0$. Thus as sets
\[\operatorname{GL}_2(F)\longrightarrow\operatorname{Aut}_F(K).\]We can directly check that this is a group homomorphism—I don’t know a better way—and its kernel is just $F$. Therefore $\operatorname{Aut}_F(K)=\operatorname{PGL}_2(F)$.
Post date: 2024/12/04
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