Let $K(x)$ be the rational function field over a field $K$.

Show that for a subgroup $U \leq Aut(K(X)/K)$ the following conditions are equivalent:

$(1)$ $(K(x) : F(U))$ is a Galois extension;

$(2)$ $U$ is finite;

$(3)$ there is a non constant rational function with $U= \{\sigma \in G \mid \sigma * \phi = \phi\}$.

A field extension $(L : K)$ is called a Galois extension, if $F(Aut(L/K)) = K$, i.e., if for any $a \in L- K$ there is an automorphism of $L$ which leaves $K$ pointwise fixed, but actually moves $a$.

$Aut(K(X)/K) = \{\phi \mid \phi : K(X) \to K(X) , \phi(k) = k \ \ \forall k \in K\}$ where $\phi$ is an automorphism.

  • $\begingroup$ Do you mean "Non-constant rational function $\phi$"? $\endgroup$
    – Arthur
    Jan 9, 2017 at 8:07

1 Answer 1


If $\mathbb{F}$ is any field and $G$ is a finite group of automorphisms of $\mathbb{F}$ then $\mathbb{F}/\mathbb{F}^G$ is Galois with group $G$ - this is just Artin's theorem which more or less implies that $(2)\Rightarrow (1)$.

Assuming (3), you get that $U$ contains all the automorphisms that keep $K(\phi(x))$ invariant where $\phi(x)$ is not constant. Since $[K(x):K(\phi(x))]$ is finite (of degree $\deg(\phi)$), then $U$ must be finite which is exactly (2).

Finally, assume (1), so that $[K(x):K(x)^U]$ is Galois. Since $K\leq K(x)^U \leq K(x)$, you can use Luroth theorem which states that $K(x)^U=K(f(x))$ for some rational function $f$. This function is not going to be constant, since $K(x)/K$ is not Galois (it is not an algebraic extension). You then get that $f(x)$ is a nonconstant rational function that is invariant under $U$. Moreover, since $U$ is the Galois group of $K(x)/K(x)^U=K(x)/K(f(x))$ (again by using Artin's theorem), any other automorphism that keeps $f(x)$ invariant will keep $K(f(x))$ invariant, and therefore will be in $U$, so we get that $(1)\Rightarrow (3)$.

By the way, your definition of Galois is not true for infinite extensions. For example, $K(x)/K$ is not Galois because it is not an algebraic extension, but for every $f(x)\in K(x)-K$ there is a $K$-automorphism which moves $f(x)$. More precisely, you have the automorphisms defined by $x\mapsto ax+b$ for any $a,b \in K$ where $a\neq 0$. Writing $f(x)=f_1(x)/f_2(x)$ with $(f_1,f_2)=1$, if it is invariant under such automorphism, then you get that $f_1(ax+b)f_2(x)=f_1(x)f_2(ax+b)$ as polynomials. If $f(x)$ is invariant under all the automorphisms above, then comparing the leading coefficients show that we must have $\deg(f_1)=\deg(f_2)$. Suppose that $f$ is not constant, so that both $f_1,f_2$ are not constant and since they are coprime we can find a root $\alpha$ for $f_1$ which is not a root for $f_2$, but then we get that $f_1(a\alpha+b)f_2(\alpha)=0$ for all $a,b$ which implies that $f_1=0$ - contradiction. It follows that $K=(K(x))^{Aut(K(x)/K)}$ .


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .