When does the fraction field of a ring have a non-trivial Galois extension I have read this previous question on existence of a non-trivial Galois extension. I was wondering about the following situation. Suppose, $R$ is a domain that is not a field. When does the fraction field of $R$ have a non-trivial Galois extension?
 A: If $R$ is a (commutative) noetherian domain and is not a field, and suppose that its integral closure is also noetherian. Then $K=\mathrm{Frac}(R)$ has a non-trivial Galois extension. 
First note that the integral closure of $R$ is not a field: otherwise, for any non-zero $b\in R$, $1/b$ is integral over $R$. Writing an integral relation of $1/b$ over $R$, we see easily that $b$ is a unit in $R$, hence $R=K$. 
So we can suppose that $R$ is integrally closed. As said Ted in the comments, we have to show that $K$ is not separably closed. Suppose the contrary. Let $p\ge 2$ be an integer, prime to the characteristic of $K$ if the latter has positive characteristic. By hypothesis, there exists $b\in R$ non-zero and non-invertible. For any $n\ge 1$, $b^{1/p^n}$ is separable over $K$. So $b^{1/p^n}\in K$. As it is clearly integral over $R$, it belongs to $R$. Then we get an increasing sequence of ideals  $b^{1/p^n}R$ in $R$. By noetherianity, for some $n$, we have $b^{1/p^{n+1}}R=b^{1/p^n}R$, hence $b^{1/p^{n+1}}\in b^{1/p^n}R$. This immediately implies that $b$ is a unit. Contradiction.
Note that the noetherian condition on the integral closure of $R$ is satisfied for a large class of noetherian rings (called excellent rings). For example this is true if $R$ contains $\mathbb Q$. I think that with some more effort, one shoul be able to prove the statement without any condition on the integral closure of $R$. (maybe not so easy....).  
