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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?

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First, this question only makes sense when $R$ is a (commutative) domain. Now I think this is too trivial to make into an answer, but you are really asking when the fraction field of $R$ is not algebraically closed. Based on concrete examples I have on hand (e.g. rings of integers in number fields and $p$-adic fields, algebras of regular functions on irreducible varieties, etc.) the answer is "almost always." – Justin Campbell Nov 13 '11 at 17:58
The algebraic closure of a field is always a Galois extension of the field, and if you want the extension to be finite you can always take the splitting field of any separable polynomial. This will be nontrivial unless the field is already separably closed. – Ted Nov 13 '11 at 18:00
@Justin: Thanks, I did mean a domain. – Vince Cairns Nov 13 '11 at 18:11
Oops, I should have said "separably closed" as in Ted's comment. Actually, maybe I should retract my claim that the fraction field is "almost always" not separably closed in interesting examples. For example, consider the ring of integers in $\overline{\mathbb{Q}}_p$. So I guess it really depends. – Justin Campbell Nov 13 '11 at 18:12
@Ted: Isn't the algebraic closure a Galois extension iff the field is perfect? – Vince Cairns Nov 13 '11 at 18:12
up vote 2 down vote accepted

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....).

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Thank you. I was particularly interested in complete local rings, so this should work for those. – Vince Cairns Nov 14 '11 at 17:45
Complete noetherian rings are excellent. – user18119 Nov 14 '11 at 22:20

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