Appropriate notion of localization of a Galois ring extension Earlier, I had a asked a question for a notion of Galois ring extension. I was particularly interested in Peter Patzt's answer. So, given an integral domain $R$ with field of fractions $F$ and a Galois extension $L$ of $K$, let $T$ be the integral closure of $R$ in $L$. We will say the extension $T/R$ is Galois. 
Now, I was wondering what an appropriate notion of "localization" of this extension would be. Explicitly, given a prime ideal $P$ of $T$, I would like to find a subring $S$ of $T$ so that $T_P / S$ is Galois. My first guess was $R_{P\cap R}$ would work. But the integral closure of $R_{P\cap R}$ is $(R-P)^{-1} T$. 
 A: Added. Thinking about it a bit more, in general it will not be the case that you can find an appropriate localization.
To see this, consider the case of a number field (a finite extension of $\mathbb{Q}$), and $T$ the ring of integers, and let $p\in\mathbb{Z}$ be a prime that does not ramify and is not inert: that means that the prime ideal $(p)$ can be written (uniquely) as a product of prime ideals of $T$, $(p) = \mathfrak{p}_1\cdots\mathfrak{p}_r$, and $\mathrm{Gal}(L/K)$ acts transitively on the set $\{\mathfrak{p}_1,\ldots,\mathfrak{p}_r\}$. Now say that you localize $T$ at $\mathfrak{p}_1$, and that $M$ is a multiplicative subset of $R$ such that $M^{-1}R$ has $T_{\mathfrak{p}_1}$ as its integral closure in $L$. 
Take an element $t\in \mathfrak{p}_2-\mathfrak{p}_1$. Then $\frac{1}{t}\in T_{\mathfrak{p}_1}$, and so must be integral over $M^{-1}R$. That means that the monic irreducible polynomial of $t$ has coefficients in $M^{-1}R$. But since the action of the Galois group is transitive on the $\mathfrak{p}_i$, there is a conjugate $t'$ of $t$ that lies in $\mathfrak{p}_1$. Therefore, $1/t'$ is conjugate of $1/t$, so they have the same monic irreducible polynomial, and so $1/t'$ is integral over $M^{-1}R$. But $1/t'\notin T_{\mathfrak{p}_1}$, because $t'\in\mathfrak{p}_1$.
Thus, there is no $M$ such that $M^{-1}R$ has $T_{\mathfrak{p}_1}$ as an integral closure. 
The argument shows that, at least in the number field case, if you try to localize at $P$, then the only case in which you'll be able to obtain $T_P$ as the integral closure of a localization of $R$ will be if $P\cap R$ is either inert (so that $P = (P\cap R)T$), or $P\cap R$ is totally ramified (so that the prime factorization of $P\cap R$ in $T$ is $P^e$ for some $e$).
And for the general case, you are going to need $P$ to be closed under the action of the Galois group, else the same argument as above show that no localization can be found. 
If that is the case, then perhaps the construction I proposed below will work.

As I recall from algebraic number theory, the localization usually goes "the other way": if $T$ is integral over $R$, and $S$ is a multiplicative subset of $R$, then $S^{-1}T$ is integral over $S^{-1}R$.
But let's say you want to go the other way: what is going on is that you are allowing some elements that were algebraic but not integral into your $T$. In order for $T_P$ to be integral, you need all the irreducible monic polynomials of elements of $T-P$ to have coefficients in the ring $S$.
If $r\in T$, then the leading coefficient of the monic irreducible polynomial of $\frac{1}{r}$ is the constant term of the monic irreducible polynomial of $r$ over $K$. But this is precisely the product of all the conjugates of $r$, and the norm, $N_{L/K}(r)$ will give you a power of this, and it lies in $R$. If you localize so that $N_{L/K}(r)$ is invertible, that will ensure that the constant term of the monic irreducible of $r$ is itself invertible, and so that $1/r$ is integral over the localization.
So I believe that what you want to do is localize at $N_{L/K}(T-P)$, the image of the norm map restricted to $T-P$. This should work for localization over any set: if $M$ is a multiplicative subset of $T$, then $M^{-1}(T)$ should be integral over $(N_{L/K}(M))^{-1}(R)$. Note that $N_{L/K}(M)$ is a multiplicative set, since the norm is multiplicative.
