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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 $(R-P)^{-1} T$.

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@B M: Dear B M, I understand (based on looking at your previous question) why you say that $T/R$ is Galois, but let me advise you not to adopt this terminology, since it is not at all standard, whereas the situation you are interested in, of an integral extension of domains inducing a Galois extension of the corresponding fraction fields, is very standard. (Unfortunately, there is not a standard pithy term that is used to describe it.) I am speaking here from my experience as an algebraic number theorist. Regards, –  Matt E Sep 17 '11 at 11:52
    
@B M: I think you are looking at the problem the wrong way around (see my answer): you can localize $R$ and look at the closure of that in $L$; this is especially nice when you localize at a prime ideal $P$ of $R$, because the localization will give you $T$ localized at all primes that lie above $P$; but trying to go "down" will generally not work. –  Arturo Magidin Sep 17 '11 at 20:20
    
@Arturo: Thanks for your answer. I was already aware that studying the integral closure of $R$ in $L$ is nice, as indicated in the last line to my question. I actually needed to derive some properties of the localization of $T$ at it's prime ideals and having it as this kind of an extension would have certainly helped, but it seems (from your answer and otherwise), that this approach might not work. –  B M Sep 17 '11 at 20:52
    
@B M: Well, glad I could help. I was so sure as I was writing that $N_{L/K}(P)$ would work, but as I was typing it up it became apparent that you need at least that, but this would localize away from $P$ and all its Galois conjugates. You can do that, I think, but not if your set is not closed under the action of the Galois group (which I think makes sense, if you think about it, because whatever it is you construct should be invariant under the action of the Galois group). –  Arturo Magidin Sep 17 '11 at 20:55
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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.

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