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Let $R$ be an integrally closed integral domain with fraction field $K$. Let $L$ be a finite Galois extension of $K$ and let $\sigma_1,\dots,\sigma_n$ be the elements of $Gal(L/K)$. Let $S$ be $R$'s integral closure in $L$ and let $\mathfrak{a}$ be an ideal of $S$. Consider the following statement:

The product ideal $\sigma_1(\mathfrak{a})\dots\sigma_n(\mathfrak{a})$ is generated over $S$ by its intersection with $R$.

How general is this statement? Does it always hold? Does it hold whenever $R,S$ are Dedekind domains? Is there a counterexample even when they are Dedekind? Does it depend on the Galois group?

If $R,S$ are Dedekind domains, and if $L/K$ is either quadratic or biquadratic (i.e. if $Gal(L/K)$ is either $\mathbb{Z}/2$ or the Klein 4-group) then I have rather grisly, computation-heavy proofs of the above claim. However, it feels to me like something much more general is going on, and I wonder if I have overlooked a much cleaner argument.

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    for Dedekind domains it's true - if $\mathfrak{a}$ is prime then your product is a power of $(\mathfrak{a}\cap R)S$ (by the theorem about factorization of prime ideals in Galois extensions).2012-09-11
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    Where do you use the fact they're both Dedekind? Because I just read Lang's chapter about this stuff and although all the statements take maximal ideals, he says one just need to localize to make the proofs work for prime ideals as well, so I was wondering if one really need the Dedekind condition "prime=maximal". Thanks for your insights.2012-09-11
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    @Niccolò - my proofs used in an essential way the theorem that any ideal in a Dedekind domain is 2-generated. I took the generators for $\mathfrak{a}$ and wrote down generators for $\sigma_1(\mathfrak{a})\dots\sigma_n(\mathfrak{a})$ and its intersection with $R$, and showed that the former could be expressed in terms of the latter. It was very ugly. The arguments below are much more satisfying.2012-09-28

2 Answers 2

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  1. This doesn't hold in general as thought Makoto Kato.
  2. This holds if $\mathfrak a=0$ or if $\mathfrak a$ is invertible (e.g. if $R$ is a Dedekind domain).

Proof of (1). Consider $S=\mathbb C[X,Y]$ with the involution $\sigma$ permuting the variables $X, Y$. Let $$R:=S^{\langle \sigma \rangle}=\mathbb C[X+Y, XY].$$ Then $L/K$ is Galois whose Galois group is generated by $\sigma$, and $S$ is the integral closure of $R$ in $L$ because it is already integrally closed.Let $\mathfrak a=(X, Y)$. Then its norm is $(X^2, Y^2, XY)$, and its intersection with $R$ consists in symmetric polynomials in $((X+Y)^2, XY)R$. Now it is easy to see that in $S$, $X^2$ doesn't belong to the ideal generated by $(X+Y)^2$ and $XY$: if $$X^2=(X+Y)^2F(X,Y)+XYG(X,Y),$$ decompose $F, G$ as sums of their homogeneous components, then $X^2$ would be a linear combination of $(X+Y)^2, XY$, which is impossible.

Proof of (2). If $\mathfrak a=0$, there is nothing to prove. If $\mathfrak a$ is principal, generated by $f\in S$, then its norm is generated by $\prod_{\sigma\in \mathrm{Gal}(L/K)} \sigma(f)$ which belongs to $R$. We are done in this case.

In general, $\mathfrak a$ is only supposed to be invertible. We are comparing two ideals in $S$: $$N(\mathfrak a):=\sigma_1(\mathfrak a)...\sigma_n(\mathfrak a), \quad J:=(N(\mathfrak a)\cap R)S.$$ To prove they are equal, it is enough to prove they are equal locally. So we can localize $R$ at prime ideals (we can't localize at maximal ideals of $S$ because we would then lose the Galois setting). Notice that the construction of the ideals in question is compatible with localization in $R$. So we can suppose $R$ is local. As the extension $L/K$ is finite separable, $S$ is finite over $R$, therefore $S$ is semi-local. As $\mathfrak a$ is invertible and $\mathrm{Pic}(S)$ is trivial, $\mathfrak a$ is principal and we are reduced to the previous case.

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    Great. Can the following be proved asumming only $\mathfrak{a}$ is invertible? $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) = (\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) \cap R)S$2012-09-12
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    @MakotoKato, this is what I was supposed to prove in (2).2012-09-12
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    I think you proved that $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a})$ is generated by an ideal of $R$. It is not clear to me that $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a})$ is generated by the ideal $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) \cap R$.2012-09-12
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    Dear @MakotoKato, I edit the answer to make clear the ideals I was comparing. On the otherhand, an ideal $I$ of $S$ generated by an ideal $J$ of $R$ is necessarily generated by $I\cap R$: $JS=I$ implies that $J\subseteq I\cap R$, hence $I=JS\subseteq (I\cap R)S\subseteq I$. QED.2012-09-12
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    Thanks. I think I should edit my answer.2012-09-12
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    @QiL - Why is $\operatorname{Pic}(S)$ trivial? Is this a theorem about semilocal rings?2012-09-28
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    @BenBlum-Smith: Yes, this is a standard fact.2012-09-30
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    @QiL - So I can understand the nuances of the argument: you are considering $M^{-1}S\supset M^{-1}R$ where $M$ is the complement of a prime ideal of $R$, right? If so, why is localizing at all prime ideals of $R$ sufficient to conclude equality between ideals of $S$?2012-10-10
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    @BenBlum-Smith:because two modules over a same ring are equal if they are equal over the localizations at all prime (or maximal) ideals. This is another standard fact in commutative algebra. You could think about case of Dedekind domains.2012-10-11
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    Let's see... $S$ is a module over $R$ and I have two submodules $M,N\subset S$ such that localization at all maximal ideals of $R$ yields equality. Without loss of generality, $M\subset N$ (if not, consider both in relation to their intersection). Then $0\rightarrow M\rightarrow N\rightarrow N/M\rightarrow 0$ is exact, and localizing it at all maximal ideals I find $(N/M)_\mathfrak{m}=0$; conclusion, $N/M=0$. Okay I buy it!2012-10-17
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I think the question is not generally true.

Suppose $S$ is a Dedekind domain. This condition is satisfied if $R$ is a Dedekind domain.

Let $S[X_1, ..., X_m]$ be the polynomial ring over $S$. Let $F ∈ S[X_1, ..., X_m]$. We denote by $I(F)$ the ideal of $S$ generated by all the coefficients of $F$. Given $F, G ∈ S[X_1, ..., X_m]$, it can be proved by a simple argument which goes back to Gauss that $I(FG)$ = $I(F)I(G)$(for example, see theorem 13 in Hilbert's "The theory of algebraic number fields"). Note that the proof uses essentially that $S$ is a Dedekind domain.

Let $\mathfrak{a}$ be an ideal of $S$. Since $S$ is a Dedekind domain, $\mathfrak{a}$ is finitely generated. Suppose $\alpha_1,\dots,\alpha_r$ generates $\mathfrak{a}$. Let $F = \alpha_1X_1 + \cdots + \alpha_rX_r \in S[X_1,\dots,X_r]$. For $\sigma \in Gal(L/K)$, we write $\sigma(F) = \sigma(\alpha_1)X_1 + \cdots + \sigma(\alpha_r)X_r$.Clearly $\sigma_1(F)\cdots\sigma_n(F) \in R[X_1,\dots,X_r]$. Let $\mathfrak{b}$ be the ideal of $R$ generated by the coefficients of $\sigma_1(F)\cdots\sigma_n(F)$.

Then $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) = I(\sigma_1(F))\cdots I(\sigma_n(F)) = I(\sigma_1(F)\cdots\sigma_n(F)) = \mathfrak{b}S$. Hence $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) = (\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) \cap R)S$.

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    Good idea to consider $F$ and $I(F)$!2012-09-12