Galois theorem for ideals? 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.
 A: 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 $\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) = (\sigma_1(\mathfrak{a})\cdots\sigma_n(\mathfrak{a}) \cap R)S$.
A: 
  
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*This doesn't hold in general as thought Makoto Kato. 
  
*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. 
