Galois principle for ideals Let $L/K$ be a finite Galois extension of number fields with Galois group $G$. Determine a necessary and sufficient condition on $L/K$ to
ensure that 
$$\{I\in \text{Id}_L,\text{ such that }\sigma (I)=I, \forall \sigma \in G\}=\text{Id}_K$$
where $\text{Id}_K$ is seen as a subgroup of $\text{Id}_L$ through the extension
map $I\rightarrow IO_{L}$.
So here is my argument. One can show that if ${Q_i}\subset O_L$ are all the primes lying over a prime $P\subset O_K$, then $\forall \sigma \in G$, we have $\sigma (Q_i)$ is one of the $Q_j$'s. Now we know that there must exist a prime ideal $P$ which ramifies in $O_L$, say $\displaystyle P=\prod Q_i^{e}$, where $e>1$. Then I can take $I=\prod Q_i$ which clearly is fixed by every automorphism and not in $\text{Id}_K$, and thus nothing else besides $L=K$ works. 
However, because of the way the question is asked I have suspicions that the above is not completely true. Any help?
EDIT: I already see where this failed, thanks to Mathmo123, but I guess I will leave it here if someone wants to show how this is done.
 A: Your mistake is this:

It's not true that there must exist a prime ideal of $\mathcal O_K$ which ramifies in $\mathcal O_L$. 

For example, the extension $\mathbb Q({\sqrt{-5}},i)/\mathbb Q(\sqrt{-5})$ is unramified at all primes. Note that the class group of $\mathbb Q({\sqrt{-5}})$ is cyclic of order $2$.
In general, one of the key results of class field theory is that there is a maximal extension $L$ of a number field $K$ that is unramified at all primes, such that $\mathrm{Gal}(L/K)$ is isomorphic to the (narrow) class group of $K$. In particular, if $K=\mathbb Q$, you are correct that we must have $L=\mathbb Q$.
Therefore a necessary condition is that $L$ must be unramified at all primes.
To see why this is sufficient, let $I$ be any fractional ideal of $L$ and write
$$I=\mathfrak{p}_1^{a_1}\cdots\mathfrak{ p}_k^{a^k}$$where the $a_i$ are non-zero integers and the $\mathfrak p_i$ are primes. Suppose that $\sigma(I)=I$ for all $\sigma\in G=\mathrm{Gal}(L/K)$.
Since the $G$ acts transitively on the primes above any prime of $K$ and $L/K$ is unramified at all primes,  we deduce that if $\mathfrak p^{a}\mid I$, then $\mathfrak q^a\mid I$, where $\mathfrak q = (\mathfrak p\cap \mathcal O_K)\mathcal O_L=\prod_{\text{primes $\mathfrak p'$ above }\mathfrak p\cap\mathcal O_L}\mathfrak p'$.
A: As a complement to Matmo123's answer, I would like to take a cohomological approach, not for the sake of « abstract non sense », but in order to show that your problem is just an introduction the so called « genus theory » for class groups. Notations : $L/K$ is finite Galois with group $G$, $I_K$ and $I_L$ are the respective groups of fractional ideals. There are two canonical maps relating $I_K$ and $I_L$ , the extension map $\epsilon : I_K -->I_L^{G}$ and the norm map $\nu : I_L-->I_L^G$. In the Galois situation, they are defined as follows on prime ideals : 
1). If $P$ is a prime ideal of $K$, $\epsilon (P)$ is the product of the powers $Q^{e_P}$ for all the primes $Q$ of $L$ above $P$ . Note that $G$ permutes transitively all the $Q$’s, and the ramification index $e_P$ depends only on $P$
2). If $Q$ is a prime ideal of $L$ above $P$, $\nu (Q)$ is the product of all the $Q$’s
Let us first determine $I_L^G /\epsilon (I_K)$ . The key point is that by definition $I_L$ is a free $\mathbf Z [G]$-module over the prime ideals of $L$, which implies that its Tate cohomology vanishes, in particular $\hat H^0(G, I_L) = 0 $ (see e.g. Serre’s « Local Fields », chapter 8). This means exactly that $I_L^{G} = \nu (I_L)$, and this gives immediately that $ I_L^{G}/\epsilon (I_K) = \nu (I_L) / \epsilon (I_K) = \Sigma  {\mathbf Z / e_P \mathbf Z}$ (the latter sum makes sense  because almost all ${e_P}$ are 1). In particular, $ I_L^{G} = \epsilon (I_K)$ iff $L/K$ is unramified.
Now let us tackle a much more interesting problem concerning the class groups $C_K$ and $C_L$, more specifically the determination of the kernel and cokernel, denoted resp. $Cap(L/K)$ and $Cocap(L/K)$, of the natural map $C_K --> C_L^{G}$ induced by $\epsilon$. Starting from the definition of the class groups and manipulating cohomology, we can get an exact sequence  $1 -->Cap(L/K)--> Pr_L^{G}/Pr_K--> I_L^{G}/\epsilon (I_K)--> Cocap (L/K)-->H^1(G, Pr_L)-->1$, where Pr(.) is the group of principal ideals. In particular cases the calculations can be pushed further : if $L/K$ is unramified, $Cap (L/K) = H^1(G, U_L)$, where $U (.)$ is the group of units ; if $G$ is cyclic, one can express the quotient #$ C_L^{G}$/#$C_K$ in normic terms (this is Chevalley’s « ambiguous class formula »), etc.
