Suppose that $K$ is a number field and $\mathfrak p$ is a prime ideal non-zero.
Suppose that for each $n\in\mathbb N$ there exist a finite Galois extension $K\subseteq L$ such that $$\mathfrak p=(\mathfrak P_1\cdots \mathfrak P_n)^r.$$
The Galois group $\text{Gal}(L/K)$ acts transitively on $\{\mathfrak P_1,\ldots,\mathfrak P_n\}$, and this gives a homomorphism $$\varphi:\text{Gal}(L/K)\to S_n.$$
Furthermore we have that the image of $\varphi$ is a transitive subgroup of $S_n$.
If $n=3$, the symmetric group $S_3$ has two transitive subgroups, $G=\langle (123) \rangle$ and $S_3$.
Do you know two examples of number fields $L$ satisfying the following requirements:
(1) $L$ is a finite Galois extension of $K$,
(2) $\mathfrak p=(\mathfrak B_1\mathfrak B_2\mathfrak B_3)^r, r\in\mathbb N$
(3) the imagen of the homomorphism $\varphi:\text{Gal}(L/K)\to S_3$ is $G$, and another example such that the image is $S_3$.
++Thanks you all
If you extend $K=\mathbb{Q}$ to the splitting field $L$ of $x^3-5$, you get a Galois extension of degree $6$. Letting $r=\sqrt[3]{5}\in \mathbb{R}$ and $z=e^{2\pi i/3}$, you get $[\mathbb{Q}(r):\mathbb{Q}]=3$ via minimal polynomial $x^3-5$ and $[\mathbb{Q}(r,z):\mathbb{Q}(r)]=2$ via minimal polynomial $x^2+x+1$, with $L=\mathbb{Q}(r,z)$ the splitting field. You can show that the Galois group $Gal(L/K)$ must be isomorphic to $S_3$, since it must have at least $6$ automorphisms but can act transitively on no more than $3$ elements.
The prime ideal generated by $13$ has three distinct prime ideal factors by Dedekind's Theorem because $x^3-5$ factors as $(x + 2) (x + 5) (x + 6)$ mod $13$. The image of the homomorphism $\varphi$ must be all of $S_3$, because no map from $S_3\to C_3$ preserves the group structure.
Now, if you extend $K$ to the splitting field $L$ of $x^7-1$, you get a Galois extension of degree $6$ with a different group structure. Letting $w=e^{2\pi i/7}$, you get $[\mathbb{Q}(w):\mathbb{Q}]=6$ via minimal polynomial $x^6+x^5+x^4+x^3+x^2+x+1$, with $L=\mathbb{Q}(w)$ the splitting field. You can show that the Galois group $Gal(L/K)$ must be isomorphic to $C_6$. Any automorphism $\sigma$ in the group must map $\sigma(w)=w^k$ for some $k$ in $1,\ldots,6$. Once this image is fixed, all of the other images are fixed as well as a cyclic permutation of order $6$ in the powers of $w$. Since all of $1,\ldots,6$ are coprime with $7$, they form a group under multiplication mod $7$, and satisfy $k^6=1$ mod $7$, making $\sigma^6=id$.
The prime ideal generated by $13$ has three distinct prime ideal factors by Dedekind again because $x^6+x^5+x^4+x^3+x^2+x+1$ factors as $(x^2 + 3 x + 1) (x^2 + 5 x + 1) (x^2 + 6 x + 1)$ mod $13$. The image of the homomorphism $\varphi$ must be the subgroup in $S_3$ isomorphic to $C_3$, because if it were all of $S_3$ it would be an isomorphism, and $C_6$ is not isomorphic to $S_3$.
Edit: Whether this same construction can be done for an arbitrary $\mathfrak p$? The first trick of finding a cubic extension whose Galois group is exactly $S_3$ works if $\mathfrak p=1$ mod $3$. The second trick of finding a cyclotomic extension works if you can find another prime $p'=1$ mod $3$ so that the order of $\mathfrak p$ mod $p'$ is exactly $n'=(p'-1)/3$, that is, $\mathfrak p^{n'}=1$ mod $p'$. So I was able to construct both extensions for $\mathfrak p=13$. The second trick can work when the first trick fails. For example, $2$ has exactly three prime ideal factors over it in the field extension that splits $x^{43}-1$ (check the factors of the $43$rd cyclotomic polynomial mod $2$) with Galois group isomorphic to $C_{42}$ and $\varphi$ mapping it into $C_3$.
You might look at Class Field Theory, Kummer Theory, and the Artin reciprocity laws. I am not deeply familiar with these, I just thought it was interesting to try to construct extensions of the type you described to see if I could do it. There seem to be 'obvious' patterns of the kind that are easy to see but very hard to pin down and prove generally, which I believe make for some of the most interesting mathematics.