This is a solution to Exercises 4.6 and a partial solution to exercise 4.7 in Daniel A. Marcus, Number Fields.
Let \(K = \mathbb{Q}[\sqrt{m}, \sqrt{n}]\) be a field extension of the rational numbers \(\mathbb{Q}\). \(K\) has degree 4 and contains three quadratic subfields: \(\mathbb{Q}[\sqrt{m}]\), \(\mathbb{Q}[\sqrt{m}]\), and \(\mathbb{Q}[\sqrt{k}]\), where \(k = \frac{mn}{\text{gcd}(m, n)^2}\). The Galois group of this field extension is the Klein 4-group, which has three normal subgroups. Let \(p\) be a prime of \(\mathbb{Z}\). Take \(R\) to be the ring of integers of \(K\).
Let \(P\) be a prime ideal with \(Q\) a prime lying over \(P\). If \(K\) is a normal extension of the rationals with Galois Group \(G\), then the decomposition group \(D\) of \(P\) is the subgroup of \(G\) containing all \(\sigma\) such that \(\sigma(Q) = Q\). The inertia group \(E\) for a prime ideal \(P\) is the subgroup of \(G\) containing all \(\sigma\) such that \(\sigma(\alpha) \equiv \alpha\ \text{mod}\ Q\) for \(\alpha\) an algebraic integer of \(K\).
Background: Factoring in Quadratic Fields
Let \(m\) be a squarefree integer.
Let \(p = 2\). \(p\) ramifies if \(m \equiv 0, 2, 3\ (4)\) since the discriminant of the quadratic subfield is even in these cases. \(2R\) splits if \(m \equiv 1\ (4)\).
Let \(p\) be an odd prime. If \(p \mid m\), then \(p\) ramifies. If \(m\) is a quadratic residue mod \(p\), then \(p\) splits in \(\mathbb{Q}[\sqrt{m}]\). Otherwise \(p\) remains prime (inert).
\(p\) ramified in every subfield
If \(p\) is ramified in each of the subfields then \(p\) is totally ramified in \(K\). \(p\) being totally ramified implies that the both the decomposition and inertia groups of \(p\) over \(R\) are the entire Galois group. To find an example we need to find \(m, n\) such that \(p\) divides both \(m\) and \(n\) along with the \(k\) expression above.
Take \(K = \mathbb{Q}[\sqrt{2}, \sqrt{3}]\), so \(m = 2\) and \(n = 3\). \(K\) then contains the subfields \(K_1 = \mathbb{Q}[\sqrt{2}]\), \(K_2 = \mathbb{Q}[\sqrt{3}]\), and \(K_3 = \mathbb{Q}[\sqrt{6}]\) with corresponding ring of integers \(R_1, R_2, R_3\) and discriminants \(8, 12, 24\). The prime \(p = 2\) ramifies in each of these subfields.
We can confirm this in Sage:
sage: for i in [2, 3, 6]:
....: print(QuadraticField(i).ideal(2).factor())
....:
(Fractional ideal (a))^2
(Fractional ideal (a + 1))^2
(Fractional ideal (-a + 2))^2
sage: L = QQ.extension(x^2 - 2, 'a').extension(x^2 - 3, 'b')
sage: L.absolute_polynomial()
x^4 - 10*x^2 + 1
sage: L = QQ.extension(x^4 - 10*x^2 + 1, 'a')
sage: L.ideal(2).factor()
(Fractional ideal (-1/2*a*b + 1/2*a - 1))^4
sage: G = L.galois_group()
sage: L.primes_above(2)
[Fractional ideal (-1/4*a^3 + 1/4*a^2 + 9/4*a - 9/4)]
sage: G.decomposition_group(L.primes_above(2)[0]).order()
4
sage: G.inertia_group(L.primes_above(2)[0]).order()
4
\(p\) splits completely in every subfield
If \(p\) splits completely in each subfield, then it will split completely in the composite field (Marcus, Theorem 31). The decomposition group of \(p\) is the empty subgroup - every element of the Galois group permutes a prime lying over \(p\) it to a different prime lying over \(p\), so no permutations fix the primes over \(p\).
To find an example, we note that for \(p\) to split completely in every subfield, it must not divide the discriminant of each subfield, and each subfield must be a quadratic residue mod \(p\).
Let’s take \(p = 131\) (I just wanted to try a larger number than the usuals ;)), and observe using Sage that \(15, 21,\) and \(35 = \frac{15 \cdot 21}{\text{gcd}(15, 21)^2}\) are each quadratic residues mod \(131\), so \(131\) will split completely in each subfield.
We then confirm through Sage that 131 splits completely each subfield and then over the biquadratic field \(\mathbb{Q}[\sqrt{15}, \sqrt{21}]\):
sage: for m in [15, 21, 35]:
....: print(QuadraticField(m).ideal(131).factor())
....:
(Fractional ideal (-3*a + 2)) * (Fractional ideal (3*a + 2))
(Fractional ideal (5/2*a + 1/2)) * (Fractional ideal (5/2*a - 1/2))
(Fractional ideal (-2*a + 3)) * (Fractional ideal (2*a + 3))
sage: L = QQ.extension(x^2 - 15, 'a').extension(x^2 - 21, 'b')
sage: L.ideal(131).factor()
(Fractional ideal ((1/3*a - 1)*b + 2*a - 7)) * (Fractional ideal ((-1/6*a - 1)*b + 3/2*a + 7)) * (Fractional ideal ((1/3*a - 3/2)*b + a - 11/2)) * (Fractional ideal ((1/3*a + 1)*b - a - 2))
\(p\) inert in every subfield
This scenario can never occur. If \(p\) were inert in every subfield, then the decomposition group of \(p\) would be the entire Galois group, while the inertia group would be the empty subgroup. There would then be a surjective mapping from \(D / E = G\) onto a cyclic group of order 4 (the inertial degree of \(p\)), but this contradicts what we know about the structure of the Galois group (the Klein 4-group is not cyclic).
\(p\) splits into two primes \(p_1 p_2\)
We need \(p\) to split in one subfield and be inert in another. Take \(p = 13, m = 3, n = 5\). Here \(m = 3\) is a quadratic residue mod 13, but \(n = 5\) is not. Therefore 13 stays inert in \(\mathbb{Q}[\sqrt{5}]\) but splits in \(\mathbb{Q}[\sqrt{3}]\). Sage confirms that 13 has the desired form in the biquadratic field:
sage: for m in [3, 5]:
....: print(QuadraticField(m).ideal(13).factor())
....:
(Fractional ideal (a + 4)) * (Fractional ideal (a - 4))
Fractional ideal (13)
sage: QQ.extension(x^2 - 5, 'a').extension(x^2 - 3, 'b').ideal(13).factor()
(Fractional ideal (b + 4)) * (Fractional ideal (b - 4))
Note that since \(15 \equiv 2\ (13)\) is not a quadratic residue mod 13, 13 stays inert in \(\mathbb{Q}[\sqrt{15}]\), so this provides a scenario where a prime can be inert in two fields (here the quadratic subfields \(m = 5, n = 15\)) but not be inert in their composite (Marcus, Exercise 4.7 (c)).
\(p\) splits into two primes \(p_1^2 p_2^2\)
We need \(p\) to split in one subfield and ramify in another. Take \(p = 7, m = 2, n = 14\). Here \(m = 2\) is a quadratic residue mod \(7\), and the quadratic field \(\mathbb{Q}[\sqrt{14}]\) will have a discriminant divisible by 7, since 7 divides 14.
sage: for m in [2, 14]:
....: print(QuadraticField(m).ideal(7).factor())
....:
(Fractional ideal (-2*a + 1)) * (Fractional ideal (2*a + 1))
(Fractional ideal (-2*a + 7))^2
sage: QQ.extension(x^2 - 14, 'a').extension(x^2 - 2, 'b').ideal(7).factor()
(Fractional ideal ((1/2*a - 3/2)*b + 1/2*a - 1))^2 * (Fractional ideal (1/2*b - 1/2*a + 2))^2
The third quadratic subfield of \(K\) is \(\mathbb{Q}[\sqrt{7}]\), where \(7\) is totally ramified. This provides an example where a prime can be totally ramified in two fields (here the quadratic subfields \(m = 7, n = 14\)) but not totally ramified in their composite (Marcus, Exercise 4.7 (a)).
\(p\) ramifies as \(p_1^2\)
We need \(p\) to ramify in one subfield and be inert in another. Take \(p = 13, m = 5, n = 13\). 5 is not a quadratic residue mod 13, so 13 will stay inert in that quadratic subfield. Sage confirms:
sage: for m in [5, 13]:
....: print(QuadraticField(m).ideal(13).factor())
....:
Fractional ideal (13)
(Fractional ideal (-a))^2
sage: QQ.extension(x^2 - 5, 'a').extension(x^2 - 13, 'b').ideal(13).factor()
(Fractional ideal (-b))^2