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]:
....:
(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]:
....:
(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]:
....:
(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]:
....:
(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]: