I've seen the following exercise from an old problem sheet:
For $\zeta:=\zeta_{24}$ a primitive $24$-th root of unity and $\mathcal{O}:=\mathbb{Z}[\zeta]$, determine the prime decomposition of $3$. Determine the decomposition and inertia fields of the primes above $3$.
[Hint: show that there is a unique $4$-subextension $F$ of $\mathbb{Q}(\zeta)|\mathbb{Q}$ in which $3$ does not ramify, and that $F$ is the inertia field. Describe $F$ explicitly, then determine all quadratic fields $E$ under $F$ and find one where $3$ splits]
Using a famous theorem on the decomposition of primes in cyclotomic fields, we find easily that $3\mathcal{O}=(\mathfrak{p}\mathfrak{q})^2$ for some primes $\mathfrak{p}, \mathfrak{q}$.
For $G:=\text{Gal}(\mathbb{Q}(\zeta)|\mathbb{Q})$, we have $G\simeq(\mathbb{Z}/(24))^\times=\{\overline{1},\overline{5},\overline{7},\overline{11},\overline{13},\overline{17},\overline{19},\overline{23}\}$. Since $\overline{d}^2=\overline{1}$ for all $\overline{d}\neq \overline{1}$, then all subgroups $H<G$ with order $2$ are of the form $\langle\overline{d}\rangle$ with $\overline{d}\in G\setminus\{\overline{1}\}$. By the Galois correspondence, $F$ must have the form $\mathbb{Q}(\zeta)^H$ for some $H$ as above.
My questions are:
1) How do we know whether or not $3$ ramifies in $\mathbb{Q}(\zeta)^H$ for a given $H$?
2) Once we have $F$, how do we find $E$?
Use the fact that $\mathbb{Q}(\zeta_{24}) = \mathbb{Q}(\zeta_3)\mathbb{Q}(\zeta_8)$. Then $3$ won't ramify in $\mathbb{Q}(\zeta_8)$, as $3$ doesn't divide the discriminant of the field. This is your wanted subfield $F$. Obviously $F$ is the inertia field, as it's the biggest subfield in which ramification doesn't occur.
Moreover, using the fact that: $\text{Gal}(\mathbb{Q}(\zeta_{24})/\mathbb{Q}) \cong \text{Gal}(\mathbb{Q}(\zeta_{8})/\mathbb{Q}) \times \text{Gal}(\mathbb{Q}(\zeta_{3})/\mathbb{Q})$ we get that $\mathbb{Q}(\zeta_8)$ corresponds to $H = \{1,17\}$ in $(\mathbb{Z}/(24))^\times$
Now the quadratic subfields of $F$ are $\mathbb{Q}(i), \mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(i\sqrt{2})$. It's not hard to see that $3$ is inert in $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$, while it splits in $\mathbb{Q}(i\sqrt{2})$. Hence the decomposition field is $\mathbb{Q}(i\sqrt{2})$.