# Decomposition and inertia fields in the factorization of $3$ in $\mathbb{Q}(\zeta_{24})$

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 $$H with order $$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 $$E$$?

• By the way, you can find $\mathfrak{p}$ and $\mathfrak{q}$ explicitly in the factorization of $(3)$ using Proposition (8.3) in Neukirch's ANT book. The 24-th cyclotomic polynomial is $x^8 - x^4 + 1$, and its factorization mod $3$ is $(x^2 + x + 2)^2(x^2 + 2x + 2)^2$. The proposition then says that the primes are $\mathfrak{p} = \langle 3, \zeta_{24}^2 + \zeta_{24} + 2\rangle$ and $\mathfrak{q} = \langle 3, \zeta_{24}^2 + 2\zeta_{24} + 2\rangle$. – Tob Ernack May 18 at 22:04
• For the decomposition group, maybe you can find the explicit subgroup of $G$ (in terms of maps $\zeta_{24} \to \zeta_{24}^i, \gcd(i, 24) = 1$) that sends $\zeta_{24}^2 + \zeta_{24} + 2$ back to $\mathfrak{p}$. – Tob Ernack May 18 at 22:11
• To be honest, I just used WolframAlpha. But given the theorem that you mentioned about prime factorizations in cyclotomic fields, you can already guess the form of the factorization, and use a bit of brute force on the irreducible quadratics mod $3$. – Tob Ernack May 18 at 22:28
• I think the approach hinted at in the problem statement might be more elegant actually. I haven't thought it through yet but it might spare you these computations. – Tob Ernack May 18 at 22:33
• Ok looking at their approach, one idea could be that the fixed field of $\langle \overline{d}\rangle$ is $\mathbb{Q}\left(\zeta_{24} + \zeta_{24}^{d}\right)$ (I haven't proved that). The minimal polynomial of $\zeta_{24} + \zeta_{24}^d$ can be computed for each $d$ in $\{1, 5, 11, ..., 23\}$ (incidentally this would prove that the fixed fields really are what I said, by checking that the degree is $4$). Then you can check whether $3$ ramifies by checking whether it divides the discriminant. This approach should work although there might be a smarter way to avoid the computations. – Tob Ernack May 18 at 23:43

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 $$\mathbb{Q}(i\sqrt{2})$$.
• How did you find out that $\mathbb{Q}(\zeta_8)$ corresponts to $H=\{1,17\}$ from the fact that $\text{Gal}(\mathbb{Q}(\zeta_{24})|\mathbb{Q})\simeq \text{Gal}(\mathbb{Q}(\zeta_8)|\mathbb{Q})\times\text{Gal}(\mathbb{Q}(\zeta_3)|\mathbb{Q})$? – rmdmc89 May 31 at 0:00
• And how did you conclude that the quadratic subfields of $F$ are $\mathbb{Q}(i)$, $\mathbb{\sqrt{2}}$, $\mathbb{Q}(i\sqrt{2})$? I'm sure there are many ways to do it, but I'm curious to know how you did it – rmdmc89 May 31 at 0:09
• @rmdmc89 From the Chinese Remainder's Theorem we have that $(\mathbb{Z}/(24))^\times \cong (\mathbb{Z}/(8))^\times \times (\mathbb{Z}/(3))^\times$, where the isomorphism is given by $n \to (n \mod 8, n\mod 3)$. Now the group fixing $\mathbb{Q}(\zeta_8)$ is given by $\{1\} \times (\mathbb{Z}/(3))^\times$, which under the isomorphism corresponds to elements of $(\mathbb{Z}/(24))^\times$ having remainder $1$ modulo $8$. They are exactly $1$ and $17$. – Stefan4024 May 31 at 8:00
• @rmdmc89 One way is to use the Galois group of $\mathbb{Q}(\zeta_8)$ and see what elements are fixed by the subgroups of order $2$. However this method is tedious. The easier method would be to use the explicit form od $\zeta_8$, i.e. $\frac{1+i}{\sqrt{2}}$. We have $\zeta_8^2 = \frac{1+2i-1}{2} = i$. Thus $\mathbb{Q}(i) \subset F$. Also we have that $\zeta_8 + \zeta_8^{-1} = \frac{1+ i}{\sqrt{2}} + \frac{1-i}{\sqrt{2}} = \sqrt{2}$. Thus $\mathbb{Q}(\sqrt{2}) \subset F$. From above we also have that $\mathbb{Q}(i\sqrt{2}) \subset F$. Since there are 3 quadratic fields we have found them all. – Stefan4024 May 31 at 8:08