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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 $E$?

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    $\begingroup$ 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 $\mathfrak{q} = \langle 3, \zeta_{24}^2 + 2\zeta_{24} + 2\rangle$. $\endgroup$
    – Tob Ernack
    May 18 '19 at 22:04
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    $\begingroup$ 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 $\mathfrak{p}$. $\endgroup$
    – Tob Ernack
    May 18 '19 at 22:11
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    $\begingroup$ 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 $3$. $\endgroup$
    – Tob Ernack
    May 18 '19 at 22:28
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    $\begingroup$ 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. $\endgroup$
    – Tob Ernack
    May 18 '19 at 22:33
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    $\begingroup$ 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. $\endgroup$
    – Tob Ernack
    May 18 '19 at 23:43
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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 $\mathbb{Q}(i\sqrt{2})$.

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  • $\begingroup$ How did you find out that $\mathbb{Q}(\zeta_8)$ corresponts to $H=\{1,17\}$ from the fact that $\text{Gal}(\mathbb{Q}(\zeta_{24})|\mathbb{Q})\simeq \text{Gal}(\mathbb{Q}(\zeta_8)|\mathbb{Q})\times\text{Gal}(\mathbb{Q}(\zeta_3)|\mathbb{Q})$? $\endgroup$
    – rmdmc89
    May 31 '19 at 0:00
  • $\begingroup$ 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 $\endgroup$
    – rmdmc89
    May 31 '19 at 0:09
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    $\begingroup$ @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 $17$. $\endgroup$
    – Stefan4024
    May 31 '19 at 8:00
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    $\begingroup$ @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. $\endgroup$
    – Stefan4024
    May 31 '19 at 8:08

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