Let $F$ be a number field and consider the cyclotomic extension $E = F(\zeta_{10})$ where $\zeta_{10}$ is a primitive 10th root of unity. Why is it true that the only primes of $F$ that ramify in in $E$ lies above 2 and 5?
[Math] Primes that ramify in a cyclotomic extension
abstract-algebraalgebraic-number-theory
Related Solutions
The answer is certainly yes, with the caveat that it depends on how you specify $L$.
Let me just give one example, that of "ramification calculus," the game of figuring out ramification indices by using the fact that such indices are multiplicative in towers of extensions. The idea here is to use your knowledge of previously-determined ramification in other fields to determine the ramification in your given field.
For example, if $L\subset \mathbb{Q}(\zeta_p)$ (for $\zeta_p$ a primitive $p$-th root of unity) is of degree $n\mid p-1$, then $p$ is totally ramified (i.e., with ramification index $n$) in $L$ just by the multiplicativity \begin{equation} p-1=e_p(\mathbb{Q}(\zeta_p):\mathbb{Q})=e_p(\mathbb{Q}(\zeta_p):L)e_p(L:\mathbb{Q}) \end{equation} and the observation that $e_p(K_2:K_1)\leq [K_2:K_1]$ for any extension $K_2/K_1$. (Of course, one should really be careful about taking a prime $\mathfrak{p}$ above $p$ in $L$ and writing $e_{\mathfrak{p}}(\mathbb{Q}(\zeta_p):L)$, etc., but never mind that.)
In some sense, this particular example is sneakily replacing the "ramifies if and only if divides the discriminant" property with the "ramifies if and only if it divides the conductor" property -- in fact, this conductor approach might provide a more global "yes" to your original question. While the conductor and discriminant are certainly related, they come from different places (in some vague and not particularly rigorous sense) . In any case, note that you didn't have to compute the discriminant of $L$ for this to work -- the one-time overhead of computing the discriminant of $\mathbb{Q}(\zeta_p)$ suffices to give the ramification information for all $L\subset \mathbb{Q}(\zeta_p)$. A similar analysis can be done for $\mathbb{Q}(\zeta_n)$, so if you know $L$ is abelian, you've got a pretty good headstart.
This sort of calculus can get fancier and fancier (and honestly, is pretty fun to work out). For example, I frequently encounter the following situation: Suppose you know an extension $K/\mathbb{Q}$ such that $KL/L$ and $K/\mathbb{Q}$ have easily-determined ramification. Then the two tower-multiplicativity statements
\begin{equation} e_p(KL:L)e_p(L:\mathbb{Q})=e_p(KL:\mathbb{Q})=e_p(KL:K)e_p(K:\mathbb{Q}) \end{equation}
might let you solve for $e_p(L:\mathbb{Q})$, i.e., determine the ramification in $L$, from known information about ramification in $K$.
One final thought is that there are a local-global relationships for ramification, coming from (for example) that the conductor is a product of local conductors. Local information can come in handy for determining seemingly-unrelated global ramification --Abhyankar's Lemma comes to mind as a first instantiation of this.
Hope that helps.
Notations
We denote by $|S|$ the number of elements of a finite set $S$.
Let $A$ be a ring. We denote by $A^{\times}$ the group of invertible elements of $A$.
Let $f(X) \in \mathbb{Z}[X]$ and $p$ be a prime number. We denote by $\bar f(X)$ the reduction of $f(X)$ (mod $p$).
Fixed symbols
We fix the following symbols.
Let $l$ be a prime number(possibly 2).
Let $h \geq 1$ be an integer.
Let $\zeta$ be a primitive $l^h$-th root of unity in $\mathbb{C}$.
Let $K = \mathbb{Q}(\zeta)$.
Lemma 1 Let $\Omega$ be an algebraically closed field of characteristic $p$($p$ may be 0). Suppose $p \neq l$. Let $S$ be the set of roots of $X^{l^h} - 1$ in $\Omega$. Then $S$ is a cyclic subgroup of $\Omega^{\times}$ of order $l^h$. Let $P$ be the set of generators of $S$. Then $|P| = l^{h-1}(l - 1)$. Let $\Phi(X) = \prod_{\omega\in P}(X - \omega)$. Then $\Phi(X) = (x^{l^{h-1}})^{l-1} + (x^{l^{h-1}})^{l-2} +\cdots+ x^{l^{h-1}} + 1$.
Proof: Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$ and $l \neq p$, $X^{l^h} - 1$ has no multiple roots in $\Omega$. Hence $|S| = l^h$. It is well known that $S$ is a cyclic subgroup of $\Omega^{\times}$. Let $P$ be the set of generators of $S$. Then $|P| = l^h - l^{h-1} = l^{h-1}(l - 1)$. $\Phi(X) = (X^{l^h} - 1)/(X^{l^{h-1}} - 1) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$. QED
Definition $\Phi(X)$ of Lemma 1 is called the cyclotomic polynomial of order $l^h$.
Lemma 2 Let $\zeta', \zeta$ be primitive $l^h$-th roots of unity in $\mathbb{C}$. Then $(1 - \zeta')/(1 - z)$ is a unit of $K$.
Proof: There exists an integer $a \geq 1$ such that $\zeta' = \zeta^a$. Hence $(1 - \zeta')/(1 - \zeta) = \zeta^{a-1} +\cdots+ \zeta + 1$. Hence $(1 - \zeta')/(1 - \zeta)$ is an algebraic integer in $K$. Similarly $(1 - \zeta)/(1 - \zeta')$ is an algebraic integer in $K$. Hence $(1 - \zeta')/(1 - \zeta)$ is a unit of $K$. QED
Lemma 3 The following assertions hold.
(1) The cyclotomic polynomial $\Phi(X)$ of order $l^h$ is irreducible in $\mathbb{Q}[X]$.
(2) $\mathfrak{l} = (1 - \zeta)$ is a prime ideal of degree 1 of $K$.
(3) $l = \mathfrak{l}^{l^{h-1}(l - 1)}$.
Proof: Let $P$ be the set of primitive $l^h$-th roots of unity in $\mathbb{C}$. By Lemma 1, $\Phi(X) = \prod_{\zeta\in P}(X - \zeta) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$. Substituting $X = 1$, we get $l = \prod_{\zeta\in P}(1 - \zeta)$. By Lemma 2, $(l) = \mathfrak{l}^{l^{h-1}(l - 1)}$, where $\mathfrak{l} = (1 - \zeta)$. This proves (3).
Hence $l^{h-1}(l - 1) \leq [K : \mathbb{Q}]$. On the other hand, since $\Phi(\zeta) = 0$, $[K : \mathbb{Q}] \leq l^{h-1}(l - 1)$. Hence $l^{h-1}(l - 1) = [K : \mathbb{Q}]$. Hence $\Phi(X)$ is irreducible. This proves (1).
Since $[K : \mathbb{Q}] = l^{h-1}(l - 1)$, $\mathfrak{l}$ is a prime ideal of degree 1 by (3). This proves (2). QED
Lemma 4 The notations are the same as Lemma 3.
$(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$
Proof: $X^{l^h} - 1 = \Phi(X)(X^{l^{h-1}} - 1)$. Taking derivatives of the bothe sides, we get $l^hX^{l^h-1} = \Phi'(X)(X^{l^{h-1}} - 1) + l^{h-1}\Phi(X)X^{l^{h-1}-1}$. Substituting $X = \zeta$, we get $l^h\zeta^{l^h-1} = \Phi'(\zeta)(\zeta^{l^{h-1}} - 1)$.
Since $\zeta^{l^{h-1}}$ is a primitive $l$-th root of unity, $(l) = (\zeta^{l^{h-1}} - 1)^{l-1}$ in $\mathbb{Q}(\zeta^{l^{h-1}})$. Since $(l) = \mathfrak{l}^{l^{h-1}(l - 1)}$ in $K$, $(\zeta^{l^{h-1}} - 1) = \mathfrak{l}^{l^{h-1}}$ in $K$.
Hence $\mathfrak{l}^{hl^{h-1}(l - 1)} = \Phi'(\zeta)\mathfrak{l}^{l^{h-1}}$. Hence $(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$. QED
Lemma 5
The notations are the same as Lemma 3.
Let $d$ be the discriminant of $\Phi(X)$. Then $|d| = l^{l^{h-1}(h(l - 1) - 1)}$.
Proof: By Lemma 4, $(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$. Taking the norms of the both sides, we get $|d| = |N_{K/\mathbb{Q}}(\Phi'(\zeta))| = N(\mathfrak{l})^{l^{h-1}(h(l - 1) - 1)}$. Since $N(\mathfrak{l}) = l$, $|d| = l^{l^{h-1}(h(l - 1) - 1)}$. QED
Lemma 6 $\mathbb{Z}[\zeta]$ is the ring of algebraic integers of $K$.
Proof: This follows from Lemma 3, Lemma 5 and the proposition of my answer to this question.
Lemma 7 Let $\Phi(X)$ be the cyclotomic polynomial of order $l^h$ in $\mathbb{Q}[X]$. Let $p$ be a prime number such that $p \neq l$. Let $X^{l^h} - 1 \in \mathbb{Z}[X]$. Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$, $X^{l^h} - 1$ has no multiple irreducible factor mod $p$. Since $X^{l^h} - 1 = \Phi(X)(X^{l^{h-1}} - 1)$, $\Phi(X)$ has no multiple irreducible factor mod $p$. Let $\Phi(X) \equiv f_1(X)\cdots f_r(X)$ (mod $p$), where $f_i(X)$ are distinct monic irreducible polynomials mod $p$. Let $f$ be the smallest positive integer such that $p^f \equiv 1$ (mod $l^h$). Then the degree of each $f_i(X)$ is $f$.
Proof: Let $F = \mathbb{Z}/p\mathbb{Z}$. Let $\Omega$ be the algebraic closure of $F$. Let $S$ be the set of roots of $X^{l^h} - 1$ in $\Omega$. Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$, $X^{l^h} - 1$ has no multiple roots in $\Omega$. Hence $|S| = l^h$. It is well known that $S$ is a cyclic subgroup of $\Omega^{\times}$. Let $P$ be the set of generators of $S$. $|P| = l^h - l^{h-1} = l^{h-1}(l - 1)$. Let $\bar \Phi(X) = \prod_{\omega\in P}(X - \omega) \in F[X]$. Then $\bar \Phi(X) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$. Hence $\bar \Phi(X) = \bar f_1(X)\cdots\bar f_r(X)$.
Let $\omega$ be a root of \bar f_i(X) in $\Omega$. Since $\omega$ is a root of $\bar \Phi(X)$, $\omega \in P$. Let $E$ be the unique subfield of $\Omega$ such that $|E| = p^f$. It is well known that $E^{\times}$ is a cyclic group. Since $|E^{\times}| = p^f - 1$ and $l^h|p^f - 1$, $E^{\times}$ has a unique cyclic subgroup of order $l^h$. Hence $\omega \in E$.
Let $L$ be a proper subfiled of $E$. Let $[L : F] = p^r$. Suppose $\omega \in L$. Then $l^h|p^r - 1$, i.e. $p^r \equiv 1$ (mod $l^h$). Since $r < f$, this is a contradiction. Hence $E = F(\omega)$. Hence the minimal polynomial of $\omega$ over $F$ has degree $f$. Since $\bar f_i(X)$ is irreducible, $\bar f_i(X)$ is the minimal polynomial of $\omega$. This completes the proof. QED
Proposition The only prime number which ramifies in $K$ is $l$, except $l = 2$ and $h = 1$.
Proof: By Lemma 3, $l$ ramifies in $K$ except $l = 2$ and $h = 1$.
Let $p$ be a prime number such that $p \neq l$. Let $\Phi(X) \equiv f_1(X)\cdots f_r(X)$ (mod $p$) be as in Lemma 7. By this question, $P_i = (p, f_i(\zeta))$ is a prime ideal of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$. It is easy to see that $\mathbb{Z}[\zeta]/P_i$ is a finite extension of $\mathbb{Z}/p\mathbb{Z}$ of degree $f$. By Lemma 6, $\mathbb{Z}[\zeta]$ is the ring of algebraic integers in $K$. It is well known that each $P_i$ has the same ramification index $e$ and $l^{h-1}(l - 1) = efg$, where $g$ is the number of prime ideals of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$. Since $l^{h-1}(l - 1) = fr$, $r = g$ and $e = 1$. Hence $p$ does not ramify in $K$. This completes the proof. QED
Best Answer
In fact, the only primes that can ramify in $E/F$ are primes above 5.
This is because $E$ is the compositum of $F$ and $\mathbb{Q}(\zeta_{10})=\mathbb{Q}(\zeta_5)$ (this is true more generally for $n\equiv 2\pmod{4}$), and only 5 ramifies in $\mathbb{Q}(\zeta_5)/\mathbb{Q}$.
Added in response to the comment:
Here's the general principle in play: If $K$ is a number field, and $F$ and $G$ are two extensions of $K$ with compositum $E=FG$, and $\mathfrak{p}$ is a prime of $F$ above $p$ in $K$, then there is the following relationship between the ramification indices: $$ e_{\mathfrak{p}}(E/F)\leq e_p(G/K). $$
In your example (with $K=\mathbb{Q}$ and $G=\mathbb{Q}(\zeta_5)$) gives $$ e_{\mathfrak{p}}(F(\zeta_5)/F)\leq e_p(\mathbb{Q}(\zeta_5)/\mathbb{Q})=1 $$ for all $p\neq 5$.
The "$\leq$" occurs because $F/K$ can eat up/absorb/render redundant some of the ramification that occurs in $L/K$. Most notably, you could take $F=\mathbb{Q}(\zeta_5)$ itself when $p=5$ to get an obvious counter-example to equality.