Let $K$ be a number field, let $\mathscr{O}_K$ denote the ring of algebraic integers of $K$ and let $d_K$ be the discriminant of $\mathscr{O}_K$, i.e. the discriminant of a $\mathbb{Z}$-basis for $\mathscr{O}_K$. I know that if a rational prime $p$ ramifies in $\mathscr{O}_K$, then $p$ divides $d_K$. Is it true also the converse? If a prime $p$ divides $d_K$, then $p$ ramifies in $\mathscr{O}_K$?
[Math] Primes ramifying in a number field
algebraic-number-theory
Related Solutions
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
This looks good to me. Your concern about different ramification indices is not an issue in this case because every extension in sight is Galois. cf Corollary 2.15 of http://www.math.umass.edu/~weston/cn/notes.pdf .
Best Answer
Yes, this is true.
More generally, if you define the discriminant ideal $\Delta_{L/K}$ of an extension of number fields $L/K$ to be the ideal of $\mathcal{O}_K$ generated by elements of the form $\text{disc}(\omega_1,\ldots,\omega_n)$ where $\{\omega_1,\ldots,\omega_n\}$ is a basis of $L/K$ contained in $\mathcal{O}_L$, then a prime $\mathfrak{p}$ of $\mathcal{O}_K$ ramifies in $L$ if and only if $\mathfrak{p}\mid \Delta_{L/K}$.
In the special case when $K=\mathbb{Q}$, you have that $\Delta_{L/\mathbb{Q}}=d_L\mathbb{Z}$, and so your result follows.
You can look up a proof of the above, but I think the most informative case is when $L/\mathbb{Q}$ is monogeneic. In other words, $\mathcal{O}_L=\mathbb{Z}[\alpha]$. In this case, the Dedekind-Kummer theorem tells you that a prime $p$ ramifying in $L$ is the same thing as $\overline{m_{\alpha,\mathbb{Q}}(T)}$ having multiple roots in $\mathbb{F}_p[T]$ (where $m_{\alpha,\mathbb{Q}}\in\mathbb{Z}[T]$ is the minimal polynomial for $\alpha$ and the bar denotes passing to $\mathbb{F}_p[T]$). But, $\overline{m_{\alpha,\mathbb{Q}}(T)}$ having multiple roots means that the polynomial discriminant $\text{disc}(\overline{m_{\alpha,\mathbb{Q}}(T)})\in\mathbb{F}_p$ is zero. But, it's not hard to check that $\text{disc}(\overline{m_{\alpha,\mathbb{Q}}(T)}=\overline{\text{disc}(m_{\alpha,\mathbb{Q}}(T)})$. Thus, $p$ should ramify if and only if $\text{disc}(m_{\alpha,\mathbb{Q}}(T))$ is in $(p)$ ,or that $(p)\mid (\text{disc}(m_{\alpha,\mathbb{Q}}(T)))$. But, a quick check shows that $d_L=\pm \text{disc}(m_{\alpha,\mathbb{Q}}(T))$, and so we see that $p$ ramifies if and only if $(p)\mid (d_L)$.