1
$\begingroup$

Let $K$ be a number field and $\mathcal{O}_K$ be its ring of integers. Also let $p$ be a prime number, $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ and $\zeta_{m}$ be a primitive mth root of unity. I have two questions:

  1. When we work over $‎\mathbb{Q}$, the (finite) abelian extensions of $‎\mathbb{Q}$ which are unramified outside prime $p$ are subfields of $L=‎\mathbb{Q}(\zeta_{p^{r}})$ for some integer $r$. In fact, the ray class field of $‎\mathbb{Q}$ associated to the modulus $p^r$ is $L=‎\mathbb{Q}(\zeta_{p^{r}})$. Then, by a generalization of classical principal ideal theorem for Hilbert class field (see for example this page of MO), any unramified prime of $\mathbb{Q}$ (all primes other than $p$) becomes principal in $L=‎\mathbb{Q}(\zeta_{p^{r}})$. But for $p$, which is ramified, we have $p\mathcal{O}_L = (1-\zeta_{p^{r}})^{[L:‎\mathbb{Q}]}$, i.e. $p\mathcal{O}_L$ is principal and totally ramified.

    I think in any number field $K$ instead of $\mathbb{Q}$, for any prime power modulus $\mathfrak{p}^r$, the ray class field modulo $\mathfrak{p}^r$, $L=K(\mathfrak{p}^r)$, has this property: the unique ramified prime of $K$ becomes principal and totally ramified in $L$, i.e. $\mathfrak{p}\mathcal{O}_L=(a)^{[L:K]}$ for some $a \in L$.
    But I can't prove this, or find a theorem that say this.

  2. When we work over $‎\mathbb{Q}$, the ray class field for any modulus $m=p_1^{r_1}p_2^{r_2}...p_n^{r_n}$ is $L=\mathbb{Q}(\zeta_m)$. In this case ramified primes $p_i$ maybe don't become principal in $L$, but the product of all prime ideals of $L$ above $p_i$ is principal. In fact this product is equal to $(1-\zeta_{p_i^{r_i}})\mathcal{O}_L$. Remember that we saw unramified primes become principal in $L$.

    I think in any number field $K$ instead of $\mathbb{Q}$, for any modulus $\mathfrak{m}$, the ray class field modulo $\mathfrak{m}$, $L=K(\mathfrak{m})$, has this property: for any (finite) $\mathfrak{p}|\mathfrak{m}$, the product of all prime ideals of $L$ above $\mathfrak{p}$ is principal.
    But I can't prove this, or find a theorem that say this.

$\endgroup$
  • $\begingroup$ @Franz Lemmermeyer: Thank you. I have read the page of MO that you mentioned and in my question a have sited to it. For Iyanaga's paper, I can't read German but it seems that Iyanaga prove the theorem for unramified primes, as you have said in mentioned page of MO. But, my question is about ramified primes. $\endgroup$ – Ehsan Shahoseini Jan 18 at 1:22
  • 1
    $\begingroup$ My mistake. But if all primes except finitely many are principal, then all ideals are principal by the standard density theorems. $\endgroup$ – Franz Lemmermeyer Jan 18 at 13:25
  • $\begingroup$ @Franz Lemmermeyer: Yes. But my question is about principalization in extensions. By Iyanaga's generalization of PIT, all but finitely many of primes of ground field become principal in any of its ray clas fields, because all but finitely many of primes of ground field are unramified. For ramified primes, if we take the ground field to be the rationals, when in a cyclotomic field at leat two primes ramified, then the prime ideals over them are not in general principal (but their product is, see my second question). Also for my first question, totally ramifiedness is not solved for me. Thanks $\endgroup$ – Ehsan Shahoseini Jan 18 at 14:20
  • $\begingroup$ @Franz Lemmermeyer: Also if we adjoin the square root of -5 to rationals, then the only nonprincipal prime in its ring of integers is the one above 2, I think. $\endgroup$ – Ehsan Shahoseini Jan 19 at 15:20
1
$\begingroup$

It is not true that the prime ${\mathfrak p}$ is totally ramified in the ray class field of a number field defined modulo ${\mathfrak p}$. For example, the ray class field modulo $3$ over the rationals is trivial, and the ray class field of $K = {\mathbb Q}(\sqrt{-6})$ modulo the prime ideal $(2, \sqrt{-6})$ above $2$ is equal to the Hilbert class field $K^1 = {\mathbb Q}(\sqrt{2}, \sqrt{-3})$ of $K$: it has conductor $1$, and $(2, \sqrt{-6})$ is not ramified.

For the second question, choose a number field $K$ and a prime ideal ${\mathfrak p}$ such that the prime ideal ${\mathfrak P}$ above ${\mathfrak p}$ in the Hilbert class field of $K$ is not principal. You are claiming that ${\mathfrak P}$ becomes principal in the ray class field modulo ${\mathfrak P}$ of $K$; to me it seems unlikely to be true for every choice of $K$ and ${\mathfrak P}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think ray class field modulo 3 of rationals is cyclotomic field by 3rd root of unity. For $K$ and $K^1$ you'r right. So I think I should correct my question. I think I should say "conductor" instead of "modulus". In fact in my first question, I want unique ramified prime, and in second question I want arbitrary ramified primes (not unramified). Because in unramified case the Iyanaga's theorem (and other things related to my work) solve any thing that I want. $\endgroup$ – Ehsan Shahoseini Jan 24 at 16:09
  • $\begingroup$ So, should I edit my question? (excuse me for my unknowing and my poor English). $\endgroup$ – Ehsan Shahoseini Jan 24 at 16:10
  • $\begingroup$ Replacing modulus by conductor does not save the first question. Even in the class field with conductor ${\mathfrak p}^r$, the prime ideal {\mathfrak p}$ need not be totally ramified since the ray class field modulo any ideal contains the Hilbert class field. $\endgroup$ – Franz Lemmermeyer Jan 24 at 17:46
  • $\begingroup$ Sorry for my inaccuracy. Yes and thank you very much. But if it be principal, it is still very good for me. $\endgroup$ – Ehsan Shahoseini Jan 24 at 18:29
  • $\begingroup$ In fact I think this sentence is not true: "since the ray class field modulo any ideal contains the Hilbert class field". For example, in your example $K=\mathbb{Q}(\sqrt{-6})$, any extension (with prime power conductor, or not) of $K$ that its modulus is prime to ideals above $2$ and $3$ is not contains the Hilbert class field of $K$. $\endgroup$ – Ehsan Shahoseini Jan 24 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.