What are the prime ideals in rings of cyclotomic integers? - MathOverflow

What are the prime ideals in rings of cyclotomic integers?

2010-01-02T03:58:26Z

Is a good characterization of Spec $\mathbb{Z}[\zeta_n]$ known? Same question for its unit group.

Answer by Chandan Singh Dalawat for What are the prime ideals in rings of cyclotomic integers?

2010-01-02T04:19:09Z

The extension $\mathbb{Q}(\zeta_n)|\mathbb{Q}$ is abelian of group $(\mathbb{Z}/n\mathbb{Z})^\times$ so class field theory tells you everything about the prime ideals in $\mathbb{Z}[\zeta_n]$, the ring of integers of $\mathbb{Q}(\zeta_n)$.

You should try to do the cases $n=3,4$ by hand.

As for the group $\mathbb{Z}[\zeta_n]^\times$, an explicit subgroup of "cyclotomic units" can be constructed which has finite index.

Any book on Cyclotomic Fields (Lang, Washington) should help. For a start, you can look up Chapter VI of Fröhlich-Taylor.

Answer by Chandan Singh Dalawat for What are the prime ideals in rings of cyclotomic integers?

2010-01-02T06:48:54Z

Let me summarise what Hilbert says in his Zahlbericht about the behaviour of rational primes in the cyclotomic field $\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $l$-th root of $1$ and $l$ is an odd prime. You can read the original at the Göttingen site or a French translation at the Grenoble site.

Satz 117. The ideal $\mathfrak{l}=(1-\zeta)\mathbb{Z}[\zeta]$ is prime of residual degree $1$, and $l\mathbb{Z}[\zeta]=\mathfrak{l}^{l-1}$.

Satz 118. The discriminant of the field $\;\mathbb{Q}(\zeta)$ is $(-1)^{(l-1)/2}l^{l-2}$.

Satz 119. If $p\neq l$ is a rational prime, $f>0$ is the smallest exponent such that $p^f\equiv1\pmod l$, and $e$ is defined by $ef=l-1$, then $$ p\mathbb{Z}[\zeta]=\mathfrak{p}_1\ldots\mathfrak{p}_e, $$ where the $\mathfrak{p}_i$ are distinct prime ideals of residual degree $f$.

These results go back to Kummer (1847). All this was much before anyone dreamt of Class Field Theory.

Answer by Qiaochu Yuan for What are the prime ideals in rings of cyclotomic integers?

2010-01-03T01:29:34Z

Theorem: Let $\alpha$ be an algebraic integer such that $\mathbb{Z}[\alpha]$ is integrally closed, and let its minimal polynomial be $f(x)$. Let $p$ be a prime, and let

$\displaystyle f(x) \equiv \prod_{i=1}^{k} f_i(x)^{e_i} \bmod p$

in $\mathbb{F}_p[x]$. Then the prime ideals lying above $p$ in $\mathbb{Z}[\alpha]$ are precisely the maximal ideals $(p, f_i(\alpha))$, and the product of these ideals (with the multipicities $e_i$) is $(p)$. (Theorem 8.1.3.)

In this particular case we have $f(x) = \Phi_n(x)$. When $(p, n) = 1$, its factorization in $\mathbb{F}_p[x]$ is determined by the action of the Frobenius map on the elements of order $n$ in the multiplicative group of $\overline{ \mathbb{F}_p }$, which is in turn determined by the minimal $f$ such that $p^f - 1 \equiv 0 \bmod n$ as described in Chandan's answer. (This $f$ is the size of every orbit, hence the degree of every irreducible factor.) When $p | n$ write $n = p^k m$ where $(m, p) = 1$, hence $x^n - 1 \equiv (x^m - 1)^{p^k} \bmod p$. Then I believe that $\Phi_n(x) \equiv \Phi_m(x)^{p^k - p^{k-1}} \bmod p$ and you can repeat the above, but you'd have to check with a real number theorist on that. (Edit: Indeed, it's true over $\mathbb{Z}$ that $\Phi_n(x) = \frac{ \Phi_m(x^{p^k}) }{ \Phi_m(x^{p^{k-1}}) }$.)