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Let $K$ be an algebraic number field, and suppose its ring of integers is $\mathcal{O}_K = \mathbb{Z}[\theta]$ for some $\theta \in \mathcal{O}_K$. Let $f \in \mathbb{Z}[X]$ be the minimal polynomial of $\theta$, and suppose $\overline{f} = \overline{f_1}^{e_1} \cdots \overline{f_r}^{e_r} \in \mathbb{F}_p [X]$, where $p$ is a rational prime and $\overline{f_1}, \ldots, \overline{f_r}$ are distinct monic irreducible polynomials. Then Dedekind's theorem on the factorisation of rational primes tells us that the ideal $(p) \triangleleft \mathcal{O}_K$ factors as ${\mathfrak{p}_1}^{e_1} \cdots {\mathfrak{p}_r}^{e_r}$, where $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ are distinct prime ideals and moreover if $f_j \in \mathbb{Z}[X]$ is monic and reduces to $\overline{f_j} \in \mathbb{F}_p[X]$, then $\mathfrak{p}_j = (p, f_j(\theta))$.

In Lang's Algebraic Number Theory (1968, p.28) there is a proof of this theorem, which I can follow right up to the penultimate line. Then there's a mysterious equation which seems to come from nowhere, and it's literally the only obstruction to my understanding of the proof. The gist of the proof is as follows:

  1. First we construct the prime ideals $\mathfrak{p}_j$ as the kernels of homomorphisms $\mathbb{Z}[\theta] \to \mathbb{F}_p(\alpha)$, where $\theta$ is mapped to a root $\alpha$ of $\overline{f_j}$.
  2. Then, we show that $\mathfrak{p}_j = (p, f_j(\theta))$.
  3. Suppose $(p) = {\mathfrak{p}_1}^{e'_1} \cdots {\mathfrak{p}_r}^{e'_r}$. (I presume this is by unique prime factorisation in Dedekind domains? But then how do we know there aren't additional factors, which aren't the $\mathfrak{p}_j$ we've constructed?) Then, we note that $f - {f_1}^{e_1} \cdots {f_r}^{e_r} \in p \mathbb{Z}[X]$, so $f_1(\theta)^{e_1} \cdots f_r(\theta)^{e_r} \in (p)$.
  4. So it follows that ${\mathfrak{p}_1}^{e_1} \cdots {\mathfrak{p}_r}^{e_r} \subseteq (p) = {\mathfrak{p}_1}^{e'_1} \cdots {\mathfrak{p}_r}^{e'_r}$, and thus $e_j \le e'_j$.
  5. Let $d_j = \operatorname{deg} f_j$. Then clearly $n = \operatorname{deg} f = d_1 e_1 + \cdots d_r e_r$. I can see that if we can show that $d_1 e'_1 + \cdots + d_r e'_r = n$ then we are done. But why is the equation true? Does it follows from unique factorisation?

It feels like the proof shouldn't be this long though, and the Chinese remainder theorem doesn't seem to be invoked anywhere even though it seems to be the obvious thing to do. Is there a simpler proof, using the Chinese remainder theorem?

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@lhf : I am thinking of a generalised Chinese remainder theorem, where we deal with coprime ideals instead of coprime moduli. –  Zhen Lin May 11 '11 at 14:08
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3 Answers

up vote 2 down vote accepted

HINT $\rm\ \ e_i \le\: e_i'\ $ and $\rm \ d_1\ e_1 +\:\cdots\:+ d_r\ e_r\ =\ d_1\ e_1'+\:\cdots\:d_r\ e_r'\ \Rightarrow\ e_i = e_i'\:.$

Edit $\ $ To answer the clarified question, the equation $\rm\:n\ =\ \sum d_i e_i'$ follows from Proposition 21, namely $\rm\:n\ =\ [L,K]\ =\ \sum_{P|p}\ e_P\:f_P\:.$

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Yes, that's what I meant by "if we can show ... we are done". Why the equation $d_1 e_1 + \cdots + d_r e_r = d_1 e'_1 + \cdots + d_r e'_r$ hold? This must be something obvious but it escapes me. –  Zhen Lin May 11 '11 at 18:19
    
@Zhen: I read your query as asking why the proof is done if it holds, not why it holds. You may wish to edit the question to remove the ambiguity. –  Bill Dubuque May 11 '11 at 18:28
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You have to reason on the size of $\mathcal{O}_K / (p)\mathcal{O}_K$ here :

$p^n = \#(\mathcal{O}_K / (p)\mathcal{O}_K) = \Pi \# (\mathcal{O}_K / (\mathfrak{p}_i^{e'_i})\mathcal{O}_K)$ thanks to the Chinese Remainder Theorem.

For any relevant $i,j$, the inclusion $\mathcal{O}_K / (\mathfrak{p}_i^{j+1}\mathcal{O}_K) \to \mathcal{O}_K / (\mathfrak{p}_i^{j}\mathcal{O}_K)$ has kernel $(\mathfrak{p}_i^{j}\mathcal{O}_K) / (\mathfrak{p}_i^{j+1}\mathcal{O}_K)$, which is a $(\mathcal{O}_K/\mathfrak{p}_i \mathcal{O}_K)$-vector space. Since $\mathfrak{p}_i^j \neq \mathfrak{p}_i^{j+1}$, the kernel has dimension at least 1, so its cardinal is at least $p^{d_i}$.

Plugging that information, you get $p^n \ge \Pi p^{d_ie'_i}$, thus $n \ge \Sigma d_ie'_i$.

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Here's the proof I used for class. It is mostly just from KCd's blurb on the topic.

Let $K=\Bbb Q(\theta)$ with $\theta\in{\frak O}_K$ and $p\in\Bbb Z$ a rational prime. Construct the obvious homomorphism $$\frac{\Bbb Z[\theta]}{p\Bbb Z[\theta]}\to\frac{{\frak O}_K}{p{\frak O}_K}:~x+p{\Bbb Z}[\theta]\mapsto x+p{\frak O}_K.$$

Suppose $p\nmid m=[{\frak O}_K:{\Bbb Z}[\theta]]$ (finite by structure theory of free abelian groups; they are equal rank) and pick an $\tilde{m}$ for which $\tilde{m}m\equiv1$ mod $p\Bbb Z$ (hence mod $p{\frak O}_K$ too). If $x\in{\frak O}_K$ is arbitrary then we know $mx\in\Bbb Z[\theta]$ (consider the finite quotient group ${\frak O}_K/\Bbb Z[\theta]$) hence $\tilde{m}mx+p\Bbb Z[\theta]\mapsto x+p{\frak O}_K$, so we know the map is surjective. Both quotients are finite groups (elementary abelian, size $p^n$) so the map must be an isomorphism. Therefore we have

$$\frac{{\frak O}_K}{p{\frak O}_K}\cong\frac{\Bbb Z[\theta]}{p\Bbb Z[\theta]}\cong\frac{\Bbb Z[T]}{(p,f(T))}\cong\frac{\Bbb F_p[T]}{(f(T))}$$

where $f(T)$ is the minimal polynomial of $\theta$ (monic and integer coefficients).

Suppose $p{\frak O}_K$ and $f(T)\in{\Bbb F}_p[T]$ factor into prime ideals and irreducibles respectively as

$$p{\frak O}_K={\frak P}_1^{e_1}\cdots{\frak P}_g^{e_g},\qquad f(T)=\pi_1(T)^{r_1}\cdots\pi_h(T)^{r_h}.$$

By Sun-Ze (aka the Chinese Remainder Theorem),

$$\frac{{\frak O}_K}{p{\frak O}_K}\cong\prod_{i=1}^g\frac{{\frak O}_K}{{\frak P}_i^{e_i}},\qquad \frac{\Bbb F_p[T]}{(f(T))}\cong\prod_{j=1}^h\frac{{\Bbb F}_p[T]}{(\pi_i(T)^{r_j})}.$$

A maximal ideal of a direct product is one in which all but one of the summands may contain anything, and that one coordinate contains elements from a maximal ideal of that summand's ring. Furthermore a maximal ideal of $R/P^v$ lattice-corresponds to a maximal ideal of $R$ containing $P^v$, which must be $P$ for $R={\frak O}_K,{\Bbb F}_p[T]$ and $P={\frak P}_i,(\pi_i(T))$ resp. Therefore

$$\{{\frak P}\mid p\}\cong{\rm MaxSpec}\left(\frac{{\frak O}_K}{p{\frak O}_K}\right)\cong{\rm MaxSpec}\left(\frac{{\Bbb F}_p[T]}{(f(t))}\right)\cong\{\pi\mid f\}$$

is a natural bijection. In particular this means $g=h$ (taking cardinalities above). Furthermore, the data $e_i$ and $r_i$ can respectively be read off of the factors ${\frak O}_K/{\frak P}_i^{e_i}$ and ${\Bbb F}_p[T]/(\pi_i(T))^{r_i}$ as the nilpotence of their unique maximal ideals, and the data $N({\frak P}_i)$ and $\deg\pi_i$ can be read off of the size of their unique residue fields. Yet further, if we pull back $(\pi_i(T))$ through the isomorphism and then lift back to ${\frak O}_K$ we obtain ${\frak P}_i=(p,\Pi_i(\theta))$, where $\Pi_i(T)\in\Bbb Z[T]$ is any representative of $\pi_i(T)\in{\Bbb F}_p[T]$. In summary, we have proved:

Theorem (Dedekind, Zolotarev). If $\theta$ is an integral primitive element of $K/\Bbb Q$ and a rational prime $p$ does not divide $\Bbb Z[\theta]$'s index in $K$'s integers, then $p$'s factorization in $K$ is the same shape as that of $\theta$'s minimal polynomial $f$ mod $p$. That is, there is a natural bijection between primes ${\frak P}\mid p$ and irreducibles $\pi\mid f$, and the ramification and residue data match up. Finally, ${\frak P}_i=(p,\Pi_i(\theta))$ for any representative $\Pi_i$ of $\pi_i$.

Sometimes ${\frak O}_K$ is not a simple extension of $\Bbb Z$ in which case no matter which $\theta\in{\frak O}_K$ we pick to work with, there will be primes (dividing the index) which this method does not tell us how to factor. The next thing to do would be to just pick a different $\theta'\not\in\Bbb Z[\theta]$, which is often helpful. But it turns out there are fields $K$ for which there exists a prime $p$ dividing every index $[{\frak O}_K:\Bbb Z[\theta]]$ for all $\theta\in{\cal O}_K$. Still, this method factors cofinitely many primes successfully, and establishes a fairly elegant correspondence between the two parallel worlds.

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