Dedekind's theorem on the factorisation of rational primes

Solution 1:

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 $\rm\:n\ =\ [L,K]\ =\ \sum_{P|p}\ e_P\:f_P\:.$

Solution 2:

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 $n \ge \Sigma d_ie'_i$.

Solution 3:

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.