Proof of Dedekind-Kummer theorem

Question

I will write the statement and then ask my query about part of the proof.

Statement Let $p$ be a rational prime. Let $K=\mathbb{Q}(\theta ) $ be a number field where $\theta $ is an algebraic integer. Suppose $p \nmid [\mathcal{O}_K : \mathbb{Z}[\theta ]] $. Let $$ \mu _{\theta }\equiv f_1^{e_1}...f_r^{e_r} \pmod{p} $$ where the polynomials $f_i \in \mathbb{Z}[x] $ are monic, irreducible modulo $p$ and pairwise coprime modulo p. Let $\mathfrak{p}_i =\langle p, f_i(\theta ) \rangle $.

These are all the assumptions needed for my question.

In the proof we let $J=p\mathbb{Z}[\theta ] +f_i(\theta )\mathbb{Z} [\theta ] $ and show that the map $$\phi : \mathbb{Z}[\theta ] /J \rightarrow \mathcal{O}_K / \mathfrak{p}_i , \\ \phi (g(\theta )+J) = g(\theta ) + \mathfrak{p}_i .$$

We show that $\phi $ is a surjective ring homomorphism but then we show that it is also injective since $\mathbb{Z}[\theta ] /J $ is a field. This means that $\ker \phi$ is trivial or is the whole field. The proof then somehow assumes that $\ker \phi $ is trivial.

My question is How do they assume or deduce that the kernel is trivial. Why can’t we have $\ker \phi = \mathbb{Z}[\theta ]/J $??

Answer 1

It might be helpful to work mod $p$ instead of $J$ or ${\frak p}_i$ first. Note that the inclusion $\Bbb Z[\theta]\hookrightarrow {\cal O}_K$ induces an isomorphism $\psi:\Bbb Z[\theta]/p\Bbb Z[\theta]\cong {\cal O}_K/p{\cal O}_K$ because:

• By assumption there are $a,b\in\Bbb Z$ such that $ap+b[{\cal O}_K:\Bbb Z[\theta]]=1$. Thus if $x\in {\cal O}_K$, we have $x-apx = b[{\cal O}_K:\Bbb Z[\theta]]x\in \Bbb Z[\theta]$, hence the map $\Bbb Z[\theta] \to {\cal O}_K/p{\cal O}_K$ is surjective.
• If $x\in\Bbb Z[\theta]$ such that $x=cp\in p{\cal O}_K$ where $c\in{\cal O}_K$, we already have $c\in \Bbb Z[\theta]$ because otherwise $c$ would have order $p$ in ${\cal O}_K/\Bbb Z[\theta]$, i.e. $p\mid [{\cal O}_K:\Bbb Z[\theta]]$. Hence $\Bbb Z[\theta]\cap p{\cal O}_K=p\Bbb Z[\theta]$.

As $\psi$ is an isomorphism we have $\psi(\langle f_i(\theta)+p\Bbb Z[\theta]\rangle)=\langle \psi(f_i(\theta)+p\Bbb Z[\theta])\rangle=\langle f_i(\theta)+p{\cal O}_K\rangle$, hence $\psi(J/p\Bbb Z[\theta])={\frak p}_i/p{\cal O}_K$. Thus by the third isomorphism theorem $\psi$ induces an isomorphism $$\Bbb Z[\theta]/J\cong(\Bbb Z[\theta]/p\Bbb Z[\theta])/(J/p\Bbb Z[\theta])\cong ({\cal O}_K/p{\cal O}_K)/({\frak p}_i/p{\cal O}_K)\cong {\cal O}_K/{\frak p}_i$$ which is exactly the map $\phi$.