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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 $??

Arturo Magidin
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Anonmath101
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  • That's because it is a ring homomorphism, hence $1 \mapsto 1$. The kernel cannot be the whole ring. – Crostul Aug 19 '21 at 13:36
  • Why couldn’t we have 1=0 in the latter ring? This means that $\mathfrak{p}_i = \mathcal{O}_K $. Why isn’t this possible? – Anonmath101 Aug 19 '21 at 13:53
  • $\mathfrak p_i$ is a prime ideal by definition, hence it is not the whole ring. – Crostul Aug 19 '21 at 16:24
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    How do we know that $\mathfrak{p}_i $ is a prime ideal? In fact in the proof I’m reading the fact that the ideal is prime comes from the fact that $ \phi \ $ is an isomorphism and so $ \mathcal{O}_K / \mathfrak{p}_i $ is a field. – Anonmath101 Aug 19 '21 at 17:22
  • Use \pmod{p} to automatically get the space, the parentheses, and the roman typeface mod. The p stands for "parenthetical". – Arturo Magidin Aug 19 '21 at 19:20

1 Answers1

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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 $\phi$.

leoli1
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