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If $f(x)$ is a monic, irreducible polynomial in $\mathbb{Z}[x]$ with $\theta\in\mathbb{C}$ as root, why $\mathbb{Z}[\theta]\,/\,\mathcal{P}$ is an algebraic extension over $\mathbb{Z}/p\mathbb{Z}$?

I'm interested in a simple explanantion, if there exists, without using the Krull dimension.

I already proved that $\mathbb{Z}/p\mathbb{Z} \subset \mathbb{Z}[\theta]\,/\,\mathcal{P}$.

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  • $\begingroup$ Seems like $\mathcal{P}$ is an ideal of $\mathbb{Z}[\theta]$ lying over $p$ in $\mathbb{Z}$? $\endgroup$ – sharding4 May 29 '17 at 14:48
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Note that if $R \subset S$ is an inclusion of integral domains, then for any prime ideal $\mathcal{P} \in S$, $\mathcal{P} \cap R$ is a prime ideal of $R$.

Here $R = \mathbb{Z}, S = \mathbb{Z}[\theta]$ and $\mathcal{P} \cap \mathbb{Z} = p \mathbb{Z}$ for some prime number $p$, since it is a prime ideal of $\mathbb{Z}$.

Thus $\mathbb{Z}/p\mathbb{Z} \subset \mathbb{Z}[\theta]\,/\,\mathcal{P}$, which is almost a proof of what you want.

All you need is saying that since $\theta$ is an algebraic integer, $\mathbb{Z}[\theta]$ is a finitely generated free $\mathbb{Z}$-module (its elements are of the form $\sum_{n=0}^{deg(f)-1} a_n \theta^n, a_n \in \mathbb{Z}$),

so that $$\mathbb{Z}[\theta]\,/\,p \mathbb{Z}[\theta]= \{ \sum_{n=0}^{deg(f)-1} a_n \theta^n, a_n \in \mathbb{Z}/p \mathbb{Z} \}$$ is a finite ring. Clearly it contains $\mathbb{Z}[\theta]\,/\,\mathcal{P}$,

therefore $k = \mathbb{Z}[\theta]\,/\,\mathcal{P}$ is a finite integral domain, thus a finite field.

Since $k$ contains $\mathbb{F}_p$, you get that $k/\mathbb{F}_p$ is a finite (thus algebraic) field extension.

Indeed $k = \mathbb{F}_p[\overline{\theta}]$ where $\overline{\theta}$ is one of the root of $f \bmod p$ (and there is one prime ideal above $p$ for each of those different root)

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  • $\begingroup$ The same argument can be used to prove that $\mathcal{O}_{\mathbb{Q}(\theta)}/\mathcal{P}$ is finite, where $\mathcal{P}$ is a prime ideal of $\mathcal{O}_{\mathbb{Q}(\theta)}$ (the algebraic integers of $\mathbb{Q}(\theta)$ ), right? I need also this fact in the proof that $\mathcal{O}_{\mathbb{Q}(\theta)}$ is a Dedekind domain. $\endgroup$ – ilmarchese May 30 '17 at 12:27
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    $\begingroup$ @ilmarchese Yes, if $K$ is a number field then $\mathcal{O}_K = \mathbb{Z}[\alpha_1,\ldots, \alpha_n]$ for some algebraic integers $\alpha_1$. So the proof is the same $\endgroup$ – reuns May 30 '17 at 16:50

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