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Show that the norm of a prime ideal in a number field $K$ is a power of some prime number, i.e., if $P$ is a prime ideal in $O_K$ for some number field $K$, then $N_\mathbb{Q}^K(P)=p^n$ for some prime number $p$ and some positive integer $n$.

Here is my approach:

Any prime ideal lies over some prime number $p$. If we consider the ideal decomposition of $pO_K$, and apply the norm operator, we get the following:

$pO_K=p_1^{e_1} \cdots p_r^{e_r}$ for some $r$ since $O_K$ is a Dedekind domain. Applying the norm operator to this, we get

$N(pO_K)=N(p_1^{e_1} \cdots p_r^{e_r}) = N(p_1^{e_1})\cdots N(p_r^{e_r})$ since the norm has the multiplicative property.

This is where I am unsure if I have completely answered the question because I found a list of primes as opposed to the suggested $p^n$ in the problem statement.

Thanks in advance, any help is greatly appreciated.

user26857
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Jamil_V
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2 Answers2

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By definition, the norm $N(P)$ is the cardinality of the field $\mathcal{O}_K/P$. Since this is a finite field (the ideal norm is always finite in the ring of integers $\mathcal{O}_K$), it has characteristic $p$ with a prime $p$. It follows that $$ N(P)=p^{[(\mathcal{O}_K/P):\mathbb{F}_p]}. $$ Here $n=[(\mathcal{O}_K/P):\mathbb{F}_p]$ is the degree of the field extension. For the element norm $N_{\mathbb{Q}}^{K}(p)$ we have, with $P=(p)$, that $N(P)=|N_{\mathbb{Q}}^{K}(p)|$.

hbghlyj
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Dietrich Burde
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  • So in this context, the $\mathbb{F}_p$ corresponds to $\mathbb{Q}$ in my problem statement? – Jamil_V Nov 10 '14 at 17:09
  • The element norm is $N_{\mathbb{Q}}^{K}(p)$, and it agrees with the ideal norm $N(P)$ up to sign, where $P=(p)$. So it is enough to compute $N(P)$. – Dietrich Burde Nov 10 '14 at 18:37
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For a prime ideal $\mathfrak{P}$ in $\mathcal{O}_K$, $\mathfrak{P} \cap \mathbb{Z} = p\mathbb{Z}$, where $p$ is a prime in $Z$. So, naturally, there exists a rational prime $p \in \mathfrak{P}$. Then, $(p) \subset \mathfrak{P}$. Since $\mathcal{O}_K$ is a Dedekind domain, this yields $(p) = \mathfrak{P} I$ for some other ideal $I$ in $\mathcal{O}_K$. Considering the norms on both sides, we could see that $N(\mathfrak{P})$ divides $N((p))$, which is $p^n$ where $n = [K:\mathbb{Q}]$. Therefore, the possible values of $N(\mathfrak{P})$ are $p, p^2,...,p^n$ .

Musk
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  • I upvote but I think the claim really needs to be proven before the proof that $\mathfrak{a} \subset \mathfrak{P} \implies \mathfrak{a} = \mathfrak{b} \mathfrak{P} $ whenever $\mathfrak{P}$ is a maximal ideal, which is the core of the proof of unique factorization of ideals – reuns Dec 18 '17 at 09:19
  • Could you provide me with a more elementary approach or a reference to it? ( I could not follow this part in the first answer: $N(P)=p^{[(\mathcal{O}_K/P) : \mathbb{F}_p]}$ and how $[(\mathcal{O}_K/P) : \mathbb{F}_p]$ is equal to the degree of the extension.) – Musk Dec 18 '17 at 09:51
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    Definition : $N(\mathfrak{a}) = # \mathcal{O}_K/\mathfrak{a}$. For every $a\ne 0$, $\mathcal{O}_K/(a)$ is a finite ring. If $\mathfrak{P}$ is a proper prime ideal then $\mathcal{O}_K/\mathfrak{P}$ is a finite integral domain, thus a finite field, thus of characteristic $p$ prime and a $\mathbb{Z}/(p)$ vector-space, so it has $p^k$ elements. – reuns Dec 18 '17 at 10:05
  • Thanks a lot! But, how do we view $\mathcal{O}_K/\mathfrak{P}$ as an extension of $\mathbb{Z}/(p)$? And, is $k$ easy to be determined for a given $\mathcal{O}_K$ and $\mathfrak{P}$? It has to be one of the values: $p,...,p^n$. For the determination of $k$, I believe ramification theory of primes in $\mathcal{O}_K$ would be helpful. – Musk Dec 18 '17 at 10:55
  • $\mathcal{O}_K/\mathfrak{P}$ is a finite field, thus of characteristic $p$ prime, thus it contains $\mathbf{F}_p=\mathbb{Z}/(p)$ as the subring generated by $1$. Therefore it is a $k$-dimensional $\mathbf{F}_p$ vector space and it has $p^k$ elements. Yes $k$ is given by the splitting of $(p)$ in a product of prime ideals, ie. by the factorization of $I \bmod p$ where $\mathcal{O}_K \cong \mathbb{Z}[x_1,\ldots,x_m]/I$. – reuns Dec 18 '17 at 11:59
  • In your answer, why $N((p)) = p^n$ where $n=[K :\mathbb{Q}]$? – Plantation Apr 20 '24 at 12:16