Relating prime spots and prime ideals in number fields

Question

I'm reading through O'Meara's book on quadratic forms. In the first chapter he describes this beautiful result:

Let $E/F$ is a separable extension where $E=F[\alpha]$ and $f$ is the minimal polynomial of $\alpha$ over $F$. If $\mathfrak{p}$ is a prime spot on $F$ then $F_{\mathfrak{p}}$ denotes the completion of $F$ at $\mathfrak{p}$. Now suppose that in $F_{\mathfrak{p}}$ the polynomial $f$ splits into irreducibles as

$$f(x)=f_1(x)\dots f_r(x).$$ Then the prime spots $\mathfrak{P}$ in $E$ that divide $\mathfrak{p}$ are in bijection with the irreducible factors of $f$.

Now this means that if $F=\mathbb{Q}$ and $p$ is a prime, I can find the prime spots dividing $p$ by looking at the polynomial $f$ modulo $p$ and using Hensel's Lemma. For example, if $E=\mathbb{Q}(i)$ then $f=x^2+1$ and $f$ splits into linear factors in $\mathbb{Z}_p$ for $p$ odd if and only if $p\equiv 1 \pmod{4}$. Therefore when $p\equiv 1\pmod{4}$ there are exactly two prime spots dividing $p$. This suggests that we are retrieving the decomposition of $(p)$ in $\mathbb{Z}[i]$.

For the rationals Ostrowski's Theorem tells me that all non-archimedean spots are in bijection to the prime ideals of its ring of integers $\mathbb{Z}$.

My questions are the following (I am happy with a full answer or a good reference):

• Is there a characterisation on a general number field that all non-archimedean spots are in bijection to the prime ideals of its ring of integers?
• How does the division of prime spots (defined entirely in terms of restriction of valuations) relate precisely to the expression of ideals as a product in a larger ring of integers?

Finally this should probably belong to another question, but I would like to know more about what the local degree of a spot represents. For example if $E=\mathbb{Q}[\sqrt[3]{2}]$ taking the prime $5$ we have that modulo $5$ $$x^3-2 = (x-3)(x^2+3x+4),$$ and similarly in $\mathbb{Z}_5$ the polynomial splits into two irreducibles, one linear and one quadratic. That should mean that $(5)$ splits in the ring of integers of $E$ as the product of two prime ideals, but what does the local degree tells us about those prime ideals? It would be a very good example to see the precise decomposition of $(5)$!

Answer 1

For the first question:
Actually something more general is true:
Lemma: Let $R$ be any Dedekind domain and $v$ a non-trivial non-archimedian valuation on $K=\operatorname{Frac}R$. Assume that $R\subseteq \mathcal O_v$ where $\mathcal O_v$ denotes the valuation ring of $v$. Then $v$ is the valuation associated to some prime ideal $\mathfrak p$ of $R$.
Proof: Let $\mathfrak m_v$ be the maximal ideal of $\mathcal O_v$. Then $\mathfrak p_v:=R\cap \mathfrak m_v$ is a prime ideal of $R$. Note that $R_{\mathfrak p_v}\subseteq \mathcal O_v$. If $v$ were trivial on $R$ it would also trivial on $K$, hence $\mathfrak p_v\ne0$. Therefore $R_{\mathfrak p_v}$ is a discrete valuation ring. It easy to show that DVRs are maximal subrings of their field of fractions (see e.g. here) so that $R_{\mathfrak p_v}=\mathcal O_v$ and $\mathfrak p_vR_{\mathfrak p_v}=\mathfrak m_v$. Hence $v$ is equivalent to the valuation associated to $\mathfrak p_v$.

Now for the case of number fields $K$ we only need to verify that valuations satisfy $\mathcal O_K\subseteq \mathcal O_v$. Clearly $\Bbb Z\subseteq \mathcal O_v$. As valuation rings are integrally closed this implies $\mathcal O_K\subseteq \mathcal O_v$. Hence we can apply the lemma above and see that any non-archimedian place on $K$ corresponds to a (non-zero) prime ideal of $\mathcal O_K$.

For the second question:
Let $L/K$ be an extension of number fields and $\mathfrak P\subseteq\mathcal O_L,\mathfrak p\subseteq\mathcal O_K$ non-zero prime ideals and $v_{\mathfrak P},v_{\mathfrak p}$ their corresponding valuations. Then $\mathfrak P$ lies over $\mathfrak p$ (i.e. $\mathfrak P\mid \mathfrak p\mathcal O_L\Leftrightarrow \mathfrak p\subseteq\mathfrak P\Leftrightarrow \mathfrak P\cap \mathcal O_K=\mathfrak p$) iff $v_{\mathfrak P}$ lies over $v_{\mathfrak p}$ (i.e. $v_{\mathfrak P}$ restricts to $v_{\mathfrak p}$ on $K$ (actually one should say that it restricts to an equivalent valuation since we didn't do any normalization)). This follows almost from the definitions and the the first point: Assume that $\mathfrak P$ lies over $\mathfrak p$. Consider the restriction of $v_{\mathfrak P}$ to $K$. By the first part it is a valuation associated to some prime $\mathfrak q$ of $\mathcal O_K$. This prime has to be $\mathfrak p$ since $\mathfrak q=\{x\in\mathcal O_K\mid v_{\mathfrak P}(x)>0\}=\{x\in\mathcal O_K\mid x\in\mathfrak P\}=\mathcal O_K\cap\mathfrak P=\mathfrak p$. Conversely assume that $v_{\mathfrak P}$ lies over $v_{\mathfrak p}$. Then $v_{\mathfrak P}(\mathfrak p)>0$, hence $\mathfrak p\subseteq \mathfrak m_{v_{\mathfrak P}}=\mathfrak P\mathcal O_{L,\mathfrak P}$ and therefore $\mathfrak p\subseteq\mathfrak P$.