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Let $K$ be an algebraic number field. Let $A$ be its ring of integers. Let $\mathfrak{p}$ be some nonzero prime ideal of $A$. I have some very basic questions concerning the relation between the localization of $A$ at $\mathfrak{p}$ and the completion of $K$ at $\mathfrak{p}$.

  1. Let $\pi \in \mathfrak{p}A_{\mathfrak{p}}$ be a uniformizer of the DVR $A_{\mathfrak{p}}$. It defines a discrete valuation $\nu$ on the field of fractions of $A_{\mathfrak{p}}$, which is $K$. On the other hand, the prime ideal $\mathfrak{p}$ defines a discrete valuation $\mu$ on $K$ determined by $\mu(x) = e$, if the exponent of $\mathfrak{p}$ in the factorization of the principal ideal $xA$ is $e$. Is $\mu = \nu$?
  2. Given the valuation $\nu$ (or $\mu$) from 1, we can complete the field $K$ with respect to an assosicated absolute value $|x| = q^{-\nu(x)}$ for some $q > 1$ and obtain a complete non-archimedean field $K_{\nu}$. Does the residue field of that completion $K_v$ identify with the one of $A_{\mathfrak{p}}$?
  3. Suppose $\mathfrak{p} = \mathfrak{p}_1$ lies above the rational prime $p$. Let $pA = \mathfrak{p}_1^{e_1} \dots \mathfrak{p}_n^{e_n}$ be the factorization of $p$ into prime ideals of $A$. Let $f_1$ be the degree of the extension $A/\mathfrak{p} \supset \mathbb{Z}/p\mathbb{Z}$. With notation as in 3, we obtain an extension of complete non-archimedean fields $K_v/\mathbb{Q}_p$. Does $p$ factor in in $K_v$ with the same parameters $e_1, f_1$? I.e., if $\pi$ is a uniformizer of $K_v$, do we have $\pi^{e_1}u = p$ for some unit $u$ in the ring of integers of $K_v$? Is $[K_v : \mathbb{Q}_p] = e_1f_1$?

Thanks.

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Regarding question $1$: Yes, $\nu=\mu$. Indeed, if $xA=\mathfrak{p}^{e} \mathfrak{p}_1^{e_1}\ldots \mathfrak{p}_r^{e_r}$ for pairwise distinct prime ideals $\mathfrak{p}_i$ not equal to $\mathfrak{p}$, then passing to the localization, you obtain $xA_{\mathfrak{p}}=(\mathfrak{p}A_{p})^{e}=\pi^{e} A_{\mathfrak{p}}$, i.e. $x=u\cdot \pi^{e}$ for a unit $u$ in $A_{\mathfrak{p}}$.

What you write in your second point is also correct: If $(R,\mathfrak{p},v)$ is a DVR with fraction field $K$, then $R_v/\mathfrak{p}_v \cong R/\mathfrak{p}$, where $R_v$ is the valuation ring of $K_v$ and $\mathfrak{p}_v$ its maximal ideal. For proving this it is enough to prove that the natural map $R\to R_v/\mathfrak{p}_v$ is surjective, which follows from the fact that every element in $R_v$ is the limit of elements in $R$.

Finally, given a finite extension $L/K$ of complete fields with valuation $v$, one defines the ramification index $e$ to be the index $(v(L^\ast):v(K^\ast))$ and the inertia degree $f$ to be the degree of the extension of the residue fields. If the valuation $v$ is discrete and if $L/K$ is separable, then $[L:K]=ef$ (in general one only has $ef \leq [L:K]$), see Neukirch's $\textit{Algebraic Number Theory}$ for details.
Back to the notations in your question, if $K_v$ is the completion of $K$ with respect to $v$, then $v(K_v^\ast)=v(K^\ast)$. Now we can normalize the valuation $v$ on $K$ in such a way that $v(p)=1$, namely by replacing $v$ by $\frac{1}{e_1} v$. Then we see that $$ e=(v(K_v^\ast):v(\mathbb{Q}_p^\ast))=(v(K^\ast):v(\mathbb{Q}^\ast))=(\frac{1}{e_1}\mathbb{Z}:\mathbb{Z})=e_1. $$ It follows from 2. that $f=f_1$. Moreover, we have $v(p)=1=\frac{1}{e} \cdot e=v(\pi^{e})$, i.e. $p=u\cdot \pi^{e}$ for $u$ a unit in the ring of integers of $K_v$.
In conclusion: what you write in 3 is also correct.

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  • $\begingroup$ Thank you for this nice answer. $\endgroup$ – m.s Jul 16 '17 at 5:42

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