Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A proof I'm trying to understand refers to the set of all finite places dividing an algebraic integer x. What does this mean? I can't seem to find a definition in any of the texts I've looked at.


[I am looking at the proof of Theorem 5, starting at the end of page 8, in the paper]

share|cite|improve this question
Are you familiar with valuation theory of fields? – Bill Dubuque Feb 6 '14 at 15:49
Not overly! In my search for an answer to this question, I've come across the definition of a valuation on a field: $$v:f^x \rightarrow R$$ such that $v(xy)=v(x)+v(y)$ and $v(x+y)\geq \min {v(x),v(y)}$. But that is the extent of my knowledge on the subject. – Briony Feb 6 '14 at 15:59
See for example p. 146 in these notes. – Bill Dubuque Feb 6 '14 at 16:03
In these notes they refer to places dividing the "modulus" of a place. I don't really understand how to relate this to numbers in general. Must the algebraic number I'm considering automatically be the modulus of places? – Briony Feb 6 '14 at 16:29
If you don't already know valuation theory then it does not make much sense to try to jump into the middle of it and attempt to understand concepts which depend on prior objects that you have not yet studied. What is the proof you wish to understand? It may be easily expressible without the language of valuation theory. If you mention it in the question then someone may help you do so. – Bill Dubuque Feb 6 '14 at 17:13
up vote 4 down vote accepted

An absolute value on an integral domain $D$ is a function $|\cdot|$ on it which satisfies

  • Trivial vanishing: $|x|=0\Leftrightarrow x=0$
  • Positivity: $|x|\ge0$ for all elements $x$
  • Multiplicativity: $|xy|=|x||y|$ for all $x,y$
  • Triangle inequality: $|x+y|\le |x|+|y|$

An absolute value $|\cdot|$ induces a metric $d(a,b):=|a-b|$, and a metric induces a topology. A basis for the topology is the set of all open balls $\{x\in D:|x-a|<r\}$ as $r>0$ and $a\in D$ vary. If the stronger inequality $|x+y|\le \max\{|x|,|y|\}$ holds then $|\cdot|$ is ultrametric or non-archimedean, otherwise $|\cdot|$ is archimedean (equivalently, $|1+\cdots+1|$ is eventually bigger than $1$).

On a field $F$ a valuation is a function $v(\cdot):F^\times\to G$ satisfying

  • $G$ is a totally ordered abelian group (the value group)
  • Multiplicativity: $v(xy)=v(x)+v(y)$
  • Triangle inequality: $v(x+y)\ge \min\{v(x),v(y)\}$

An ordering $\le$ on a set $X$ induces a topology with basis given by intervals $\{x\in X:a< x< b\}$, here if $G$ is discrete we call $v(\cdot)$ a discrete valuation. Rings that admit discrete valuation functions like this are called discrete valuation rings, or DVRs (with numerous equivalent definitions).

An equivalent definition of valuation appends a symbol "$\infty$" to the value group with the stipulation that $\infty+a=\infty$ for all elements $a\in G\cup\{\infty\}$ and $a<\infty$ for all $a\in G$. Then we make $v$ a function $F\to G\cup\{\infty\}$ with the property that $F^\times\to G$ is a homomorphism and $v(0)=\infty$. In this context, the triangle inequality still holds but arguments of $v(\cdot)$ can now be $0$.

One of the first examples of a valuation is the $p$-adic valuation $v_p(x)$ which gives the exponent of $p$ (positive or negative) in the prime factorization of a rational $x\in\Bbb Q$. One may check that this function satisfies the properties for being a valuation with some elementary number theory. It makes some sense that $v_p(0)=\infty$ because $0$ is "infinitely divisible" by $p$ in the integers $\Bbb Z$.

If $K$ is a field, any valuation $v$ induces an nonarchimedean absolute value $|x|:=c^{-v(x)}$ for any real number $c>1$. In fact, every nonarchimedean absolute value arises in this way, and conversely one applies logarithms to obtain valuations from nonarchimedean absolute values.

If $K$ is a global field (finite extension of $\Bbb Q$ or $\Bbb F_p(T)$) with valuation $v$ or nonarchimedean absolute value $|\cdot|$, its valuation subring is the subring of elements with nonnegative valuation, or equivalently, absolute value $\le1$. The set of elements with positive valuation, or absolute value $<1$, will form the unique maximal ideal $\frak m$ of the valuation subring, and elements of this ring will have valuation zero if and only if they are units.

In the case of $\Bbb Q$ and $v_p$, the valuation subring is $\Bbb Z_{(p)}$ (the set of rationals with no $p$ in their denominator), and the maximal ideal is $p\Bbb Z_{(p)}$. More generally, if $K$ is a global field and $\cal O$ its ring of integers and $\frak p$ a prime ideal, there is a $\frak p$-adic valuation $v_{\frak p}(x)$ which tells us the exponent of $\frak p$ in the prime factorization of a fractional ideal $(x)$. This definition makes sense because number rings $\cal O$ are Dedekind and have unique prime factorization of ideals. All valuations arise in this way.

If we metrically complete $\Bbb Q$ with respect to the familiar Euclidean absolute value we obtain the field of real numbers $\Bbb R$. If we algebraically complete this we obtain the complex numbers $\Bbb C$. If instead we metrically complete $\Bbb Q$ with respect to $|\cdot|_p:=p^{-v_p(\cdot)}$ (the $p$-adic absolute value), we obtain the field of $p$-adic numbers $\Bbb Q_p$. If we algebraically close we obtain $\overline{\Bbb Q_p}$, to which the absolute value extends, and if we metrically complete again we obtain $\Bbb C_p$.

All archimedean absolute values can be obtained by composing an embedding $K\hookrightarrow \Bbb C$ with $\Bbb C$'s Euclidean absolute value, and all nonarchimedean absolute values can be obtained by composing an embedding $K\to\Bbb C_p$ with $\Bbb C_p$'s $p$-adic absolute value. Thus, all absolute values on global fields are obtained via embedding them into the appropriate metrically complete algebraically closed field.

Here is a roadmap of the territory so far:

$$\begin{array}{ccccccc} \{\text{primes }{\frak p}\} & \leftrightarrow & \{\text{valuations }v\} & \leftrightarrow & \{\text{nonarchimedean }|\cdot|\} & \leftrightarrow & \{K\hookrightarrow \Bbb C_p\} \\ & & & & {\large\cap} & & \\ & & & & \{\text{absolute values}\} & & \\ & & & & {\large\cup } & & \\ & & & & \{\text{archimedean }|\cdot|\} & \leftrightarrow & \{K\hookrightarrow\Bbb C\} \end{array} $$

If $c_1,c_2>0$ and $v$ is a valuation on $K$, the topologies induces by the absolute values $|x|_1=c_1^{-v(x)}$ and $|x|_2=c_2^{-v(x)}$ are the same. Equivalently if $|x|$ is an absolute value and $\alpha>0$ then $|\cdot|$ and $|\cdot|^\alpha$ induce the same topology. The property of two absolute values inducing the same topology is an equivalence relation on the class of absolute values. These equivalence classes of absolute values are called the places of $K$. The "roadmap" above in fact only speaks of valuations and absolute values up to topological equivalence.

The finite places are the nonarchimedean absolute value classes or the embeddings $K\hookrightarrow\Bbb C_p$ and the infinite places are the archimedean absolute value classes or the embeddings $K\hookrightarrow\Bbb C$.

If $L/K$ is a field extension and $\frak p$ a prime of $K$, we have a factorization ${\frak p}={\frak P}_1^{e_1}\cdots{\frak P}_g^{e_g}$ for some primes ${\frak P}_1,\cdots,{\frak P}_g$ of $L$ lying over $\frak p$ and exponents $e_1,\cdots,e_g$. We say $\frak p$ ramifies (in $L$) if any exponent $e_i$ is $>1$, and also say the corresponding place extends. An infinite place ramifies if it is a real place on $K$ (i.e. the embedding has image in $\Bbb R$) extending to a complex place on $L$ (i.e. the image is not in $\Bbb R$). [I think there is a more universal way to treat finite and infinite places simultaneously in terms of Galois actions degenerating, but I don't remember.]

Back in the ring of integers $\cal O$ of a global field $K$, we have $x\mid y$ as elements if and only if $(x)\mid(y)$ as ideals. Since finite places correspond to primes, it would make sense to say a finite place divides $x$ if the corresponding prime divides $(x)$, or equivalently if it appears in $(x)$'s factorization, or also equivalently if $v(x)>0$ with respect to the given finite place's valuation, or equivalently again if $|x|<1$ with respect to any absolute value representing the place.

share|cite|improve this answer
One way to treat the case of finite and infinite places more uniformly is in the subject of Arakelov theory. There one thinks about adjoining to some arithmetic scheme over $\text{Spec}\mathcal{O}_K$ "fibers at infinity" which correspond to the archimedean valuations. Then, the valuations correspond to metrizations of the bundles over these various fibers. – Alex Youcis Feb 7 '14 at 8:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.