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Is there a way to extract the Archimedean absolute value of $\mathbb{Q}$ from its field structure in a way analogous to its non-archimedean absolute values?

Here is some context: Given a valuation ring $A$ of a field $K$ (i.e. $x\in A$ or $x^{-1}\in A$ for all $x\in K^{\times}$), let $v$ be the quotient map $K^{\times}\rightarrow K^{\times}/A^{\times}$ and let $\Gamma=K^{\times}/A^{\times}$. If we define $\overline{x}\geq\overline{y}$ for $\overline{x},\overline{y}\in\Gamma$ whenever $xy^{-1}\in A$, we get a total ordering on $\Gamma$ that is compatible with the group structure. Moreover, if we write the group operation in $\Gamma$ additively, then $v:K\rightarrow\Gamma\cup\{\infty\}$ is an additive valuation on $K$, where $0\mapsto\infty$.

If we start with $\mathbb{Q}$, and apply this process, we get each of the $p$-adic valuations and the trivial valuation ($v(q)=0$ unless $q=0$). So I want to know if there is a similar construction that yields the archimedean valuation (or absolute value).

Attempt 1: If we start out looking for the norm on $\mathbb{C}=K$, and are aiming for $\Gamma$ to be the additive Real numbers, then by analogy with the above construction, we want a subring $A$ of $\mathbb{C}$ such that $\mathbb{C}^{\times}/A^{\times}$ is isomorphic to the positive Real numbers under multiplication. Intuitively, that would mean that $A^{\times}$ is the somewhat like the unit circle. However, every subring of $\mathbb{C}$ containing $S^1$ is actually just $\mathbb{C}$.

Attempt 2: If we try using $\mathbb{Z}=A$ in $\mathbb{Q}=K$ in spite of it failing miserably to be a valuation ring, then we get $v:\mathbb{Q}^{\times}\rightarrow\mathbb{Q}^{\times}/\mathbb{Z}^{\times}=\Gamma$. Since $\Gamma$ is isomorphic to the mulitplicative group of positive rational numbers, I initially had some hope. In this case, we get a partial order on $\Gamma$ where $\overline{x}\geq\overline{y}$ whenever $x$ is an integral multiple of $y$. Restricted to the multiplicative monoid $v(\mathbb{Z})$, this partial order is just the division structure of $\mathbb{Z}$.

One option is to look at total orders on $\Gamma$ extending this partial ordering. A priori, it seems that there is a different such total order for each permutation of the primes.

Another option is to figure out how to "complete" $\mathbb{Q}$ with respect to this partial ordering and see what results. If I continue the analogy with the case of $A$ being a valuation ring, I should consider an element $q\in\mathbb{Q}^{\times}$ to be close to $0$ if $v(q)\geq v(N)$ for some integer $N$ with many prime factors (counting multiplicity). Another approach is noticing that $\Gamma\cong\bigoplus_{p\text{ prime}}\mathbb{Z}$ and thinking of the latter group as a subgroup of $\prod_{p\text{ prime}}\mathbb{Z}$ where the latter group is given the product topology where $\mathbb{Z}$ has the cofinite topology. I'm not sure if these end up giving the same things, but I do know that $n!$ is a non-constant Cauchy sequence with respect to the former, albeit null. I am getting the feeling that non-constant Cauchy sequences must be null, but I have not managed to prove that either.

Anyway, I would appreciate any feedback or suggestions on what to try or what to read.

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No there is not: the archimedean place is defined inherently by the ordering, which is not an algebraic property. In fact, this can easily be seen because equivalent absolute values generate the same topology, and you know that the standard topology on $\Bbb R$ is generated by it's total ordering (this is in general how one can define topologies on totally ordered sets by declaring open sets to be generated by sets of the form $a<x<b$. But in particular, the absolute value generates a positive cone, and is defined completely by it, that is to say choosing $S$ which are the "positive" elements defines the absolute value since either $x$ or $-x$ is positive (here we include 0 for formality's sake). But then you can define $|x|=x$ if $x\in S$ and $|x|=-x$ if $x\not\in S$ and conversely $S=|\Bbb Q|$ is the image of the absolute value function. So no matter how you do it you're stuck appealing to the ordering on some level since the absolute value works in this way to be an archimedean absolute value.

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  • $\begingroup$ Do you think it unlikely that there would be an intrinsic way of identifying the positive rationals as a subgroup of index 2? For instance, it is some maximal subgroup not containing $-1$. If it could be found intrinsically, then the quotient map is essentially the sign function, and thus the absolute value $x\mapsto sign(x)x$. $\endgroup$ – J. David Taylor Apr 29 '16 at 6:27
  • $\begingroup$ How are you getting the index to be $2$? the positive rationals are not a group unless you poke out $0$ in which case it's under a completely different binary operation. $\endgroup$ – Adam Hughes Apr 29 '16 at 6:32
  • $\begingroup$ As in my post, I am talking about using the multiplicative group of rational numbers to get the valuations. $\endgroup$ – J. David Taylor Apr 29 '16 at 6:39
  • $\begingroup$ @J.DavidTaylor You will always need $+$ to accomplish the triangle inequality, which is an essential part of any absolute value. $\endgroup$ – Adam Hughes Apr 29 '16 at 6:44
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    $\begingroup$ It's worth noting that the nonnegative rational numbers are precisely the sums of squares, so in this case the ordering can be recovered from algebraic properties. (also, the nonnegative elements of $\mathbb{R}$ are precisely the squares) $\endgroup$ – Hurkyl May 5 '16 at 11:32

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